Deeper than primes

[doronshadmi]

Ok, so now it is Cantor's construction method for P(). Fine. Just where do you see that? It is not in the proof anywhere. We all think you are just making stuff up.

Your reasoning just making stuff up and as a result can't comprehend http://forums.randi.org/showpost.php?p=6951142&postcount=14492 .

[jsfisher]

That would be the post where you provided two separate mappings from S to P(S). The first was A->{A}; the second was A->{}. Neither mapping is a bijection between the members of {A} and P({A}).
That is again your reasoning, which simply can't get the simple fact that natural numbers are not limited to any particular collection and therefore, there is a 1-to-1 correspondence between the natural numbers and any given collection of different objects, whether the given collection has finite or infinite amount of objects.

You have reading comprehension issues again, Doron. Bijective, not injective (one-to-one), and between the members of {A} and P({A}), not the natural numbers.

Be that as it may, there is not injective mapping from the natural numbers to P({A}), either, so even accepting your non-sequitur response, you are still wrong.

In other words, by using Cantor's construction method...
That being the construction method that exists nowhere in reality
...on {{},{A}} case, there is a bijedction between the natural numbers and {{},{A}} as follows:

Case 1:
-----------------------
A ↔ {A} | Provides 1 ↔ {}
-----------------------

Not a bijection.

Case 2:
-----------------------
A ↔ { } | Provides 2 ↔ {A}
-----------------------

Also, not a bijection.

...and this is one and only one mapping.

And 1 ↔ {} / 2 ↔ {A} is not a bijection. Not an injection, either.

Doron, you have failed yet again and in multiple ways.

[jsfisher]

Your reasoning just making stuff up and as a result can't comprehend http://forums.randi.org/showpost.php?p=6951142&postcount=14492 .


The comprehension failure is yours and yours alone. Nowhere in that post to you point out there this so-called construction method occurs in the proof to Cantor's Theorem.

[doronshadmi]

Doron, you have failed yet again and in multiple ways.
jsfisher, since your reasoning prevents from you to get Cantor's construction method you are failing to get

A ↔ {A} | Provides 1 ↔ {}

A ↔ { } | Provides 2 ↔ {A}

as a one mapping.

[doronshadmi]

The comprehension failure is yours and yours alone. Nowhere in that post to you point out there this so-called construction method occurs in the proof to Cantor's Theorem.
That is only the result of your imaginary reasoning on your mind.

You simply can't get out of it, and therefore do not understand any possible reasoning beyond it.

[doronshadmi]

You have reading comprehension issues again, Doron. Bijective, not injective (one-to-one), and between the members of {A} and P({A}), not the natural numbers.
Wrong.

You have already failed there http://forums.randi.org/showpost.php?p=6934009&postcount=1443 .

[jsfisher]

jsfisher, since your reasoning prevents from you to get Cantor's construction method you are failing to get

A ↔ {A} | Provides 1 ↔ {}

A ↔ { } | Provides 2 ↔ {A}

as a one mapping.

"One mapping"? Inventing yet more terms, I see. That sort of fits with the construction method you simply invented. There is no such construction method provided by Cantor.

Be that as it may, neither of the mappings you present are injective nor bijective.

[doronshadmi]

Jsfisher, along this thread you are missing a very simple fact, which is:

Any given set it is first of all a collection of different objects.

In other words, a set has more than one object only if it is a simultaneous result of connectivity AND isolation.

Imbalance between connectivity AND isolation is resulted by sets with different amount of objects w.r.t each other, but by using balanced methods, it is possible to show a bijection between any two sets.

Cantor's construction method is a straightforward technique, which demonstrates the balance between any two sets, as shown in http://forums.randi.org/showpost.php?p=6951142&postcount=14492, and this balance can't be understood if the used reasoning gets it as isolated cases ( exactly as jsfisher does by isolate the mappings, for example, in http://forums.randi.org/showpost.php?p=6951775&postcount=14502 ).

[jsfisher]

Jsfisher, along this thread you are missing a very simple fact, which is:

Any given set it is first of all a collection of different objects.

Why do you belabor this trivial thing? Has anyone said otherwise?

In other words, a set has more than one object only if it is a simultaneous result of connectivity AND isolation.

This statement, on the other hand, is meaningless doronetics.

Imbalance between connectivity AND isolation is resulted by sets with different amount of objects w.r.t each other, but by using balanced methods, it is possible to show a bijection between any two sets.

And yet you have repeated failed in all attempts to present a bijection.

Cantor's construction method...
...which exists only in the imagination of Doron and not in anything Cantor presented...
...is a straightforward technique, which demonstrates the balance between any two sets, as shown in http://forums.randi.org/showpost.php?p=6951142&postcount=14492, and this balance can't be understood if the used reasoning gets it as isolated cases ( exactly as jsfisher does by isolate the mappings, for example, in http://forums.randi.org/showpost.php?p=6951775&postcount=14502 ).

We understand it just fine. It is awash with the typical gibberish, contradiction, and nonsense that is your trademark, Doron. That's very clear.

[doronshadmi]

This statement, on the other hand, is meaningless doronetics.
This statement, on the other hand, cannot be comprehended by your rigid isolating reasoning.



And yet you have repeated failed in all attempts to present a bijection.
And yet you have repeated failed in all attempts to get the bijection by using
your rigid isolating reasoning.


...which exists only in the imagination of Doron and not in anything Cantor presented...
...which you share with many other rigid minds like you that have no ability to comprehend Cantor's construction method.




We understand it just fine.
No, you don't. You get only the isolated aspect of Cantor's construction method.

[doronshadmi]

"One mapping"?
Yes jsfisher, it is one bijective mapping, and your isolated reasoning can't comprehend it.

[jsfisher]

...which you share with many other rigid minds like you that have no ability to comprehend Cantor's construction method.

So, why are you unable to show us where Cantor presented his construction method? Surely it's there somewhere. Just point to the place in the proof where it occurs.

[jsfisher]

Yes jsfisher, it is one bijective mapping, and your isolated reasoning can't comprehend it.


Let's review. Doron presented us with two mappings:

A ↔ {A} | Provides 1 ↔ {}
A ↔ { } | Provides 2 ↔ {A}

We have first A ↔ {A} and A ↔ { }. That's not bijective. Heck, that's not even functional since A has two possible mappings. No bijection here.

Then we have 1 ↔ {} and 2 ↔ {A}. Also not bijective, at least not between the natural numbers and the members of P({A}). There'd need to be a mapping for all of the natural numbers, not just two of them. No bijection here, either.

Another Doron masterpiece of failure.

[epix]

By using a cross-contexts reasoning, let us use Cantor's construction method...

http://img10.imageshack.us/img10/3570/treeax.png

...By using Cantor's construction method with the members of {1,2,3} and the same amount of members taken from P({1,2,3}), one enables to explicitly define each P({1,2,3}) member...
http://img695.imageshack.us/img695/6035/tercka.png
1 <----> thiefL
2 <----> "i'll be back"
3 <----> thiefR


Note: Doron, this is an atheist board so I would recommend using that cross-context reasoning sparingly.

[doronshadmi]

Let's review. Doron presented us with two mappings:

A ↔ {A} | Provides 1 ↔ {}
A ↔ { } | Provides 2 ↔ {A}

We have first A ↔ {A} and A ↔ { }. That's not bijective. Heck, that's not even functional since A has two possible mappings. No bijection here.

Then we have 1 ↔ {} and 2 ↔ {A}. Also not bijective, at least not between the natural numbers and the members of P({A}). There'd need to be a mapping for all of the natural numbers, not just two of them. No bijection here, either.

Another Doron masterpiece of failure.
jsfisher you simply repeat on your separation approach of the two mappings (you get only one mapping at a time), so by using this separation method, there is no wonder that you can't define the bijection between 1,2 and {},{A} objects.

Your problem is found right there:


Jsfisher, along this thread you are missing a very simple fact, which is:

Any given set it is first of all a collection of different objects.
Why do you belabor this trivial thing? Has anyone said otherwise?
jsfisher, it is a trivial thing for you exactly because you are not using your reasoning in order really understand what enables the existence of a collection of different objects.

For you an expression like "connectivity AND isolation" is meaningless exactly because your reasoning is not involved in any further research of the fundamental conditions that enable the existence of sets (you take their existence obviously).

A direct result of this (indeed) trivial approach, prevents from you to understand the non-trivial conditions that enable the existence of sets, in the first place.

In other words, jsfisher, What You See Is What You Get, and in this case you see triviality and therefore there is no wonder that you indeed get triviality.

-------------------------------------

By understanding sets as a result of connectivity AND isolation, one enables to use methods, which are inaccessible to any one that does not research the fundamental conditions that enable the existence of sets.

One of these methods is Cantor's construction method, which is based on the fact that sets are the results of connectivity AND isolation.

For example, finite sets are the result of stronger isolation w.r.t connectivity under connectivity AND isolation comprehensive and one framework.

Being a finite set is being isolated by strict amount of objects, such that there can be differences of these amounts under comparison (where comparison is not possible without the connectivity accept among isolated objects, under connectivity AND isolation comprehensive and one framework).

By using Cantor's construction method as a tool of connectivity AND isolation comprehensive and one framework, one enables to define any wished degree of mapping between any two given sets, where the set of natural numbers is used as a dynamic measurement tool for any mapping degree between two sets with different objects.

This dynamic measurement is useful for both finite or infinite sets, where in the case of finite sets, the aspect of connectivity is expressed as the tendency for balance between the considered sets, where this balance is fully expressed as a bijective mapping.

In the case of finite sets, this mapping has finite amount of results, which exist between non-bijection and bijection.

In the case of infinite sets, this mapping has infinite amount of results, which exist between non-bijection and bijection.

In both cases Cantor's construction method is used as the main tool.

Furthermore, by understanding that connectivity is a building-block of sets exactly as isolation is a building-block of sets under connectivity AND isolation comprehensive and one framework, one enables to understand that both building-blocks must be true (both of them are present) otherwise "connectivity AND isolation" is a false proposition.

By understanding this logical truth, one enables to understand that no amount of isolated objects is connectivity, and as a result given any set, its amount can't be summed into connectivity, such that given any set, it is an ever increasing form of existence (abstract or not).

[epix]

By understanding sets as a result of connectivity AND isolation ...

http://wwwdelivery.superstock.com/WI/223/1830/PreviewComp/SuperStock_1830-3365.jpg

... one enables to use methods, which are inaccessible to any one that does not research the fundamental conditions...

... in Folsom prison.

http://thebsreport.files.wordpress.com/2010/07/escape.jpg

[doronshadmi]

http://wwwdelivery.superstock.com/WI/223/1830/PreviewComp/SuperStock_1830-3365.jpg

... in Folsom prison.

http://thebsreport.files.wordpress.com/2010/07/escape.jpg
There is nothing like personal experience, isn't it epix?

[jsfisher]

jsfisher you simply repeat on your separation approach of the two mappings (you get only one mapping at a time), so by using this separation method, there is no wonder that you can't define the bijection between 1,2 and {},{A} objects.

A bijection between 1, 2 and {}, {A} is trivial. It is also not what you claimed.

You claimed there was a bijection between (1) the set of natural numbers and any power set and (2) any set and its power set. You have failed in every attempt to show either of these things.

...By using Cantor's construction method...

This is another of your bogus claims. Cantor provided no such method.

[doronshadmi]

A bijection between 1, 2 and {}, {A} is trivial. It is also not what you claimed.
Again you demonstrate your trivial reasoning by get this bijection as trivial.

Furthermore, now you change your claim about the non-bijection between 1,2 and {},{A} as you wrote in http://forums.randi.org/showpost.php?p=6951775&postcount=14502 .

jsfisher, you do not have the slightest idea about what I claim exactly because you do not have the slightest idea of what a set is.


You claimed there was a bijection between (1) the set of natural numbers and any power set and (2) any set and its power set. You have failed in every attempt to show either of these things.
(1) and (2) are shown in http://forums.randi.org/showpost.php?p=6951142&postcount=14492 .

You also unable to comprehend:

http://forums.randi.org/showpost.php?p=6928575&postcount=14425

http://forums.randi.org/showpost.php?p=6916077&postcount=14361

http://forums.randi.org/showpost.php?p=6918253&postcount=14366



This is another of your bogus claims. Cantor provided no such method.
Yes, he provided such a method, exactly as shown in http://forums.randi.org/showpost.php?p=6951142&postcount=14492 .

You are unable to get what I say exactly because you do not understand what enable sets, in the first place.

[jsfisher]

Again you demonstrate your trivial reasoning by get this bijection as trivial.

You are the only one here who cannot see just how trivial it is.

Furthermore, now you change your claim about the non-bijection between 1,2 and {},{A} as you wrote in http://forums.randi.org/showpost.php?p=6951775&postcount=14502 .

Please pay attention. It is not a bijection between the natural numbers and {},{A}. The natural numbers include more than just 1 and 2.

...
Yes, he provided such a method, exactly as shown in http://forums.randi.org/showpost.php?p=6951142&postcount=14492 .

Please show us where Cantor provided such a method. Stop showing us your bogus posts; show us Cantor's work.

[doronshadmi]

You are the only one here who cannot see just how trivial it is.
Now you are proud of your trivial reasoning :rolleyes:



Please pay attention. It is not a bijection between the natural numbers and {},{A}. The natural numbers include more than just 1 and 2.
Pay attention. Bijection between two sets is a mapping between two sets, such that each object of one set is paired with exactly one object of the other set and there are no members of those sets that are not mapped with each other (it is called 1-to-1 and onto). Bijection (1-to-1 and onto) can be found between any two sets, even if the two sets have finite cardinality.

[doronshadmi]

Please show us where Cantor provided such a method. Stop showing us your bogus posts; show us Cantor's work.
EDIT:

Can you use your trivial reasoning in order to say how many D explicit objects are constructed, as shown in http://en.wikipedia.org/wiki/Cantor%27s_theorem ?

By this theorem Cantor explicitly provides |P(N)| D explicit objects that can be put in one-to-one correspondence with N objects.

Cantor's mistake is based on the fact that he concluded general conclusions by using only a one case of his theorem.

[jsfisher]

Now you are proud of your trivial reasoning :rolleyes:

You are the only one here rejoicing in the trivial. The rest of us see it as, well, trivial.

Pay attention. Bijection between two sets is a mapping between two sets, such that each object of one set is paired with exactly one object of the other set and there are no members of those sets that are not mapped with each other (it is called 1-to-1 and onto).

No one has disputed this. Why do you so prone to argue things that are not in contention?

And, as should be blatantly obvious even to you, Doron, since you just recited the basic meaning of bijection, your mapping between the natural numbers and {},{A} isn't a bijection since you provide no mapping for 3, 4, 5, etc.

Bijection (1-to-1 and onto) can be found between any two sets, even if the two sets have finite cardinality.

This part contradicts what you just posted about bijections.

[jsfisher]

Can you use your trivial reasoning in order to say how many D explicit objects are constructed, as shown in http://forums.randi.org/showpost.php?p=6916077&postcount=14361 ?


So where, exactly, did Cantor provide this construction method for these mysterious "D explicit objects" of which you speak?

[jsfisher]

EDIT:

Can you use your trivial reasoning in order to say how many D explicit objects are constructed, as shown in http://en.wikipedia.org/wiki/Cantor%27s_theorem ?

By this theorem Cantor explicitly provides |P(N)| D explicit objects that can be put in one-to-one correspondence with N objects.

Cantor's mistake is based on the fact that he concluded general conclusions by using only a one case of his theorem.

I see you have taken the time to completely re-write a previous post. You poor reasoning skills and lack of reading comprehension ability are exceeded by your rudeness.

Be that as it may, I see you'd like to use Wikipedia for your Cantor's proof reference. You may notice that the proof, amazingly enough, appears under the heading Proof. No where in that proof is there a construction method given for any "D explicit objects".

The proof given in Wikipedia is substantially the same as the proof I gave some ways back in this thread. Neither presentation provides a construction method for any existing set.

[doronshadmi]

You are the only one here rejoicing in the trivial. The rest of us see it as, well, trivial.



No one has disputed this. Why do you so prone to argue things that are not in contention?

And, as should be blatantly obvious even to you, Doron, since you just recited the basic meaning of bijection, your mapping between the natural numbers and {},{A} isn't a bijection since you provide no mapping for 3, 4, 5, etc.



This part contradicts what you just posted about bijections.
No jsfisher, this part simply does not fit to your trivial reasoning about 1-to-1 and onto mapping between two sets, because your reasoning can't get the fact that also two finite sets have a bijection, by using Cantor's construction method.

[jsfisher]

No jsfisher, this part simply does not fit to your trivial reasoning about 1-to-1 and onto mapping between two sets, because your reasoning can't get the fact that also two finite sets have a bijection, by using Cantor's construction method.


You have yet to show a bijection between the members of {A} and P({A}), with or without the use of the non-existent construction method. Substituting a trivial mapping between two sets of two elements each isn't even close.

[doronshadmi]

I see you have taken the time to completely re-write a previous post. You poor reasoning skills and lack of reading comprehension ability are exceeded by your rudeness.

Be that as it may, I see you'd like to use Wikipedia for your Cantor's proof reference. You may notice that the proof, amazingly enough, appears under the heading Proof. No where in that proof is there a construction method given for any "D explicit objects".

The proof given in Wikipedia is substantially the same as the proof I gave some ways back in this thread. Neither presentation provides a construction method for any existing set.
Call Cantor's theorem whatever you like.

It does not change the fact that by using this theorem, one enables to explicitly define |P(N)| Ds AND a bijection between N members and these Ds.

Once again, Cantor missed his own construction method, because, like you, he wrongly was focused on some single and particular case of that construction.

[jsfisher]

Cantor missed his own construction method, because, like you, he wrongly was focused on some single and particular case of that construction.


Please, point out for all of us this construction method as it appears in Cantor's proof. Surely, it must appear right there under that Proof heading in Wikipedia. We anxiously await your revelation of that that isn't there.

[doronshadmi]

You have yet to show a bijection between the members of {A} and P({A}), with or without the use of the non-existent construction method. Substituting a trivial mapping between two sets of two elements each isn't even close.
Are you kidding?

Do you really unable to understand that the same construction method also provides |P(P(N))|, |P(P(P(N)))|, ... ,|...P(P(P(N)))...| explicit D's that are explicitly paired (1-to-1 and onto) with N members, such that N ↔ P(N) ↔ P(P(N)) ↔ P(P(P(N))) ↔ ... ↔ ...P(P(P(N)))... ?

[jsfisher]

Are you kidding?

Nope, not kidding. You have been completely unable to show a bijection between the members of {A} and {{},{A}}.

Use any construction method you like, real or imaginary, then just show us the final the result: a bijection between the members of {A} and {{},{A}}.

Leave out the natural numbers (or any part of them); leave out all your fabulous multi-round step-by-step processing. All we need to see is the conclusion: a bijection between the members of {A} and {{},{A}}.

Just so you know, we are all predicting you will fail again.

[doronshadmi]

Please, point out for all of us this construction method as it appears in Cantor's proof. Surely, it must appear right there under that Proof heading in Wikipedia. We anxiously await your revelation of that that isn't there.

EDIT:

Cantor's construction method constructs explicit |{{},...,{1,2,3,...}}l {{},...,{1,2,3,...}} members as follows:

1) Some defined explicit {{},...,{1,2,3,...}} member is the result of a 1-to-1 correspondence between {1,2,3,...} members and the same amount of members taken from {{},...,{1,2,3,...}}.

2) This explicit constructed {{},...,{1,2,3,...}} member (notated as D) includes a {1,2,3,...} member only if this {1,2,3,...} member does not exist as one of the members of the {{},...,{1,2,3,...}} member that is mapped with it.

By using this construction method |{{},...,{1,2,3,...}}l times, one enables to define a bijection between {1,2,3,...} members
and {{},...,{1,2,3,...}} members, where D is actually a placeholder for |{{},...,{1,2,3,...}}l {{},...,{1,2,3,...}} members.

[doronshadmi]

Nope, not kidding. You have been completely unable to show a bijection between the members of {A} and {{},{A}}.

Use any construction method you like, real or imaginary, then just show us the final the result: a bijection between the members of {A} and {{},{A}}.

Leave out the natural numbers (or any part of them); leave out all your fabulous multi-round step-by-step processing. All we need to see is the conclusion: a bijection between the members of {A} and {{},{A}}.

Just so you know, we are all predicting you will fail again.
I am not a participator of your "running in circles on a closed box" game.

Because of your game you are unable to get, for example, the following:

http://forums.randi.org/showpost.php?p=6955135&postcount=14519

http://forums.randi.org/showpost.php?p=6955416&postcount=14521

http://forums.randi.org/showpost.php?p=6955576&postcount=14530

http://forums.randi.org/showpost.php?p=6955625&postcount=14532

[jsfisher]

EDIT:

Cantor's construction method constructs explicit |{{},...,{1,2,3,...}}l {{},...,{1,2,3,...}} members as follows:


Where is that in the Proof?

[jsfisher]

I am not a participator of your "running in circles on a closed box" game.


So, you have given up trying to show a bijection between the members of {A} and P({A}), then?

Just as well, since there isn't one.

[epix]

Again you demonstrate your trivial reasoning by get this bijection as trivial.

Trivial reasoning cannot be used to establish the cardinality of a set through bijection. Please get familiar with the meaning of various prefixes, such as "bi-" and "tri-" in this particular case:

bi = "two" => bicycle (http://bicycletutor.com/images/bike-map.jpg)
tri = "three" => tricycle (http://cdn.caughtoffside.com/wp-content/uploads/2009/07/tricycle_2.jpg)

Since 3 ≠ 2, trivial reasoning cannot contribute to the construction of bijection, as opposed to the case where erotic reasoning facilitates erection.

[The Man]

Again your limited reasoning airs its view.

The set of natural numbers, is some particular case of a collection of different objects.

“objects” that are all specifically natural numbers. Are you claiming that some member(s) of your “set of natural numbers” is/(are) not (a) natural number(s)? If that is not your claim then all the members of your “set of natural numbers” are natural numbers. Thus you have no actual problems with “all” in reference to the natural numbers or any set of natural numbers.

Are all of your natural numbers natural numbers, Doron, or are some of your natural numbers not natural numbers?

Perhaps you just don’t know or just don’t want to know? Whichever, it really doesn’t matter.

You can propose no argument to all natural numbers being natural numbers that is not simply self contradictory (some natural numbers aren’t natural numbers) or simply an argument from ignorance (I don’t know if all natural numbers are natural numbers). The former simply and directly refutes itself while the latter asserts a specific lack of knowledge about whether all natural numbers are natural numbers or not and thus just professes its own irrelevance.




Given any collection of different objects, succession (in terms of present continuous, where also parallel reasoning is used) is its essential property, whether it is finite (and in this case the succession is external w.r.t the collection) or infinite (and in this case the succession is internal w.r.t the collection).

“external w.r.t the collection”? So by your own assertion you’re just talking about extraneous elements that you specifically claim are not included in that collection.

“succession is internal w.r.t the collection”? So the successor of any member of the set is also a member of that set? Perhaps we could come up with an expression to assert this amazing notion? How about “the set is closed under an operation of succession”? Oh wait that is already being used but, hey, by some strange coincidence that is exactly what it means. If we didn’t know any better we might start thinking that you’re learning some math (or at least set theory).


You still do not understand the meaning of succession +1 in terms of cardinality as a present continuous, because parallel reasoning is beyond your only step-by-step serial reasoning.

You are still deliberately trying to confuse a list with a set. “succession +1”? So the successor of a successor? By your own assertion above it too can be a member of the set.





















Oh and again…


By all means please explain to us the difference between increasing and decreasing with “no past (before) and no future (after)”?

[epix]

|S| = |P(S)| is right, where both |S| and |P(S)| are equally ever increasing (which is present continuous).
LOL. Considering a finite case: if |S| = 3, then |P(S)| = 23. According to you, 3 = 23.

Considering infinite case S = N: The consequence of |N| = |P(N)| is that your "parallel reasoning" holds 2aleph0 equivalent to aleph0. In other words, if there is a 1-on-1 correspondence between the members of N and the members of P(N), then the cardinality of the power set P(N) is strictly aleph0. But any construction that proves the previous statement true must apply to the finite case. So show us the jolly-good construction when S = {1, 2, 3}.

[epix]

Pay attention.
http://img690.imageshack.us/img690/2442/redondobeach.jpg


Released from its prison of absolute truth, mathematics was free to roam and to develop any kind of system it wanted -- with the one restriction that these systems must be internally consistent. Whether they are useful in describing the world does not matter, for modern mathematicians are no longer concerned with depicting what they see in the most realistic way possible. Like modern painters, they have become more interested in the techniques and methods of their art than in mere description.

[doronshadmi]

So, you have given up trying to show a bijection between the members of {A} and P({A}), then?

Just as well, since there isn't one.
Not at all.

If the powerset of set A is {{},A} it means that |A|=1

It means that also |{A}|=1, and in the case P({A}) = {{},{A}}.

By using Cantor's construction method we get a 1-to-1 correspondence between 1,2 and {},{A}, which is not different than the 1-to-1 correspondence between 1,2 and {},A, such that:

1 ↔ {} ↔ {}
|
2 ↔ A ↔ {A}

and we get the same 1-to-1 correspondence for both cases, which are A and P(A) or {A} and P({A}).

Here it is again, by using Cantor's construction method:

A ↔ A | Provides 1 ↔ {}, which is equivalent to {A} ↔ {A} | Provides 1 ↔ {}
A ↔ { } | Provides 2 ↔ A, which is equivalent to {A} ↔ {} | Provides 1 ↔ {A}

[jsfisher]

Not at all.

If the powerset of set A is {{},A} it means that |A|=1

You are the first to introduce any "set A". Why do you insist on irrelevant asides?

It means that also |{A}|=1

No, it doesn't. The cardinality of {A} is 1 because that's how cardinality works. It has nothing to do with what "set A" happens to be. That was just an irrelevant aside from you.

...and in the case P({A}) = {{},{A}}

The would follow directly from the meaning of "power set". So, you have ventured from the irrelevant to the bloody obvious. Please continue.

By using Cantor's construction method
...being the method that Cantor never provided...
we get a 1-to-1 correspondence between 1,2 and {},{A}, which is not different than the 1-to-1 correspondence between 1,2 and {},A

Were you expecting a "1-to-1 correspondence between 1,2 and {},{A}" to be different from the "1-to-1 correspondence between 1,2 and {},{A}"? Seriously?

...such that:

1 ↔ {} ↔ {}
|
2 ↔ A ↔ {A}

and we get the same 1-to-1 correspondence for both cases.

The original set is {A} and its power set is {{},{A}}. Your trivial mapping, above, for "both cases" does not provide a mapping between the members of the two sets at issue. Since you didn't provide a mapping between members of {A} and {{},{A}}, it is unlikely to be a bijective mapping.

Again, Doron, you have failed. You tried but failed to show any bijection between the members of {A} and its power set.

[doronshadmi]

LOL. Considering a finite case: if |S| = 3, then |P(S)| = 23. According to you, 3 = 23.

EDIT:

You are still missing my argument about S and P(S).

According to it the cardinality of natural numbers with any collection of different objects, exists between non-bijection to bijection.

When we are dealing with finite collections of different objects, if we understand each collection in terms of fixed cardinality, one of the results is that two finite collections have different amount of objects, exactly as can be seen in the particular case of {1,2,3} and {{},{1},{2},{3},{1,2},{1,3},{2,3},{1,2,3}}.

By using the set of natural numbers together with Cantor's constriction method, we are no longer closed under the different amounts of {1,2,3} and {{},{1},{2},{3},{1,2},{1,3},{2,3},{1,2,3}}, as follows:

.
.
.
|
|
|-----------------------
|1 ↔ {1} |
|2 ↔ {2} | Provides 1 ↔ {}
|3 ↔ {3} |
|-----------------------
|
|
|-----------------------
|1 ↔ { } |
|2 ↔ {2} | Provides 2 ↔ {1}
|3 ↔ {3} |
|-----------------------
|
|
|-----------------------
|1 ↔ {1} |
|2 ↔ { } | Provides 3 ↔ {2}
|3 ↔ {3} |
|-----------------------
|
|
|-----------------------
|1 ↔ {1} |
|2 ↔ {2} | Provides 4 ↔ {3}
|3 ↔ { } |
|-----------------------
|
|
|-----------------------
|1 ↔ { } |
|2 ↔ {3} | Provides 5 ↔ {1,2}
|3 ↔ {1,3} |
|-----------------------
|
|
|-----------------------
|1 ↔ { } |
|2 ↔ {2} | Provides 6 ↔ {1,3}
|3 ↔ {1,2} |
|-----------------------
|
|
|-----------------------
|1 ↔ {1} |
|2 ↔ {3} | Provides 7 ↔ {2,3}
|3 ↔ {1,2} |
|-----------------------
|
|
|-----------------------
|1 ↔ { } |
|2 ↔ {1} | Provides 8 ↔ {1,2,3}
|3 ↔ {2} |
|-----------------------
|
|
.
.
.

We actually deal with a DNA-like code between 3 and 8 objects, and by using it are able to define any mapping degree between |S| and |P(S)| for our purpose.

Such flexibility is impossible by the rigid |S| < |P(S)| particular case, which is wrongly taken as some universal principle of the possible relations between S and P(S).


--------------------------


Furthermore, by using infinite sets we discover that the whole idea of mapping between sets is changed, because we are also able to define a bijection between S and its proper subset, as can be seen bet between the natural numbers and, for example, their proper subset of even (or odd) numbers.

We also are able to define a bijection between rational numbers and their proper subset of natural numbers.

This ability does not stop also in the case of real numbers, where natural numbers are their proper subset, and the possible bijection between real numbers (which are equivalent by their amount to the powerset of natural numbers) is shown by using Cantor's construction method, as follows:

Cantor's construction method constructs explicit |{{},...,{1,2,3,...}}l {{},...,{1,2,3,...}} members as follows:

1) Some defined explicit {{},...,{1,2,3,...}} member is the result of a 1-to-1 correspondence between {1,2,3,...} members and the same amount of members taken from {{},...,{1,2,3,...}}.

2) This explicit constructed {{},...,{1,2,3,...}} member ( notated as D. Please read more about D construction in http://en.wikipedia.org/wiki/Cantor%27s_theorem ) includes a {1,2,3,...} member only if this {1,2,3,...} member does not exist as one of the members of the {{},...,{1,2,3,...}} member that is mapped with it.

By using this construction method with amount |{{},...,{1,2,3,...}}l , one enables to define a bijection between {1,2,3,...} members
and {{},...,{1,2,3,...}} members, where D is actually a placeholder for |{{},...,{1,2,3,...}}l {{},...,{1,2,3,...}} members.

---------------------------

In both cases, whether the considered set is finite or not, by using Cantor's construction method, one enables to define any wished degree of mapping between two sets, and use the resulted mapping for some useful purpose.

[doronshadmi]

EDIT:

You are the first to introduce any "set A". Why do you insist on irrelevant asides?
Not exactly, in http://forums.randi.org/showpost.php?p=6922616&postcount=14407 you wrote:

Let's take for example S = {A}. It's power set P(S) = {{},A}.
but I show that also if the powerset is {{},A} then:

1) |A| must be 1.

2) By using Cantor's construction method, we are able to define a 1-to-1 correspondence between the members of {{},{A}} (or {{},A}) and the appropriate amount of natural numbers, and in this case, any 2 different natural numbers can be used in order to demonstrate the 1-to-1 correspondence.

jsfisher, your rigid reasoning still missing http://forums.randi.org/showpost.php?p=6963549&postcount=14542 .

[jsfisher]

By using Cantor's construction method
...which exists only in your imagination...
we are able to define a 1-to-1 correspondence between the members of {{},{A}} (or {{},A}) and the appropriate amount of natural numbers

...which was both (a) trivial to do and (b) unrelated to the required task.

The required task is now and has always been: Show a bijection between the members of {A} and its power set.


Again you fail to accomplish that which you claimed you could do.

[doronshadmi]

QUOTE=jsfisher;6963761...which was both (a) trivial to do and (b) unrelated to the required task.[/QUOTE
...which is both (a) general to do an (b) exactly related to the required task, which is the existence of collections of different objects, where the properties of the sets have no significance, by generalization.

[jsfisher]

QUOTE=jsfisher;6963761...which was both (a) trivial to do and (b) unrelated to the required task.[/QUOTE
...which is both (a) general to do an (b) exactly related to the required task, which is the existence of collections of different objects, where the properties of the sets have no significance, by generalization.


Formatting errors aside, you still haven't provided a bijection between the members of {A} and it power set.

You again fail.

[doronshadmi]

EDIT:

...which was both (a) trivial to do and (b) unrelated to the required task.
...which is both (a) general to do and (b) exactly related to the required task, which is the existence of collections of different objects, where the properties of the mapped sets have no significance, by generalization.


You again fail.
You are failing to get http://forums.randi.org/showpost.php?p=6963549&postcount=14542 .

[doronshadmi]

“objects” that are all specifically natural numbers. Are you claiming that some member(s) of your “set of natural numbers” is/(are) not (a) natural number(s)?
Since you do not understand present continuous in terms of "ever increasing", let us express it as "permanently changing", where the permanent (invariant) aspect is like pi among the collection of circles, and the change (variant) aspect is like the curvature among the collection of circles.

The present continuous state, as described above, is the fundamental property of any infinite collection of different objects.

For more details, please look at http://forums.randi.org/showpost.php?p=6963549&postcount=14542 .

[epix]

If the powerset of set A is {{},A} it means that |A|=1

Yes, it means that if A = {$}, and $ is put in 1:1 correspondence with the set of natural numbers, you get

1 <---> $

and the cardinality of set A is therefore |A| = 1


By using Cantor's construction method we get a 1-to-1 correspondence between 1,2 and {},{A}, which is not different than the 1-to-1 correspondence between 1,2 and {},A, such that (...)

1 ↔ {} ↔ {}
|
2 ↔ A ↔ {A}
Where does that 2 come from? You are matching A with P(A), and A returned only one natural number and that's 1.

[doronshadmi]

Yes, it means that if A = {$}, and $ is put in 1:1 correspondence with the set of natural numbers, you get

1 <---> $

and the cardinality of set A is therefore |A| = 1


Where does that 2 come from? You are matching A with P(A), and A returned only one natural number and that's 1.

http://forums.randi.org/showpost.php?p=6963549&postcount=14542

http://forums.randi.org/showpost.php?p=6964932&postcount=14548

[jsfisher]

http://forums.randi.org/showpost.php?p=6963549&postcount=14542

http://forums.randi.org/showpost.php?p=6964932&postcount=14548


Yep, both those links provide evidence of how wrong Doron is. Repeated failed attempts to show a bijection between the members of a set and its power set.

[epix]

http://forums.randi.org/showpost.php?p=6963549&postcount=14542

http://forums.randi.org/showpost.php?p=6964932&postcount=14548

"Yep, both those links provide evidence of how wrong Doron is. Repeated failed attempts to show a bijection between the members of a set and its power set."
jsfisher

[doronshadmi]

Yep, both those links provide evidence of how wrong Doron is. Repeated failed attempts to show a bijection between the members of a set and its power set.
You are unable (yet) to understand the generalization of sets as collections of different objects (where the property of any given set is insignificant), and how by this generalization, any given set has a 1-to-1 correspondence with some arbitrary objects of natural numbers (where, by generalization, this 1-to-1 correspondence is some particular case of many degrees of mapping between collections of different objects).

So let us do it in baby steps.

-----------------------------------------------------------

By baby steps, there is an interesting difference between mapping among finite sets and infinite sets, such that there is no 1-to-1 correspondence between a non-empty finite set and its proper subset, but there is a 1-to-1 correspondence between an infinite set and its proper subset.

Among infinite sets we are able to define a bijection between S and its proper subset, as can be seen between the natural numbers and, for example, their proper subset of even (or odd) numbers.

We are also able to define a bijection between rational numbers and their proper subset of natural numbers.

This ability does not stop also in the case of real numbers, where natural numbers are their proper subset, and the bijection between real numbers (which are equivalent by their amount to the powerset of natural numbers) is shown by using Cantor's construction method, as follows:

Cantor's construction method constructs explicit |{{},...,{1,2,3,...}}l {{},...,{1,2,3,...}} members as follows:

1) Some defined explicit {{},...,{1,2,3,...}} member is the result of a 1-to-1 correspondence between {1,2,3,...} members and the same amount of members taken from {{},...,{1,2,3,...}}.

2) This explicit constructed {{},...,{1,2,3,...}} member ( notated as D. Please read more about D construction in http://en.wikipedia.org/wiki/Cantor%27s_theorem ) includes a {1,2,3,...} member only if this {1,2,3,...} member does not exist as one of the members of the {{},...,{1,2,3,...}} member that is mapped with it.

By using this construction method with amount |{{},...,{1,2,3,...}}l , one enables to define a bijection between {1,2,3,...} members
and {{},...,{1,2,3,...}} members, where D is actually a placeholder for |{{},...,{1,2,3,...}}l {{},...,{1,2,3,...}} members.

[doronshadmi]

As for collections of different objects, let us look at Nicholas of Cusa's ( http://www.integralscience.org/cusa.html )model of infinity.

By using a collection of circles, he wished to show that there is a continuous relation between a circle and a stright line, as shown in the following diagram:

http://farm6.static.flickr.com/5255/5519866444_4c55f67312.jpg

I added a point to this diagram, which is the central point of the considered circles.

It has to be stressed that a form is considered as a circle only if the invariant pi is defined as the result of the ratio between its circumference and its diameter.

Pi is not found at point or straight line forms, so they can't be considered as objects of the collection of circles, but they are used as the inaccessible limits of this collection, such that infinitely many circles with increased curvature degrees can't be a point AND infinitely many circles with decreased curvature degrees can't be a line.

In other words, the collection of circles is a permanently changing model , where the permanent aspect is pi and the changing aspect is the circle's curvature degrees.

If the collection of circles is finite, we are able to define a complete collection of different circle's curvature degrees. But such completeness is not found in the case of infinite collection because the amount of different circle's curvature degrees is not satisfied.

Following the same principle about the collection natural, rational and irrational numbers, their amount is not satisfied if they have infinitely many objects.

[jsfisher]

You are unable (yet) to understand the generalization of sets as collections of different objects (where the property of any given set is insignificant)

It is only you, Doron, that seems to have trouble with this basic properties of sets. You belabor the trivial.

and how by this generalization, any given set has a 1-to-1 correspondence with some arbitrary objects of natural numbers (where, by generalization, this 1-to-1 correspondence is some particular case of many degrees of mapping between collections of different objects).

More trivia. No one disputes there is a bijection between any set of n objects and the first n natural numbers. That is a trivial observation. The fact you dwell on it as significant speaks to your most limited understanding of Mathematics.

However, you made a couple of claims, different from above, that are flatly untrue. You claimed that for any set, S,

(1) a bijection exists between S and its power set, P(S).
(2) a bijection exists between P(S) and the natural numbers.

S does not mean the original set S plus other things to make it the same size as P(S). Natural numbers means all of them, not just a few of them.

Your claims are wrong. You have failed. You also claimed some construction method is to be found in Cantor's Theorem. You failed to establish that, too.

In things mathematical, Doron, you do little else than fail.

[jsfisher]

This explicit constructed {{},...,{1,2,3,...}} member ( notated as D. Please read more about D construction in http://en.wikipedia.org/wiki/Cantor%27s_theorem ) includes a {1,2,3,...} member only if this {1,2,3,...} member does not exist as one of the members of the {{},...,{1,2,3,...}} member that is mapped with it.

Here it is, Doron, the entire proof from your wikipedia reference. Please highlight the constuction method for the member "notated as D".

Two sets are equinumerous (have the same cardinality) if and only if there is a one-to-one correspondence between them. To establish Cantor's theorem it is enough to show that given any set A no function f from A into P(A), the power set of A, can be surjective, i.e. to show the existence of at least one subset of A that is not an element of the image of A under f. Such a subset, B ∈ P(A), is given by the following construction:
B = { x ∈ A : x ∉ f(x) }
This means, by definition, that for all x in A, x ∈ B if and only if x ∉ f(x). For all x the sets B and f(x) cannot be the same because B was constructed from elements of A whose images (under f) did not include themselves. More specifically, consider any x ∈ A, then either x ∈ f(x) or x ∉ f(x). In the former case, f(x) cannot equal B because x ∈ f(x) by assumption and x ∉ B by the construction of B. In the latter case, f(x) cannot equal B because x ∉ f(x) by assumption and x ∈ B by the construction of B.

Thus there is no x such that f(x) = B; in other words, B is not in the image of f. Because B is in the power set of A, the power set of A has a greater cardinality than A itself.

That's the whole proof, Doron. There is no set D to be found. Your reading comprehension skills have failed you again. You have failed again.

[doronshadmi]

Jsfisher, you continue to run in circles inside your closed box.

As a result you ignore even the baby steps of my arguments on this fine subject, and therefore do not have any chance to understand:

http://forums.randi.org/showpost.php?p=6965940&postcount=14553

http://forums.randi.org/showpost.php?p=6966452&postcount=14554

Enjoy your "running in circles in a closed box" game.

[sympathic]

Enjoy your "running in circles in a closed box" game.

You are indeed the master of irony.

[doronshadmi]

That's the whole proof, Doron. There is no set D to be found. Your reading comprehension skills have failed you again. You have failed again.

You simply ignore this:


Using this idea, let us build a special set of natural numbers. This set will provide the contradiction we seek. Let D be the set of all non-selfish natural numbers. By definition, the power set P(N) contains all sets of natural numbers, and so it contains this set D as an element. Therefore, D must be paired off with some natural number, say d. However, this causes a problem. If d is in D, then d is selfish because it is in the corresponding set. If d is selfish, then d cannot be a member of D, since D was defined to contain only non-selfish numbers. But if d is not a member of D, then d is non-selfish and must be contained in D, again by the definition of D.

This is a contradiction because the natural number d must be either in D or not in D, but neither of the two cases is possible. Therefore, there is no natural number which can be paired with D, and we have contradicted our original supposition, that there is a bijection between N and P(N).

Note that the set D may be empty. This would mean that every natural number x maps to a set of natural numbers that contains x. Then, every number maps to a nonempty set and no number maps to the empty set. But the empty set is a member of P(N), so the mapping still does not cover P(N).

There is no bijection between N and P(N), only if (each D is taken separately) AND (it is not mapped with some N member).

By using this dead end "proof", one misses the ability to demonstrate the bijection between P(N) and its proper subset N.

By the way, by your provided part taken from http://en.wikipedia.org/wiki/Cantor%27s_theorem , D is called B.

In other words, you have no case.

[doronshadmi]

You are indeed the master of irony.
There is nothing new about the fact the you are one of the players of "running in circles inside a closed box" game.

As a result you are unable to comprehend the links in http://forums.randi.org/showpost.php?p=6967466&postcount=14557 .

[The Man]

Since you do not understand present continuous in terms of "ever increasing", let us express it as "permanently changing", where the permanent (invariant) aspect is like pi among the collection of circles, and the change (variant) aspect is like the curvature among the collection of circles.

Are all of your circles circles, Doron, or are some of your circles not circles?

You can try to dance around with the wording all you want but the tune still hasn’t changed.



The present continuous state, as described above, is the fundamental property of any infinite collection of different objects.

For more details, please look at http://forums.randi.org/showpost.php?p=6963549&postcount=14542 .

Since you still don't understand or perhaps just don't want to accept that all natural numbers are natural numbers please explain to us specifically which natural numbers your think are "permanently changing" into, well, natural numbers. Doron in case your haven’t got it yet, that all natural numbers are natural numbers is your so called “(invariant) aspect” of all natural numbers and the only aspect relevant to them being members of the set of all natural numbers.

It seems now that you have abandoned your “increasing” for just “changing”. Could that be because you were unable (or just unwilling) to address this question…


By all means please explain to us the difference between increasing and decreasing with “no past (before) and no future (after)”?

?

Not to worry I can rephrase to accommodate your change in lexicon.


So by all means please explain to us the difference between changing and unchanging with “no past (before) and no future (after)”?


Again you should try to learn English better. “permanently” is an adverb, a class of words that modify verbs. The verb being modified in your expression is “changing”. So "permanently changing" specifically refers to the act of changing being permanent. Not a combination of some permanent aspects and some changing aspects.

[jsfisher]

You simply ignore this:


You are the one simply ignoring that that isn't part of the proof.

You are the one simply ignoring that the "construction of D" assumes there is an existing bijection between N and P(N).


So, (1) you have been claiming it is part of Cantor's Theorem. It isn't. You fail. (2) You didn't pay attention to that fine detail, the definition for D requires an non-existence bijection. (3) You claim from repeated constructions of different D's from different non-existent bijections you are able to come up with one of these non-existent bijections.

So, you "prove" the impossible by assuming the very impossible thing you are trying to prove.


And after all that, you still cannot provide a bijection between the members of {A} and its power set.

[epix]

However, you made a couple of claims, different from above, that are flatly untrue. You claimed that for any set S,

(1) a bijection exists between S and its power set, P(S).

That amounts to a catastrophic collapse of reason and a supernova of ignorance in connection with Doron's insight into an argument in Wikipedia where the introduction clearly reads

In elementary set theory, Cantor's theorem states that, for any set A, the set of all subsets of A (the power set of A) has a strictly greater cardinality than A itself. For finite sets, Cantor's theorem can be seen to be true by a much simpler proof than that given below, since in addition to subsets of A with just one member, there are others as well, and since n < 2n for all natural numbers n.

[jsfisher]

That amounts to a catastrophic collapse of reason and a supernova of ignorance in connection with Doron's insight into an argument in Wikipedia where the introduction clearly reads

You are just missing his remarkable insight. You see, Cantor's Theorem is proved by contradiction. Assume a construction is possible; derive a contradiction. Simple. But you see, if you simply repeat that impossible construction enough times, sooner or later you must get a different result, right?

And this fully explains the brilliance that is Doron.

[epix]

You are just missing his remarkable insight. You see, Cantor's Theorem is proved by contradiction. Assume a construction is possible; derive a contradiction. Simple. But you see, if you simply repeat that impossible construction enough times, sooner or later you must get a different result, right?

And this fully explains the brilliance that is Doron.
The most remarkable thing is that Doron would not use the nature of the arguments that kept Cantor teaching in Halle and not in Berlin where he wanted to go. Doron says that there is always a bijection between members of N and P(N) with the consequence |N| = |P(N)|. He sees
http://en.wikipedia.org/wiki/Cantor's_theorem
but is unable to follow on the clue provided within the article:

N...........P(N)

1 <---> {4,5}
2 <---> {1,2,3}
3 <---> {4,5,6}
4 <---> {1,3,5}
.
.
.

Set N is an infinite source of natural numbers and P(N) is an infinite source of subsets of N. The natural numbers in the N-bag are ordered 1, 2, 3, 4 . . . (you pick them that way) and the subsets {},{},... are randomly organized in the P(N)-bag. So the example above can be continued infinitely, coz you never run out of natural numbers or the subsets -- there always will be a bijection between one member of N and one member of P(N).

This is much better and understandable argument. After some 50 trillion years of bijecting, the story is the same: Ni <---> P(N)i. The only problem is that after those 50 trillion years, the demonstration still concerns the finite case, coz the latest bijection is known, even though Albert Einstein would show reservations due to his definition of insanity:

Insanity: doing the same thing over and over again and expecting different results.


I'm not sure if Einstein and Cantor ever met.

[zooterkin]

Oops, I accidentally unsubscribed to this thread a few days ago. What did I miss? Has doronshadmi been right about anything yet?

[jsfisher]

Oops, I accidentally unsubscribed to this thread a few days ago. What did I miss? Has doronshadmi been right about anything yet?

Not a single thing.

Happy birthday, on this, the first day of March Madness.

[zooterkin]

Not a single thing.
Colour me unsurprised. :)


Happy birthday, on this, the first day of March Madness.

Thank you!

[epix]

Oops, I accidentally unsubscribed to this thread a few days ago. What did I miss? Has doronshadmi been right about anything yet?
Yes, he maybe right about quitting his prolonged discourse on the revolutionary and eye-opening science of Organic Mathematics.

In case he's not coming back, I think I found a qualified replacement next door:
http://forums.randi.org/showpost.php?p=6980720&postcount=133

[doronshadmi]

Are all of your circles circles, Doron, or are some of your circles not circles?
They are all circles, yet the exact amount of them is unsatisfied (it is permanently changing), or in other words, the set of all circles has no exact size, which is a fact that you ignore and as a result you do not understand the inherent property of "+1" (permanently next ...) of the concept of cardinality.

Here is another example of the unsatisfied exact amount of an infinite set.

{...} is the set of all differences.

{...} is a member of {...}, such that {{...},...} where {...} is different than the rest ,... of {{...},...}, but then

{{...},...} is a member of {{...},...} such that {{{...},...},{...},...} where {{...},...} is different than the rest ,{...},... of {{{...},...},{...},...} but then

... etc. ... ad infinitum ... such that the size of the set of all differences is inherently unsatisfied.

[doronshadmi]

The only problem is that after those 50 trillion years, the demonstration still concerns the finite case, coz the latest bijection is known,
There is no latest bijection in an infinite bijection.

[doronshadmi]

You are just missing his remarkable insight. You see, Cantor's Theorem is proved by contradiction. Assume a construction is possible; derive a contradiction. Simple. But you see, if you simply repeat that impossible construction enough times, sooner or later you must get a different result, right?

And this fully explains the brilliance that is Doron.
Eech contradiction provides an explicit P(N) member that is mapped with an explicit N member, such that

{} ↔ 1
{1,2,3,...} ↔ 2

and the rest of N members are mapped with the rest of the P(N) members between {} and {1,2,3,...}.

It is all done by using Cantors construction method where B
( B = { x ∈ A : x ∉ f(x) } ) ( please see http://en.wikipedia.org/wiki/Cantor%27s_theorem ) is a placeholder for explicit |P(N)| P(N) members that are mapped in a bijection with N members, which are members of N that is a proper subset of P(N).

As we all know, there is a bijection between a set and its proper subset among infinite sets, in the case of N ↔ P(N) the the proper subset is N members.

Another irrelevant conclusion, seen in http://en.wikipedia.org/wiki/Cantor%27s_theorem is:

Another way to think of the proof is that B, empty or non-empty, is always in the power set of A. For f to be onto, some element of A must map to B. But that leads to a contradiction: no element of B can map to B because that would contradict the criterion of membership in B, thus the element mapping to B must not be an element of B meaning that it satisfies the criterion for membership in B, another contradiction. So the assumption that an element of A maps to B must be false; and f can not be onto.

It is irrelevant exactly as the contradiction of Cantor's "proof" does not prevent the construction of explicit P(N) members, and the bijection between explicit P(N) members and explicit N members.

[epix]

There is no latest bijection in an infinite bijection.

1. The set of natural numbers has infinite membership, but that is not an obstacle that would prevent counting and the disemination of the latest figure of that process.
http://www.yaf.org/uploadedImages/Blogs/national_debt_clock.jpg?n=7931

2. The term "infinite bijection" is a classic Doronian expression, coz the prefix bi- means two, and "two" doesn't trail aroma of infinity -- that's for sure.

[jsfisher]

It is all done by using Cantors construction method where B ( B = { x ∈ A : x ∉ f(x) } )

The fact the function, f(), does not / cannot possibly exist is no problem for you at all, is it Doron.

[doronshadmi]

The fact the function, f(), does not / cannot possibly exist is no problem for you at all, is it Doron.

EDIT:

The fact that B is a placeholder for explicit P(N) members (exactly because of the fact that f() does not / cannot possibly exist) does not say anything to you at all, is it jsfisher?

[doronshadmi]

1. The set of natural numbers has infinite membership, but that is not an obstacle that would prevent counting and the disemination of the latest figure of that process.
"latest" or "process" have nothing to do with infinite sets.


2. The term "infinite bijection" is a classic Doronian expression, coz the prefix bi- means two, and "two" doesn't trail aroma of infinity -- that's for sure.
Two infinite sets have "trail aroma of infinity -- that's for sure."

[jsfisher]

EDIT:

The fact that B is a placeholder for explicit P(N) members (exactly because of the fact that f() does not / cannot possibly exist) does not say anything to you at all, is it jsfisher?


Poor grammar aside, yes, it does say quite a lot. Since f() does not exist, if B is defined in terms of f(), then B does not exist either. At least that's the way it works outside of Doronetics. B is not a "placeholder" for anything. It doesn't exist. Moreover, invoking it for multiple "rounds" doesn't bring it into existence.

Doron, you have constructed a trivial bijection using an imaginary construction and produced a result irrelevant for the problem at hand. You keep wandering away from the point: You claimed a bijection exists between any set and its power set. You continue to fail in all attempts to demonstrate even one such example.

[epix]

"latest" or "process" have nothing to do with infinite sets.
The idea that you are wrong begins to emerge in the mind of normally developing Homo sapiens around the age of 8. Here is why: The sequence of natural numbers defined as N doesn't have a bound and can progress infinitely according to n+1 axiom (Peano). Since every member of the progressive sequence of the naturals is distinct, the infinite sequence represents an infinite set. The natural numbers are also called "counting numbers"
http://mathworld.wolfram.com/CountingNumber.html
after the way they've been used since Homo erectus. Since counting is a process designed to answer a question regarding quantities, there surely is something called "the latest count."
http://www.swingstateproject.com/diary/7493/aksen-the-latest-count

The neural axons of Homo sapiens are suppossed to create path

/count/counting numbers/natural numbers/sequence of natural numbers/ N/set of natural numbers/infinite sets

but sometimes the neural association fails, as seen in you quote. Does it mean "back to the drawing board?" (http://www.cnsforum.com/content/pictures/imagebank/hirespng/genetics_bipolar_4.png)

But that would mean the devolution of human species back to time T-Gen0!

"What's the latest banana count?" (http://www.magnusandcheryl.com/images/reallybigtrip/26-27/med_27_74_murchison_chimps_in_tree_2_closeup.jpg)

:jaw-dropp
I thought there were more of them affected.
No, Heavenly Father. It was just Doron.

[doronshadmi]

Poor grammar aside, yes, it does say quite a lot. Since f() does not exist, if B is defined in terms of f(), then B does not exist either.
Wrong, B is defined by the opposite terms of f(), such that "B exists AND f() does not exits" is a true proposition.

You do not understand this part ( http://en.wikipedia.org/wiki/Cantor%27s_theorem ):

Thus there is no x such that f(x) = B; in other words, B is not in the image of f. Because B is in the power set of A, the power set of A has a greater cardinality than A itself.
and the fact that B is a placeholder for |P(A)| P(A) explicit members, that are mapped with explicit A members, which are members of A that is a proper subset of P(A), where both A and P(A) are infinite sets, and it is well-known that there is a bijection between infinite sets and the members of some of their proper subset.

This fact is shown between, for example, the set of natural numbers and its proper subset of, for example, even numbers,

This fact is shown between, for example, the set of rational numbers and its proper subset of natural numbers (the n/1 form).

This fact is shown between, for example, the set P(N) and its proper subset of N members, such that, by using Cantor's construction method, one explicitly defines (without exceptional) the P(N) members, which enables him\her to define the bijection between the explicit P(N) members, and the explicit N members, where N is a proper subset of P(N), and such a bijection is a fact among infinite sets and any arbitrary given proper subset of them.

[doronshadmi]

The idea that you are wrong begins to emerge in the mind of normally developing Homo sapiens around the age of 8. Here is why: The sequence of natural numbers defined as N doesn't have a bound and can progress infinitely according to n+1 axiom (Peano). Since every member of the progressive sequence of the naturals is distinct, the infinite sequence represents an infinite set. The natural numbers are also called "counting numbers"
http://mathworld.wolfram.com/CountingNumber.html
after the way they've been used since Homo erectus. Since counting is a process designed to answer a question regarding quantities, there surely is something called "the latest count."
http://www.swingstateproject.com/diary/7493/aksen-the-latest-count

The neural axons of Homo sapiens are suppossed to create path

/count/counting numbers/natural numbers/sequence of natural numbers/ N/set of natural numbers/infinite sets

but sometimes the neural association fails, as seen in you quote. Does it mean "back to the drawing board?" (http://www.cnsforum.com/content/pictures/imagebank/hirespng/genetics_bipolar_4.png)

But that would mean the devolution of human species back to time T-Gen0!

"What's the latest banana count?" (http://www.magnusandcheryl.com/images/reallybigtrip/26-27/med_27_74_murchison_chimps_in_tree_2_closeup.jpg)

:jaw-dropp
I thought there were more of them affected.
No, Heavenly Father. It was just Doron.
All you demonstrate is your serial step-by-step reasoning of the considered subject, by ignore the parallel observation of the considered subject, which is done at present continuous state, where "before" or "after" do not hold.

In other words, by using only serial step-by-step reasoning of the considered subject, one can't grasp propositions like "permanently changing" as a present continuous (parallel) observation.

The inability to use parallel observation in addition to serial observation, is the essence of the death of Evolution.

[doronshadmi]

Any attempt to completely cover an infinitely long straight line by points, is doomed to fail, because:

1) The all pairs of points (marked by red points in the following diagram) along an infinity long straight line are the result of circles with unique curvature degrees, as shown in the following diagram:

http://farm6.static.flickr.com/5134/5533739885_1b5a702131_b.jpg

2) Being a circle is based on a measurable constant, known as pi=circumference/diameter.

3) By fact (2) the common center point of the circles along the line, is inaccessible to the set of all circles along the line, because pi does not exist at the center point.

4) By fact (2) the common infinitely long straight line along the set of all circles is inaccessible to this set, because pi does not exist at the straight line state.

[laca]

Any attempt to completely cover an infinitely long straight line by points, is doomed to fail, because: <snipped garbage>


Oh my, you're back to that again? Please, do show a part of a line where there is no point.

[zooterkin]

3) By fact (2) the common center point of the circles along the line, is inaccessible to the set of all circles along the line, because pi does not exist at the center point.


So you're saying that there is no point at the common centre point of the circles? :boggled:

[doronshadmi]

So you're saying that there is no point at the common centre point of the circles? :boggled:
No, I say that this point is inaccessible to the set of circles (and theirs associated points along the line) which are based on the fact that in order to be considered as a circle, pi must be a constant property of the considered object.

As for the question about a part of a line that is not covered by points, the answer is very simple:

Any arbitrary pair of points is associated with the collection of all circles, which can't completely cover an infinitely long straight line, as clearly shown in http://forums.randi.org/showpost.php?p=6985580&postcount=14581 .

[doronshadmi]

http://www.mathacademy.com/pr/prime/articles/cantor_theorem/index.asp is a clear example of Cantor's theorem as a proof by contradiction, which leads to contradiction if one tries to define mapping between an explicit P(S) member and S member, because of the construction rules of the explicit P(S) member (the member of S must be AND can't be a member of the explicit P(S) member, according to the construction rules of the explicit P(S) member, under Cantor's theorem).

Also since P(S) is not less then S (because it is trivial to show that all S members (for example {a,b,c,d,...}) are at least mapped with {{a},{b},{c},{d},...} P(S) members), then by using this fact and the contradiction shown above, one must conclude that P(S) is a larger set than S.

----------------------------------------------

But this is not the only way to look at this case, for example, we are using Cantor's construction method to systematically and explicitly define P(S) members, for example:

By using the trivial mapping between {a,b,c,d,...} S members and P(S) {{a},{b},{c},{d},...} P(S) members, we explicitly define P(S) member {}.

Also by using the mapping between {a,b,c,d,...} S members and {{},{a},{b},{c},...} P(S) members, we explicitly define P(S) member {a,b,c,d,...}.

Actually by using Cantor's construction method independently of Cantor's theorem, we are able to explicitly define the all P(S) members between {} and {a,b,c,d,...}.

As a result, there is a bijection between S and P(S) members, as follows:

a ↔ {}
b ↔ {a,b,c,d,...}
c ↔ some explicit P(S) member, which is different than the previous mapped P(S) members

...

etc. ... ad infinitum.

Please be aware of the fact that this construction method has nothing to do with Cantor's theorem exactly because the construction is used independently of Cantor's theorem and therefore it is not restricted to the logical terms of Cantor's theorem.

jsfisher does not understand the mapping between the infinite sets S and P(S) as a bijection between a set and its proper subset, because he gets this subject only in terms of Cantor's theorem and its logical conditions, which ,again, have nothing to do with Cantor's construction method of explicit P(S) members and their bijection with S (which is a proper subset of P(S)) members.

[The Man]

They are all circles, yet the exact amount of them is unsatisfied (it is permanently changing), or in other words, the set of all circles has no exact size, which is a fact that you ignore and as a result you do not understand the inherent property of "+1" (permanently next ...) of the concept of cardinality.

Here is another example of the unsatisfied exact amount of an infinite set.

{...} is the set of all differences.

{...} is a member of {...}, such that {{...},...} where {...} is different than the rest ,... of {{...},...}, but then

{{...},...} is a member of {{...},...} such that {{{...},...},{...},...} where {{...},...} is different than the rest ,{...},... of {{{...},...},{...},...} but then

... etc. ... ad infinitum ... such that the size of the set of all differences is inherently unsatisfied.

So the set of all circles includes all of your circles? What do you mean "the exact amount"? All denotes, well, all, regardless of the amount, "exact" or otherwise.

[The Man]

Any attempt to completely cover an infinitely long straight line by points, is doomed to fail, because:

1) The all pairs of points (marked by red points in the following diagram) along an infinity long straight line are the result of circles with unique curvature degrees, as shown in the following diagram:

http://farm6.static.flickr.com/5134/5533739885_1b5a702131_b.jpg

2) Being a circle is based on a measurable constant, known as pi=circumference/diameter.

3) By fact (2) the common center point of the circles along the line, is inaccessible to the set of all circles along the line, because pi does not exist at the center point.

4) By fact (2) the common infinitely long straight line along the set of all circles is inaccessible to this set, because pi does not exist at the straight line state.


Points aren't circles. So your claim is now that an infinitely long straight line can't be completely covered by concentric circles? How about circles that aren't all, well, concentric?

[jsfisher]

Wrong, B is defined by the opposite terms of f(), such that "B exists AND f() does not exits" is a true proposition.

Doronetics may allow nonsense such as this, but Mathematics does not.

{ x ∈ A : x ∉ f(x) } lacks meaning if f() does not exist, just as { x ∈ A : x ∈ f(x) } lacks meaning when f() does not exist. Any expression that includes an undefined term is, well, undefined.

[epix]

Wrong, B is defined by the opposite terms of f(), such that "B exists AND f() does not exits" is a true proposition.

You do not understand this part ( http://en.wikipedia.org/wiki/Cantor%27s_theorem ):

and the fact that B is a placeholder for |P(A)| P(A) explicit members, that are mapped with explicit A members, which are members of A that is a proper subset of P(A), where both A and P(A) are infinite sets, and it is well-known that there is a bijection between infinite sets and the members of some of their proper subset.

This fact is shown between, for example, the set of natural numbers and its proper subset of, for example, even numbers,

This fact is shown between, for example, the set of rational numbers and its proper subset of natural numbers (the n/1 form).

This fact is shown between, for example, the set P(N) and its proper subset of N members, such that, by using Cantor's construction method, one explicitly defines (without exceptional) the P(N) members, which enables him\her to define the bijection between the explicit P(N) members, and the explicit N members, where N is a proper subset of P(N), and such a bijection is a fact among infinite sets and any arbitrary given proper subset of them.
Sure, there is a condition where it is impossible to point a finger at an instance where you have a member of a power set of an infinite set on the right side and a blank space on the left where the natural bijector should be without creating certain characteristic within the whole construct. Suppose that you ask a mathematician to show you a particular member of a power set for which there is no natural correspondent to complete the bijection. That's impossible to do. But there is a condition where it is possible and the condition depends on the organization of data in the power set. Let's start bijecting the members of N and P(N) which are organized in a comprehensive manner to rule out random distribution . . .

N............P(N)

1 <-----> {1}
2 <-----> {2}
3 <-----> {3}
4 <-----> {4}
5 <-----> {5}
6 <-----> {6}
.
.
.

After some 258 trillion years of successful bijecting, it will dawn on the thinkers, who subscribe to the school of parallel reasoning, that the chance that there will ever be a bijection between a natural number and a member of the power set that contains more than one element, such as {a,b}, is running rather slim due to the infinity of

{1}, {2}, {3}, {4}, {5}, {6}, . . .

In order to register the first instance of {#,#} = {1,2}, the last {#} must be known, which is not possible, coz there is no last {#} -- the sequence of the power set members containing just one element is infinite. So given the particular data organization, the power set is not denumerable, or is not countable. Can we conclude with

This is a contradiction[, and the existence of this contradiction shows that no element of X can be matched with this subset. Our match-up cannot be complete. And since we cannot make a one-to-one match-up between X and P(X), and since we already saw that P(X) cannot be smaller than X, the only possible conclusion is that P(X) must be larger than X. This completes the proof of Cantor’s Theorem.
???

The set of natural numbers features two well-known subsets: subset ODD and subset EVEN. In order to render N uncountable (say what?), the set is defined as

N = {1, 3, 5, 7, 9, . . .}

where the first member of the EVEN subset appears right after the sequence of the odd numbers ends. That will happen when Ruckenbaur III (http://3.bp.blogspot.com/_wW4mfl0E6gg/Sc8ILtNDbQI/AAAAAAAAAKU/11TqFmwUafY/s400/KingCobra02.jpg) spells "Cantor" backward without making a mistake.

[doronshadmi]

Doronetics may allow nonsense such as this, but Mathematics does not.

{ x ∈ A : x ∉ f(x) } lacks meaning if f() does not exist, just as { x ∈ A : x ∈ f(x) } lacks meaning when f() does not exist. Any expression that includes an undefined term is, well, undefined.
EDIT:

jsfisher, if B (the member of P(S)) does not exist, Cantor's theorem can't provide a P(S) member, which is not mapped with some S member, and therefore it can't conclude that |P(S)| > |S|, so also by your reasoning (B exists) AND (f() as "only a mapping between S and P(S)", does not exist), such that there is f() that is not a mapping between S and P(S) (where in this case B is always an explicit member of P(S), which is a fact that Cantor's theorem ignores).

Please tell to the posters of this thread why do you ignore http://forums.randi.org/showpost.php?p=6986611&postcount=14585 ?

[doronshadmi]

So the set of all circles includes all of your circles? What do you mean "the exact amount"? All denotes, well, all, regardless of the amount, "exact" or otherwise.

The set has only circles and yet it is inherently and permanently changing, such that the smallest and the biggest circles do not exist NOW at present continuous state, and as a result the amount of such a set is unsatisfied NOW at present continuous state.

[doronshadmi]

Points aren't circles.
Yet, as I show, any possible pair of points along an infinitely long straight line is associated with a circle with unique curvature, where the collection of all circles can't completely cover an infinitely long straight line, and so is the set of all points that are associated to the set of all circles.


So your claim is now that an infinitely long straight line can't be completely covered by concentric circles? How about circles that aren't all, well, concentric?
Since the smallest circle does not exist, then also the set of all non-concentric circles can't completely cover an infinitely long straight line.

Furthermore, since a set is a collection of different members (any given member is different than the rest of the members), then the set of all concentric circles and the set of all non-concentric circles, is actually the same set.

[doronshadmi]

Sure, there is a condition where it is impossible to point a finger at an instance where you have a member of a power set of an infinite set on the right side and a blank space on the left where the natural bijector should be without creating certain characteristic within the whole construct. Suppose that you ask a mathematician to show you a particular member of a power set for which there is no natural correspondent to complete the bijection. That's impossible to do. But there is a condition where it is possible and the condition depends on the organization of data in the power set. Let's start bijecting the members of N and P(N) which are organized in a comprehensive manner to rule out random distribution . . .

N............P(N)

1 <-----> {1}
2 <-----> {2}
3 <-----> {3}
4 <-----> {4}
5 <-----> {5}
6 <-----> {6}
.
.
.

After some 258 trillion years of successful bijecting, it will dawn on the thinkers, who subscribe to the school of parallel reasoning, that the chance that there will ever be a bijection between a natural number and a member of the power set that contains more than one element, such as {a,b}, is running rather slim due to the infinity of

{1}, {2}, {3}, {4}, {5}, {6}, . . .

In order to register the first instance of {#,#} = {1,2}, the last {#} must be known, which is not possible, coz there is no last {#} -- the sequence of the power set members containing just one element is infinite. So given the particular data organization, the power set is not denumerable, or is not countable. Can we conclude with

???

The set of natural numbers features two well-known subsets: subset ODD and subset EVEN. In order to render N uncountable (say what?), the set is defined as

N = {1, 3, 5, 7, 9, . . .}

where the first member of the EVEN subset appears right after the sequence of the odd numbers ends. That will happen when Ruckenbaur III (http://3.bp.blogspot.com/_wW4mfl0E6gg/Sc8ILtNDbQI/AAAAAAAAAKU/11TqFmwUafY/s400/KingCobra02.jpg) spells "Cantor" backward without making a mistake.
epix, it is well known that there is a bijection between a set and its proper subset, if the set is infinite, exactly as shown in http://forums.randi.org/showpost.php?p=6985470&postcount=14579 (which is a fact that is ignored by you).

[doronshadmi]

In http://www.tcs.tifr.res.in/~raja/publications/online/yapct08.pdf there are several ways to prove that |P(S)| > |S|, yet all of them ignore the fact that Cantor's construction method of P(S) members is not restricted to the logic that is used by each one of the given proofs.

[The Man]

The set has only circles and yet it is inherently and permanently changing, such that the smallest and the biggest circles do not exist NOW at present continuous state, and as a result the amount of such a set is unsatisfied NOW at present continuous state.

Changing? How? What specifically is changing about your set of all circles? What circle was not a circle before you changed your set of all circles to include that circle? "the smallest and the biggest circles do not exist NOW at present continuous state"? If that "present" "state" is "continuous" then it can't, well, change. You still simply have nothing to add to your set of all circles which will change that set "NOW" or at anytime.


Again the question was...

So the set of all circles includes all of your circles?

not 'does your set of all circles have only circles'?


Again as well...

So by all means please explain to us the difference between changing and unchanging with “no past (before) and no future (after)”?

[The Man]

Yet, as I show, any possible pair of points along an infinitely long straight line is associated with a circle with unique curvature, where the collection of all circles can't completely cover an infinitely long straight line, and so is the set of all points that are associated to the set of all circles.

Not all circles are concentric Doron, you have simply limited the circles you are considering to the set of all concentric circles and then seem stupefied (and expect us to be as well) as to why you can't pair the center point with another point on some circle. Please learn some basic geometry.


Since the smallest circle does not exist, then also the set of all non-concentric circles can't completely cover an infinitely long straight line.

Wrong, it has absolutely nothing to do with "the smallest circle". Try non-concentric circles of any fixed diameter (the distance between each of your pairs of points).



Furthermore, since a set is a collection of different members (any given member is different than the rest of the members), then the set of all concentric circles and the set of all non-concentric circles, is actually the same set.

Wow, that is really a bizarre claim even for you. So you don't understand the difference between concentric and non-concentric circles? Let me try to make it simple for you. The circles of the set of all concentric circles all have the same center point (that's what makes them, well, concentric). So the center point is specifically not an aspect by which any of the members of that set will differ from any other member of that set. However, the set of all non-concentric circles (even just all non-concentric circles of some fixed diameter) specifically only includes circles that do not have the same center point as any other circle in that set. So the center point is specifically an aspect by which all of the members of that set will differ from any other member of that set. You assert "since a set is a collection of different members (any given member is different than the rest of the members)" yet obviously and capriciously fail to take note of exactly where the members of each set differ from other members of that set and where they do not.

[jsfisher]

EDIT:

jsfisher, if B (the member of P(S)) does not exist, Cantor's theorem can't provide a P(S) member

Cantor's Theorem doesn't "provide a P(S) member". Cantor's Theorem establishes that |S| < |P(S)| for any set, S. The proof of the theorem relies on showing the very set you are so fascinated by does not exist.

[doronshadmi]

Not all circles are concentric Doron, you have simply limited the circles you are considering to the set of all concentric circles and then seem stupefied (and expect us to be as well) as to why you can't pair the center point with another point on some circle. Please learn some basic geometry.



Wrong, it has absolutely nothing to do with "the smallest circle". Try non-concentric circles of any fixed diameter (the distance between each of your pairs of points).




Wow, that is really a bizarre claim even for you. So you don't understand the difference between concentric and non-concentric circles? Let me try to make it simple for you. The circles of the set of all concentric circles all have the same center point (that's what makes them, well, concentric). So the center point is specifically not an aspect by which any of the members of that set will differ from any other member of that set. However, the set of all non-concentric circles (even just all non-concentric circles of some fixed diameter) specifically only includes circles that do not have the same center point as any other circle in that set. So the center point is specifically an aspect by which all of the members of that set will differ from any other member of that set. You assert "since a set is a collection of different members (any given member is different than the rest of the members)" yet obviously and capriciously fail to take note of exactly where the members of each set differ from other members of that set and where they do not.
The Man, once again your poor reasoning is demonstrated.

In this case you are unable to understand that, whether different circles are organized around a single point or not, these are the same circles, where their organization method is not significant.

Furthermore, the set of all different circles (which is a one and only one set, no matter how its members are organized) is incomplete exactly because it does not have the smallest or the biggest circle.

Indeed your poor reasoning has to be upgraded in order to deal with real Geometry or real Set theory.

[doronshadmi]

Cantor's Theorem doesn't "provide a P(S) member". Cantor's Theorem establishes that |S| < |P(S)| for any set, S. The proof of the theorem relies on showing the very set you are so fascinated by does not exist.
And in oder to establish that |S| < |P(S)| for any set, S, it must show an explicit P(S) member, which is not in the range of S.

[jsfisher]

And in oder to establish that |S| < |P(S)| for any set, S, it must show an explicit P(S) member, which is not in the range of S.

No, it does not. Even if I translate your gibberish into something meaningful like "an explicit P(S) member that is not in any mapping from S to P(S) must be shown", it is still wrong.

No explicit member is shown. All that is shown, all that needs to be shown is that for any mapping from S to P(S), any mapping at all, there must exist at least one element of P(S) not mapped by an element of S.

[epix]

epix, it is well known that there is a bijection between a set and its proper subset, if the set is infinite, exactly as shown in http://forums.randi.org/showpost.php?p=6985470&postcount=14579 (which is a fact that is ignored by you).
:confused:
Is there any conclusion limping behind your reply, like that if your statement is true, the power set of natural numbers must have greater cardinality than the set of the natural numbers?

Since by definition the cardinality of a proper subset must be less than the cardinality of the set, your statement is rather strange. Consider the infinite set of counting numbers and one of its proper subsets, such as A = {3, 1, 2}.

C..........A
1 <---> 3
2 <---> 1
3 <---> 2
4 <--->
5 <--->
6 <--->
.
.
.

As you see, there can be only a partial bijection due to aleph0 > |A|

[epix]

EDIT:

jsfisher, if B (the member of P(S)) does not exist, Cantor's theorem can't provide a P(S) member, which is not mapped with some S member, and therefore it can't conclude that |P(S)| > |[S|
B does exist, but no member of of N is onto B, given the axiomatic framework that this particular edition of the set theory runs on. Look again at the conclusion in Wiki (Cantor's Theorem) where B is defined as D.

Therefore, there is no natural number which can be paired with D, and we have contradicted our original supposition, that there is a bijection between N and P(N).

[doronshadmi]

No, it does not. Even if I translate your gibberish into something meaningful like "an explicit P(S) member that is not in any mapping from S to P(S) must be shown", it is still wrong.

No explicit member is shown. All that is shown, all that needs to be shown is that for any mapping from S to P(S), any mapping at all, there must exist at least one element of P(S) not mapped by an element of S.
In other words, this one element of P(S) that is not mapped by an element of S, is en explicit P(S) member, and by using Cantor's construction method independently of Cantor's theorem ( as shown in http://forums.randi.org/showpost.php?p=6986611&postcount=14585 ) one enables to construct explicit P(S) members without exceptional and define mapping between them and S members, again, without exceptional.

The result is a bijection between the infinite members of P(S) and S, exactly as shown in http://forums.randi.org/showpost.php?p=6986611&postcount=14585 , which you simply ignore and as a result you (the community of current formalists, Platonists and, so called, logicians of Mathematicsת, which accepts the existence of infinite sets) are closed under your own ignorance.

Shell we continue to celebrate such ignorance? I do not think so!

Real Mathematics is leaking between your ignorant fingers.

[doronshadmi]

:confused:
Is there any conclusion limping behind your reply, like that if your statement is true, the power set of natural numbers must have greater cardinality than the set of the natural numbers?

Since by definition the cardinality of a proper subset must be less than the cardinality of the set, your statement is rather strange. Consider the infinite set of counting numbers and one of its proper subsets, such as A = {3, 1, 2}.

C..........A
1 <---> 3
2 <---> 1
3 <---> 2
4 <--->
5 <--->
6 <--->
.
.
.

As you see, there can be only a partial bijection due to aleph0 > |A|

Look at this:

N..........E
1 <---> 2
2 <---> 4
3 <---> 6
4 <---> 8
5 <---> 10
6 <---> 12
.
.
.

If both sets are infinite, then there is a bijection between a set (in this case, the set of natural numbers) and (in this particular case) the set of even numbers, which are a proper subset of the set natural numbers.

[epix]

Look at this:

N..........E
1 <---> 2
2 <---> 4
3 <---> 6
4 <---> 8
5 <---> 10
6 <---> 12
.
.
.

If both sets are infinite, then there is a bijection between a set (in this case, the set of natural numbers) and (in this particular case) the set of even numbers, which are a proper subset of the set natural numbers.
I know what you meant by

epix, it is well known that there is a bijection between a set and its proper subset, if the set is infinite, exactly as shown in http://forums.randi.org/showpost.php...ostcount=14579 (which is a fact that is ignored by you).

but your assertion didn't distinguish between the finite and infinite case of a proper subset. In this type of contention, you need to be really specific otherwise goulash.

So there are two subsets of N: one is made of even numbers and the other of odd numbers. In your example above, you started to map the subset made of even numbers. How long would it take before the mapping involves the subset made of odd numbers. Any idea?

[doronshadmi]

but your assertion didn't distinguish between the finite and infinite case of a proper subset. In this type of contention, you need to be really specific otherwise goulash.
1) It was demonstrated between finite P(S) and its finite proper subset S, by using finite amount of members taken from infinite S ( as shown in http://forums.randi.org/showpost.php?p=6963549&postcount=14542 ).

2) But now we are talking about the case where both P(S) and S are infinite sets.

Any way, in both cases the existence of infinite S is involved, where in (1) case only a finite amount, taken from infinite S, is used.


So there are two subsets of N: one is made of even numbers and the other of odd numbers. In your example above, you started to map the subset made of even numbers. How long would it take before the mapping involves the subset made of odd numbers. Any idea?
At no time.

It simply doesn't happen, because we are talking about a bijection between two infinite sets of different elements, and in this case it does not matter if one of the infinite sets of different elements is a proper subset of the other infinite set of different elements.

[doronshadmi]

Not all circles are concentric Doron, you have simply limited the circles you are considering to the set of all concentric circles ...
I am talking exactly about the set of all circles with different curvatures.

In this case, each circle's curvature exists exactly once as property of a member of this set.

By using the members of this set along an infinitely long straight line, their intersections with this line do not completely cover that line, no matter how the circles are organized along this line.

The reason for that remains the same (at any organized way) which is:

The set of all circles with different curvatures does not have the smallest circle or the biggest circle, as shown in:

http://farm6.static.flickr.com/5134/5533739885_1b5a702131_b.jpg

and any attempt to be (totally curved (being a point)) AND (totally straight (being a line)) is possible only by discontinuity, and resulted by an uncovered line.

Furthermore, let the center point be any arbitrary real number along the infinitely long straight line, it does not change the facts described above.

[jsfisher]

In other words, this one element of P(S) that is not mapped by an element of S, is en explicit P(S) member

Add "explicit" to the list of terms you don't understand.

You also continue to render you thoughts as gibberish. Elements of S to not do any mapping. It is nonsense to say "mapped by an element of S", yet you persist.

The mapping would be done by some mapping function. It is this function that would map elements of S to elements of P(S). The proof for Cantor's Theorem shows that there exists no bijection (a particular type of mapping function) between elements of S and P(S). No such bijection exists.

You cannot then talk about explicit elements the mapping function does or does not map because the mapping function does not exist.

Similarly, you cannot construct any examples of the set B used in the proof, since to do that, you'd need a bijection and that simply does not existent.

[epix]

Similarly, you cannot construct any examples of the set B used in the proof, since to do that, you'd need a bijection and that simply does not existent.
I'm not 100% sure how that set B was defined by either you or Doron, but I guess that it is identical to set B in Wiki and set D used in the attempt to translate the formal logic of the proof. Set B must exist in the hypothetical form, otherwise the contradiction would not materialize. The construction of set B defines the set.

For all x the sets B and f(x) cannot be the same because B was constructed from elements of A whose images (under f) did not include themselves.


The construction is similar to the construction of infinite sequences where function f(n)=n2 constructs the infinite set of square numbers, so the hypothesis that the set includes a prime number is false due to the function. If Doron asks for an example of B, he gets its definition and not a particular or "explicit" sample of the sequence, unless a particular example is shown, like in Wiki. In that case B = {1, 3, 4, . . .}, which is pretty much a meaningless exhibit, unless its construction/definition is known.

It's a devilish stuff, coz the sight of the cases of bijection between two infinite sources of elements that obviously head for infinity is pretty convincing to junk Cantor's proof. LOL.

[jsfisher]

I'm not 100% sure how that set B was defined by either you or Doron, but I guess that it is identical to set B in Wiki and set D used in the attempt to translate the formal logic of the proof. Set B must exist in the hypothetical form, otherwise the contradiction would not materialize. The construction of set B defines the set.

Yes, given a bijection, f(), from A to P(S), B is the set, { x ∈ A : x ∉ f(x) }.

And yes, under the hypothetical that f() exists, the set B also exists. However, since the hypothetical condition leads to a contradiction, f() does not in fact exist, and by extension, neither does B.

The problem that comes up is Doron's insistence B not only exists in fact, but exists in a multitude of forms that can be used to build up a bijection already proven to not exist.

Doron is oblivious the fact what he builds is (1) not a bijection and (2) not according to any construction method attributable to Cantor. While both (1) and (2) are trivial and obvious facts, Doron remains deliberately, willfully ignorant.

[doronshadmi]

Add "explicit" to the list of terms you don't understand.

You also continue to render you thoughts as gibberish. Elements of S to not do any mapping. It is nonsense to say "mapped by an element of S", yet you persist.
Thank you for the correction (which actually is nothing but a wrong use in English).

This is a minor English problem, instead of "by" use "with", that's all.

Again you are focused on trivial and minor mistakes that have no impact on the considered subject in order to avoid the real challenge for you as written in http://forums.randi.org/showpost.php?p=6986611&postcount=14585 .

In other words, you have no case and therefore no one has to take your reasoning seriously.

[doronshadmi]

Yes, given a bijection, f(), from A to P(S), B is the set, { x ∈ A : x ∉ f(x) }.

And yes, under the hypothetical that f() exists, the set B also exists. However, since the hypothetical condition leads to a contradiction, f() does not in fact exist, and by extension, neither does B.
Another nonsense made by jsfisher.

B is a member of P(S) that is not mapped with any S member, by Cantor's theorem.

If B does not exist, as jsfisher claims, then there is no way to conclude that |P(S)| > |S| by Cantor's theorem.

Look how jsfisher contradicts himself about B:

However, since the hypothetical condition leads to a contradiction, f() does not in fact exist, and by extension, neither does B

All that is shown, all that needs to be shown is that for any mapping from S to P(S), any mapping at all, there must exist at least one element of P(S) not mapped by an element of S.
This "at least one element of P(S) not mapped by an element of S" is exactly B or D as written in http://en.wikipedia.org/wiki/Cantor%27s_theorem

Here is the relevant part taken from http://en.wikipedia.org/wiki/Cantor%27s_theorem, which explicitly says that B is a member of P(S) (and it is exactly a member of P(S) that is not mapped with any S member, by Cantor's theorem):

Thus there is no x such that f(x) = B; in other words, B is not in the image of f. Because B is in the power set of A, the power set of A has a greater cardinality than A itself.

[doronshadmi]

Add "explicit" to the list of terms you don't understand.
Add "reasoning" to the list of terms that you don't understand.

[doronshadmi]

I'm not 100% sure how that set B was defined
Maybe because you ignored http://forums.randi.org/showpost.php?p=6986611&postcount=14585 (please see the link at the top of it (in this link B is called F)).

[jsfisher]

Another nonsense made by jsfisher.

B is a member of P(S) that is not mapped with any S member, by Cantor's theorem.

If B does not exist, as jsfisher claims, then there is no way to conclude that |P(S)| > |S| by Cantor's theorem.

Proof by contradiction is another concept that escapes you, Doron.

Nonetheless, since you are so convinced this set, B, must exist in all cases, construct one for us using S = {A} in the example. All you need do, Doron, is provide a bijection, f(), from S to P(S). From that, the set B will follow.

Proceed, please.

[jsfisher]

B is a member of P(S) that is not mapped with any S member, by Cantor's theorem.

By the way, Doron, this is another demonstration of your muddled thought processes. You've been very reliable about confusing theorems, proofs, and examples, and conflating them at inappropriate opportunities.

Cantor's Theorem does nothing of the kind. It does not map elements of P(S), nor does it construct sets named B. Cantor's Theorem simply states that |S| < |P(S)|.

The reference proof for Cantor's Theorem, on the other hand, does rely on mapping functions, sets named B, and such. And an example used to explain the proof includes a set named D.

[epix]

Maybe because you ignored http://forums.randi.org/showpost.php?p=6986611&postcount=14585 (please see the link at the top of it (in this link B is called F)).
I can see it now. The fans of Cantor's theorem usually define that particular subset with B (stands for Bijection), but that dude in your reference used F.

Speaking of ignorance . . . The symbolism

# <---> @

does not automatically refer to Bijection. That function is a composition of Injection AND Surjection. In Cantor's Theorem proof, there is no bijection due to the absence of surjection. That may confuse the proof when the insufficient symbolism # <---> @ enters the intuitive logic that makes the proof improbable.

Rewrite the Wiki informal proof in such a way that the D set, which is B in the formal definition, is made of odd or even numbers. Then hit an exhaustive reference to the mapping functions. After that, it maybe easier for you to see the absence of surjection in the proof.

You need to adjust the axiomatic framework that allows the proof to exist in order to toss the proof into the garbage. But you may fall into the garbage as well, coz setting up a consistent set of axioms that could run a complex environment such as the set theory is akin to designing a bug-free computer operating system. I don't like that Cantor's proof -- it reminds me the case when a judge let a criminal go free on technicalities involving a loophole in the penal code. But unless you change the law/axioms, you need to abide by it.

[doronshadmi]

Cantor's Theorem does nothing of the kind. It does not map elements of P(S), nor does it construct sets named B. Cantor's Theorem simply states that |S| < |P(S)|.
It shows (by using a contradiction, which is demonstrated by the inability of S member to be both (a member) AND (not a member) of P(S) member) that there is B, which is a member of P(S) that is not in the range of all S members.

Independently of Cantor's theorem, by using B as a placeholder of all P(S) members

Such B is {} if:

a ↔ {a}
b ↔ {b}
c ↔ {c}
d ↔ {d}
...

where a ↔ {}


Such B is {a,b,c,d,...} if:

a ↔ { }
b ↔ {a}
c ↔ {b}
d ↔ {c}
...

where b ↔ {a,b,c,d,...}


Such B is also all the subsets of S between {} and {a,b,c,d,e,...} (in addition to {} and {a,b,c,d,...}), such that

S.....B
a ↔ {}
b ↔ {a,b,c,d,...}
c ↔ some explicit P(S) member, which is different than the previous mapped P(S) members
...

etc. ... ad infinitum, or in other words, there is a bijection between P(S) and its proper subset S.


You simply can't look at http://forums.randi.org/showpost.php?p=6986611&postcount=14585 , which is not Cantor's theorem, isn't jsfisher?

[doronshadmi]

I don't like that Cantor's proof --
If you look at http://forums.randi.org/showpost.php?p=6996924&postcount=14618 (including the link that is found in it) you are also able to understand the reason of why you don't like Cantor's proof.


You need to adjust the axiomatic framework that allows the proof to exist in order to toss the proof into the garbage.
I show that the axiomatic framework that allows the proof to exist, is itself garbage.

[jsfisher]

It shows (by using a contradiction, which is demonstrated by the inability of S member to be both (a member) AND (not a member) of P(S) member) that there is B, which is a member of P(S) that is not in the range of all S members.

You continue with your willful gibberish. Cantor's Theorem is |S| < |P(S)|, nothing more.

Independently of Cantor's theorem, by using B as a placeholder of all P(S) members

With this gibberish are you trying to say that the set B, is defined in the reference proof, must be an element of P(S)?

Such B is {} if:

a ↔ {a}
b ↔ {b}
c ↔ {c}
d ↔ {d}
...

where a ↔ {}

...except that there are two problem with this. First, your list of mappings is not a bijection, as required in the reference proof, so it cannot be used in constructing B. Second, you begin with the claim, a ↔ {a}, but then change it was something else, a ↔ {}.

Double Doron fail.

[doronshadmi]

You continue with your willful gibberish. Cantor's Theorem is |S| < |P(S)|, nothing more.



With this gibberish are you trying to say that the set B, is defined in the reference proof, must be an element of P(S)?



...except that there are two problem with this. First, your list of mappings is not a bijection, as required in the reference proof, so it cannot be used in constructing B. Second, you begin with the claim, a ↔ {a}, but then change it was something else, a ↔ {}.

Double Doron fail.
Thank you for that post, jsfisher.

It demonstrates the exact meaning of "The inability to get things beyond one's box" syndrome.


you begin with the claim, a ↔ {a}, but then change it was something else, a ↔ {}.
It is not a claim, jsfisher.

It is a systematic construction method of all P(S) members.

[jsfisher]

you begin with the claim, a ↔ {a}, but then change it was something else, a ↔ {}.
It is not a claim, jsfisher.

It is a systematic construction method of all P(S) members.

Really? So when you wrote this:

Such B is {} if:

a ↔ {a}
b ↔ {b}
c ↔ {c}
d ↔ {d}
...

where a ↔ {}

Please tell us what you meant by "a ↔ {a}" and how it isn't contradicted by "a ↔ {}".

Also, please tell us how B is constructed from this since the only "construction method" under consideration is from the reference proof and it requires a mapping with properties your mapping (either version) does not have.

Your double Doron fail remains.

[The Man]

The Man, once again your poor reasoning is demonstrated.

In this case you are unable to understand that, whether different circles are organized around a single point or not, these are the same circles, where their organization method is not significant.

It isn’t an “organization method”, Doron it is just a matter of where the center of the circle is located. Try concentric circles but with that center not on your line.


Furthermore, the set of all different circles (which is a one and only one set, no matter how its members are organized) is incomplete exactly because it does not have the smallest or the biggest circle.

Indeed your poor reasoning has to be upgraded in order to deal with real Geometry or real Set theory.


So according to you a circle with its center point on your line is no different than a circle where the center point is not on that line? While both circles may have the same radius and even be coplanar there is one specific difference between them. If you can’t figure it out, I’ll give you a hint: it has something to do with the, well, center points.

[The Man]

I am talking exactly about the set of all circles with different curvatures.

That all have the same center point, a point you are continually and intentionally ignoring as well as the fact that said center point is specifically one that is , by your own designation, on your line.


In this case, each circle's curvature exists exactly once as property of a member of this set.

So what? Again your chosen limitations don’t limit anyone but just you.



By using the members of this set along an infinitely long straight line, their intersections with this line do not completely cover that line, no matter how the circles are organized along this line.

Try concentric circles where the center point is not on your line.


The reason for that remains the same (at any organized way) which is:

Your deliberate and intentional ignorance.



The set of all circles with different curvatures does not have the smallest circle or the biggest circle, as shown in:
http://farm6.static.flickr.com/5134/5533739885_1b5a702131_b.jpg



“The set of all circles”? So “all” is no longer a Doronic no-no? Guess what? “The set of all circles with different curvatures does not have the smallest circle or the biggest circle” simply because such circles can’t be defined within that set.



and any attempt to be (totally curved (being a point)) AND (totally straight (being a line)) is possible only by discontinuity, and resulted by an uncovered line.

Well you've certainly got plenty of discontinuity, particularly in that assertion. I guess that’s why just as a circle isn’t a point a line isn’t a point either. Would you like to try using, I don’t know, say, points, as points?


Furthermore, let the center point be any arbitrary real number along the infinitely long straight line, it does not change the facts described above.

Again try non-concentric circles or concentric circle where the center point is not on your line. Deliberate ignorance will not help you Doron, even though you apparently seem to think it does.

Oh, and again…

So by all means please explain to us the difference between changing and unchanging with “no past (before) and no future (after)”?

[epix]

It is not a claim, jsfisher.

It is a systematic construction method of all P(S) members.
Your systematic mapping between the members of infinite S and P(S) is the peasant proof of P(S) being uncountable.

a <--> {}
b <--> {a}
c <--> {b}
d <--> {c}
e <--> {d}
.
.
.

The systematic construction doesn't allow a multi-element member of P(S) to enter the mapping. Since the cardinality of S is aleph0 and the cardinality of single-element members of P(S) is aleph0 as well, coz there is a bijection, as you nicely demonstrated, then it follows that the cardinality of P(S) must be higher than aleph0. But sets with C>aleph0 are uncountable, coz the cardinality of natural numbers is aleph0. This logic allowed to invent and sustain farming, so we are still here and not extinct due to famine.

I brought this particular set organization to your attention on several occasions, but you either ignored it or no comprende.

There is no N such that subset EVEN NUMBERS precedes subset ODD NUMBERS or vice versa, coz both subsets are infinite and neither of the subsets can appear after the other, coz the membership of both subsets is endless.

Your "systematic construction" of P(S) only constructs subsets of P(S) with cardinality C=1, and the construction of these subsets is heading toward infinity with no chance of the subsets with C>1 ever completing the P(S).

[epix]

Your "systematic construction" of P(S) only constructs subsets of P(S) with cardinality C=1, and the construction of these subsets is heading toward infinity with no chance of the subsets with C>1 ever completing the P(S).

The notion of "complete infinity" can materialize only in your central nervous system, epix, and it may contribute to the extinction of human species if you don't leave next door Heather alone. Try "complementing" the next time.

Aah, His Irreversible Wisdom shuffled in.
:rolleyes:

[doronshadmi]

Please tell us what you meant by "a ↔ {a}" and how it isn't contradicted by "a ↔ {}".
In general, any systematic construction method of an infinite set, such that no member of that set is omitted, allows a bijection with any infinite arbitrary proper subset of the systematically constructed set.

I provide a systematic construction method of P(S) members such that no member of P(S) is omitted, and by using this simple fact, there is a biojection between S members (where S is some proper subset of P(S)) and P(S).

So is the case with the systematic construction method of, for example, N and Q members, which allows the bijection between their members and any arbitrary proper sets of them.

No claims, assumptions or contradictions are involved by using systematic construction methods of infinite sets, and this simple fact can't comprehended by a reasoning that is based on claims, assumptions, contradictions etc. .

[doronshadmi]

The systematic construction doesn't allow a multi-element member of P(S) to enter the mapping.
There is no such thing like a set with multi-element members, because a member of a given set appears once and only once as a member of the considered set.

According to any given set theory {a,a,b}={a,b}, so your argument does not hold.

If you do not follow this restriction, you actually deal with Multi-set theory, which is defiantly not the discussed subject about mapping between sets (multi-sets are not involved), and the mapping between infinite P(S) set and S set, is done only between sets.

[doronshadmi]

So what? Again your chosen limitations don’t limit anyone but just you.
So you are also a person that do not understand that a circle with a certain curvature (where an object is considered as a circle only if pi is its essential property) exists once and only once as a member of the set of all circles with different curvatures.

Any circle not on the line or any circle along the line is already a certain member of the set of all circles with different curvatures (the location of the center point of a given circle is insignificant).

Furthermore, by using a concentric organization of the set of all circles with different curvatures, there is no point on the complex plane (where the real line is a proper subset of the complex plane) which is not associated with some circle with unique curvature.

Yet, it does not change the fact that a point (total curvature, where pi=circumference/diameter does not exist) or an infinitely long straight line (total non-curvature, where pi=circumference/diameter does not exist), are not members of the set of all circles with different curvatures, and any attempt to include them as members of the set of all circles with different curvatures, is resulted by an inherent discontinuity of that extended set, because the smallest circle or the largest circle do not exist.

This discontinuity is considered as an external fact if a point and an infinitely long straight line are not members of the set of all circles with different curvatures.

Actually, there is no difference if the discontinuity is internal or external, because in both cases it is derived from the same reason, which is:

The set of all circles with different curvatures (where an object is considered as a circle only if pi is its essential property) does not have the smallest or the largest circle, and as a result (and because of the discontinuity) the set of all circles with different curvatures can't fully cover an infinitely long straight line.

[doronshadmi]

“The set of all circles”? So “all” is no longer a Doronic no-no?
Exactly, and yet this infinite set is incomplete because its smallest or largest circles do not exist.


“The set of all circles”? So “all” is no longer a Doronic no-no? Guess what? “The set of all circles with different curvatures does not have the smallest circle or the biggest circle” simply because such circles can’t be defined within that set.
Guess what? “The set of all circles with different curvatures does not have the smallest circle or the biggest circle” simply because such circles can’t be defined at all, if this set is infinite, and the considered subject deals only with infinite sets, in case you have missed it.

[doronshadmi]

Well you've certainly got plenty of discontinuity, particularly in that assertion. I guess that’s why just as a circle isn’t a point a line isn’t a point either. Would you like to try using, I don’t know, say, points, as points?

The ability to understand X depends on one's mind ability to be simultaneously beyond AND at X's domain.

Since you do not understand that fundamental fact, you are not able to get the following diagram:

http://farm6.static.flickr.com/5134/5533739885_1b5a702131_b.jpg

which demonstrate the existence of points along an infinitely long straight line, by getting them simultaneously
(beyond (by different circle's curvatures)) AND (at the level of collection of intersecting points along an infinitely long straight line).

For example, an anthropologist really does a valuable research only if he\she is both an external observer AND an internal participator of the researched subject.

[epix]

There is no such thing like a with multi-element members, because a member of a given set appears once and only once as a member of the considered set.

According to any given set theory {a,a,b}={a,b}, so your argument does not hold.

If you do not follow this restriction, you actually deal with Multi-set theory, which is defiantly not the discussed subject about mapping between sets (multi-sets are not involved), and the mapping between infinite P(S) set and S set, is done only between sets.
This is an example of some members of P(N): {1,2,6}, {78}, {100}, {2,4,6,...}. These members are subsets of N and are made of a different number of elements. When the number of the elements exceeds just one element, it's normal to refer to these subsets as multi-element subsets as opposed to single-element subsets, such as {100}. "Multi-element members of the power set" is quite different term from your fantasmagoric translation that is highligted in the quote.

[doronshadmi]

This is an example of some members of P(N): {1,2,6}, {78}, {100}, {2,4,6,...}. These members are subsets of N and are made of a different number of elements. When the number of the elements exceeds just one element, it's normal to refer to these subsets as multi-element subsets as opposed to single-element subsets, such as {100}. "Multi-element members of the power set" is quite different term from your fantasmagoric translation that is highligted in the quote.
EDIT:

In other words, you still do not get http://forums.randi.org/showpost.php?p=6996924&postcount=14618 , which systematically provides the bijection of P(S) members (where P(S) is an infinite set) with the members of S, where S is an infinite proper subset of P(S).

[epix]

EDIT:

In other words, you still do not get http://forums.randi.org/showpost.php?p=6996924&postcount=14618 , which systematically provides the bijection of P(S) members (where P(S) is an infinite set) with the members of S, where S is an infinite proper subset of P(S).
It's obvious that you never come to grasp the fact that bijection is a union of two functions: injection AND surjection. It's like there is no kid/bijection born without function male and function female. There is no bijection when there is no surjection and Cantor's proof explicitly shows that there can be no surjection between X and P(X) due to a contradiction. And if there is no surjection there is no bijection. Like if there is no male there can't be a human fetus, apart for the Jesus fetus. Once again, a <--> b by itself doesn't represent any bijection. You need to prove the existence of both functions injection, which is apparent, and surjection, which can be far from apparent.

Your attempt to disprove Cantor's Theorem amounts to proving that human parthenogenesis is a real deal just by tossing the same photo around:

http://interdenominationaldivineorder.com/gallery/virgin%20mary.jpg

{V.M.} <---> J.

[doronshadmi]

It's obvious that you never come to grasp the fact that bijection is a union of two functions: injection AND surjection. [/size]

epix, you still do not grasp the simple fact, which is:

If one shows a systematic construction method of some infinite set, such that no member of the constructed set is omitted, then any arbitrary infinite proper subset of that set is in bijection with this set.

[doronshadmi]

Please tell us what you meant by "a ↔ {a}" and how it isn't contradicted by "a ↔ {}".

jsfisher I use a very simple principle:

If one shows a systematic construction method of some infinite set, such that no member of the constructed set is omitted, then any arbitrary infinite proper subset of that set is in bijection with this set, that's all.

In other words, it does not matter in what order the members of some infinite proper subset are mapped with systematically constructed infinite set (where "systematically constructed" means that no one of the members of the constructed set is omitted), there is a bijection between the infinite proper subset and the systematically constructed infinite set.

Some examples of infinite sets:

This simple principle is shown between N and, for example, the set of even numbers, which is a proper subset of N, and it does not matter what order is used among the proper subset members.

This simple principle is shown between Q and, for example, the set of natural numbers, which is a proper subset of Q, and it does not matter what order is used among the proper subset members.

This simple principle is shown between P(S) and, for example, S, which is a proper subset of P(S) ( as show in http://forums.randi.org/showpost.php?p=6996924&postcount=14618 ), and it does not matter what order is used among the proper subset members.

[epix]

epix, you still do not grasp the simple fact, which is:

If one shows a systematic construction method of some infinite set, such that no member of the constructed set is omitted, then any arbitrary infinite proper subset of that set is in bijection with this set.
The power set of of the set of natural numbers is not just "some infinite set." I already asked you to finish the "systematic mapping" of subsets {1}, {2}, (3}, ... and start mapping the next category {1,2}, {1,3}, {1,4}, ... , but you are not done yet with the singletons, are you?

In the case of x in R2, that means {a,b}, you can substitute a=numerator and b=denominator and see how Cantor showed that there is a surjection between the naturals and rational numbers.

http://math10blog.files.wordpress.com/2009/03/cantor_set.jpg?w=314&h=321

Cantor spent all members of N bijecting the rationals. If there were some numbers involving numerator, denominator, and one other __ator, Cantor would organize the numbers in a cube (x in R3), and spend all rational numbers on that bijection. What space is required to show surjection for x in R1 AND R2 AND R3 for example?

Your "systematic mapping" of the power set is a fantasy. Actually it is not a fantasy -- it's turbulent mayhem of neural character caused by The Delinquent Pancake of Unusual Desire dilligently spread over time T0-->always.

[doronshadmi]

The power set of of the set of natural numbers is not just "some infinite set." I already asked you to finish the "systematic mapping" of subsets {1}, {2}, (3}, ... and start mapping the next category {1,2}, {1,3}, {1,4}, ... , but you are not done yet with the singletons, are you?

You are wrong epix.

First we are talking about the general P(S);S form, where P(N);N is some particular case of it.

In the general systematic construction of P(S) members {},{a,b,c,...} and all the subsets between {} and {a,b,c,...} are systematically constructed, without omitting even a single P(S) member, and we do not need more than that in order to define a bijection between S members (where S is a proper subset of P(S)) and P(S) members.

All the countable\uncountable mambo jambo does not hold water.


Your "systematic mapping" of the power set is a fantasy.
Again you demonstrate your current inability to get http://forums.randi.org/showpost.php?p=7003057&postcount=14636 and the linked posts in it.

A natural conclusion of such current inability is understood by you as fantasy.

[jsfisher]

jsfisher I use a very simple principle:

If one shows a systematic construction method of some infinite set, such that no member of the constructed set is omitted, then any arbitrary infinite proper subset of that set is in bijection with this set, that's all.

The gibberish continues. Sets are not "in bijection" with sets. If you cannot express yourself without speaking nonsense, don't expect to be an effective communicator.

Be that as it may, you have NOT shown a systematic construction method. Moreover, the phrase "no member of the constructed set is omitted" is tautological. Finally, constructing an infinite set does not establish any sort of bijection.


How about we go back to the basic claim you made, Doron. You have said a bijection can be established between the members of ANY set and its power set. Let's stick to the case where the set is {A}.

Either retract your claim or show a bijection between the members of {A} and {{},{A}}.

[doronshadmi]

The gibberish continues. Sets are not "in bijection" with sets. If you cannot express yourself without speaking nonsense, don't expect to be an effective communicator.
The ignorance continues.

If you are unable the understand that there is a bijection between the members of ,for example, set N and the members of some proper subset of it, for example, the set of even numbers, then your replies do not hold water.


Be that as it may, you have NOT shown a systematic construction method.
Wrong jsfisher, by using your weak reasoning you are failing to see the systematic construction method, as show in http://forums.randi.org/showpost.php?p=7003057&postcount=14636 (and do not forget to follow the links of this post).


Finally, constructing an infinite set does not establish any sort of bijection.

Unless the construction of the infinite set is systematic (no object is omitted), and as a result the members of any arbitrary infinite proper subset of systematically constructed set have, a bijection (1-to-1 and onto mapping ) with the members of infinite systematically constructed set.


How about we go back to the basic claim you made, Doron. You have said a bijection can be established between the members of ANY set and its power set. Let's stick to the case where the set is {A}.

Either retract your claim or show a bijection between the members of {A} and {{},{A}}.
Before we go back, let's first see if you are able to understand that the members of any arbitrary infinite proper subset of systematically constructed set, have a bijection (1-to-1 and onto mapping) with the members of infinite systematically constructed set.

[jsfisher]

The ignorance continues.

If you are unable the understand that there is a bijection between the members of ,for example, set N and the members of some proper subset of it, for example, the set of even numbers, then your replies do not hold water.

There are many sets for which there are bijections between their members and those of specific subsets. Is this a general truth for any set and any of its proper subsets? No.


Now, about that bijection you claim exists between members of {A} and {{}, {A}}....

[epix]

You are wrong epix.

In the general systematic construction of P(S) members {},{a,b,c,...} and all the subsets between {} and {a,b,c,...} are systematically constructed, without omitting even a single P(S) member, and we do not need more than that in order to define a bijection between S members (where S is a proper subset of P(S)) and P(S) members.

Read again:

I already asked you to finish the "systematic mapping" of subsets {1}, {2}, (3}, ... and start mapping the next category {1,2}, {1,3}, {1,4}, ... , but you are not done yet with the singletons, are you?

When do you think you finish systematically bijecting the subsets with a single element and start bijecting the subsets with two elements? I had the opportunity to see the unbelievable: you actually started mapping {a}, {b}, {c}, . . . thinking that one nice day you will continue "systematic mapping" with {a,b}, {a,c}, {a,d}, . . . and then {a,b,c}, {a,b,d}, {a,b,e}, . . . Even Bernie (http://onceuponaparty.com/images/parrot_in_a_hat-sm.jpg) can figure out what will probably take a billion light years for you to comprehend . . .


All the countable\uncountable mambo jambo does not hold water.


. . . and another trillion light years to comprehend what is written bellow.
http://en.wikipedia.org/wiki/Uncountable_set


There are many equivalent characterizations of uncountability. A set X is uncountable if and only if any of the following conditions holds:

1. There is no injective function from X to the set of natural numbers.

2. X is nonempty and every ω-sequence of elements of X fails to include at least one element of X. That is, there is no surjective function from the natural numbers to X.
.
.
.


There is no bijection without injection AND surjection.

Just keep systematic mapping of {a}, {b}, {c}, . . . and I hope that just before Milky Way merges with Andromeda, God finally succeeds to stuff your head with the concept of a function called "surjection."

[epix]

The ignorance continues.

If you are unable the understand that there is a bijection between the members of ,for example, set N and the members of some proper subset of it, for example, the set of even numbers, then your replies do not hold water.

Of course there is a bijection.

1 <--> {2, 4, 6, . . .}

So is

2 <--> {4, 6, 8, . . .}
3 <--> {6, 8, 10, . . .}
4 <--> {8, 10, 12, . . .}
5 <--> {10, 12, 14 . . .}
.
.
.

When you reach the infinity by travelling alongside the even road you can start mapping the subsets made of odd numbers

infinity <--> {1, 3, 5, . . .}
infinity+1 <--> {3, 5, 7, . . .}
.
.
.

When you reach another infinity that ends the odds, you can start mapping the primes and all its infinite subsets.

2*infinity <--> {2, 3, 5, . . .}
2*infinity+1 <--> {3, 5, 7, . . .}
2*infinity+2 <--> {5, 7, 11 . . .}
.
.
.

Traveling the road of phantasmagoric multidimensional infinities, you will stop for a breakfast. A waitress named Surjection (http://i423.photobucket.com/albums/pp314/pauly_cy/pancakebunny.jpg) brings the pancakes to you. But the real Surjection (http://www.flixya.com/files-photo/a/n/g/angkiss1730250.jpg) causes two functions -- erection and ejaculation -- to join the math.

[The Man]

So you are also a person that do not understand that a circle with a certain curvature (where an object is considered as a circle only if pi is its essential property) exists once and only once as a member of the set of all circles with different curvatures.

Again that is just your self imposed limitation and once again concentric circles where the center point in not a point on your meets that limitation. Also the first of those circles that intersects your line will do so only at one point (a tangent). The rest will get your pairs.


Any circle not on the line or any circle along the line is already a certain member of the set of all circles with different curvatures (the location of the center point of a given circle is insignificant).[/qoute]

See above.



Furthermore, by using a concentric organization of the set of all circles with different curvatures, there is no point on the complex plane (where the real line is a proper subset of the complex plane) which is not associated with some circle with unique curvature.

complex plane? So now you have added a -j (-n1/2) axis?. Again see above.




Yet, it does not change the fact that a point (total curvature, where pi=circumference/diameter does not exist) or an infinitely long straight line (total non-curvature, where pi=circumference/diameter does not exist), are not members of the set of all circles with different curvatures, and any attempt to include them as members of the set of all circles with different curvatures, is resulted by an inherent discontinuity of that extended set, because the smallest circle or the largest circle do not exist.


This discontinuity is considered as an external fact if a point and an infinitely long straight line are not members of the set of all circles with different curvatures.

Actually, there is no difference if the discontinuity is internal or external, because in both cases it is derived from the same reason, which is:

The set of all circles with different curvatures (where an object is considered as a circle only if pi is its essential property) does not have the smallest or the largest circle, and as a result (and because of the discontinuity) the set of all circles with different curvatures can't fully cover an infinitely long straight line.

Again see above and once again please learn some basic geometry.



Exactly, and yet this infinite set is incomplete because its smallest or largest circles do not exist.

Exactly what? So “all” is no longer a Doronic no-no?


Guess what? “The set of all circles with different curvatures does not have the smallest circle or the biggest circle” simply because such circles can’t be defined at all, if this set is infinite, and the considered subject deals only with infinite sets, in case you have missed it.

Guess what? “The set of all circles" includes only circles, in case you have missed it.


The ability to understand X depends on one's mind ability to be simultaneously beyond AND at X's domain.

Since you do not understand that fundamental fact, you are not able to get the following diagram:

http://farm6.static.flickr.com/5134/5533739885_1b5a702131_b.jpg

which demonstrate the existence of points along an infinitely long straight line, by getting them simultaneously
(beyond (by different circle's curvatures)) AND (at the level of collection of intersecting points along an infinitely long straight line).

For example, an anthropologist really does a valuable research only if he\she is both an external observer AND an internal participator of the researched subject.

Again see above. Concentric circles where the center point is not a point on your line meets your self-imposed limitations. The first of those circles that intersects your line will do so only at one point (a tangent). The rest will get your pairs. Agian as mentioned before even with your concentric circles centered on a point of your line, as designated by you, they are superficially centered on a point of your, well, line. Again please learn some basic geometry.

[doronshadmi]

...the first of those circles that intersects your line will do so only at one point (a tangent). The rest will get your pairs.
It does not matter.

1) This first circle is omitted from the rest of the circles that get the pairs, so there is a pair of points that does not cover the line, simply becuse there are no two circles with the same curvature in the set of all circles with different curvatures.

2) Your arbitrary first circle is not the smallest and not the largest circle of the collection of all circles with different curvatures, so nothing was changed by your "tangent trick".

In other words, you still have to learn Set theory and Geometry, before you deal with:

http://forums.randi.org/showpost.php?p=6993998&postcount=14607

http://forums.randi.org/showpost.php?p=6998970&postcount=14629

or

http://forums.randi.org/showpost.php?p=6999117&postcount=14631

[epix]

Yet, it does not change the fact that a point (total curvature, where pi=circumference/diameter does not exist) or an infinitely long straight line (total non-curvature, where pi=circumference/diameter does not exist), are not members of the set of all circles with different curvatures, and any attempt to include them as members of the set of all circles with different curvatures, is resulted by an inherent discontinuity of that extended set, because the smallest circle or the largest circle do not exist.

Why is it so that CURVATURE is associated through pi with DOES NOT EXIST the same way as the opposite NON-CURVATURE? Have you been reading Loss Leader's seminal work and found it to your liking?

Why do you relate curvature to the ratio between circumference and diameter at all when the magnitude of a curvature is given by the ratio 1/radius? In other words, the curvature is a reciprocal of circle's radius.

Now I truly understand the meaning of "fire at will."

[doronshadmi]

Is this a general truth for any set and any of its proper subsets? No.

You are wrong again, the answer is Yes.

Once again jsfisher, by using your weak reasoning you are failing to see the systematic construction method, as shown in http://forums.randi.org/showpost.php?p=7003057&postcount=14636 (and do not forget to follow the links of this post).




Now, about that bijection you claim exists between members of {A} and {{}, {A}}....

In this case you are using a finite case, so Cantor's systematic construction method of {{},{A}} members is done as follows:

A ↔ {A} provides {}
A ↔ {} provides {A}

So, as can be seen, no {{},{A}} was omitted, and by using this fact the following mapping holds:

{} ↔ A
{A} ↔ B

Furthermore, also the two next cases hold:

Only one mapping:
{} ↔ A
{A} ↔

No mapping:
{} ↔
{A} ↔


In other words,there is no universal law about mapping, because we are free to define any wished degree of it for our purpose.

Any attempt to force universality on mapping is based on limited understanding of this fine subject.

Let's take for example Hilbert's Hotel ( http://en.wikipedia.org/wiki/Hilbert's_paradox_of_the_Grand_Hote ):

We actually can empty any wished number of rooms from an infinite set of rooms (where an empty room is equivalent to no-mapping), so the whole idea of mapping with some given set is defined according to our wish, and the wished degree is from no-mapping to bijection.

[sympathic]

It does not matter.



In other words, you still have to learn Set theory and Geometry, before you deal with:



Says he who learned neither, and is not willing to learn either.

[doronshadmi]

Why is it so that CURVATURE is associated through pi with DOES NOT EXIST the same way as the opposite NON-CURVATURE?
Any attempt to reach a point or a line form, starting from some arbitrary circle, is done by discontinuity between an arbitrary circle and a point, or by discontinuity between an arbitrary circle and a line.


Why do you relate curvature to the ratio between circumference and diameter at all when the magnitude of a curvature is given by the ratio 1/radius? In other words, the curvature is a reciprocal of circle's radius.
Use 2pi instead of pi, it does not matter.

[epix]

Now, about that bijection you claim exists between members of {A} and {{}, {A}}....
As it turned out, that reminder has proven itself a very powerful diagnostic tool. Don't miss the moment when Doron pulls rabbit B from the { }-hat.

[doronshadmi]

As it turned out, that reminder has proven itself a very powerful diagnostic tool. Don't miss the moment when Doron pulls rabbit B from the { }-hat.
Time is not involved here, so there is no such moment.

Actually because of focusing on time, you are missing what is written in http://forums.randi.org/showpost.php?p=7006557&postcount=14647 .

[epix]

Originally Posted by epix
Why do you relate curvature to the ratio between circumference and diameter at all when the magnitude of a curvature is given by the ratio 1/radius? In other words, the curvature is a reciprocal of circle's radius.

Use 2pi instead of pi, it does not matter.
:confused:
But 2pi doesn't equal 1/r.

First things first: Do you know what year it is?

[doronshadmi]

:confused:
But 2pi doesn't equal 1/r.

Yes it is, if the formula is circumference/radius (instead of circumference/diameter).

[epix]

Yes it is, if the formula is circumference/radius (instead of circumference/diameter).
Doron, give it some rest. Just go for a walk and try not to think about math.

[doronshadmi]

Doron, give it some rest. Just go for a walk and try not to think about math.
I am resting at the source of my mind, where no thoughts exist ( http://www.youtube.com/watch?v=wGLLDEYFAzs ).

[jsfisher]

Is this a general truth for any set and any of its proper subsets? No.

You are wrong again, the answer is Yes.

You really want to go with that answer? Let's take, for example, the set of natural numbers as "the set" and the null set as one of "its proper subsets".

There is no bijection between the members of N and {}.


Now, about that bijection you claim exists between members of {A} and {{}, {A}}....

In this case you are using a finite case, so Cantor's systematic construction
method of {{},{A}}

Whoa!!! Wait a second. This interesting. Given a set, S, and it's power set, P(S), you propose to construct P(S)? That is very adventurous of you, Doron.

...members is done as follows:

A ↔ {A} provides {}
This mapping is not a bijection between S and P(S) nor is it a surjection. So, whatever it is you are doing here, it does not correspond to anything in the reference proof. Just how do you get {} from A ↔ {A}?
A ↔ {} provides {A}
Ditto.

So, as can be seen, no {{},{A}} was omitted

So, you started with P(S) and ended up with P(S)? That is remarkable. We are all very impressed. You did include a few extra steps, though.

...and by using this fact the following mapping holds:

{} ↔ A
{A} ↔ B

Why is this diversion important? The goal was a bijection between the members of {A} and {{},{A}}. This is not such a bijection.

Again, you fail in your attempt to demonstrate your claim.

[doronshadmi]

You really want to go with that answer? Let's take, for example, the set of natural numbers as "the set" and the null set as one of "its proper subsets".

There is no bijection between the members of N and {}.



Whoa!!! Wait a second. This interesting. Given a set, S, and it's power set, P(S), you propose to construct P(S)? That is very adventurous of you, Doron.


This mapping is not a bijection between S and P(S) nor is it a surjection. So, whatever it is you are doing here, it does not correspond to anything in the reference proof. Just how do you get {} from A ↔ {A}?

Ditto.



So, you started with P(S) and ended up with P(S)? That is remarkable. We are all very impressed. You did include a few extra steps, though.



Why is this diversion important? The goal was a bijection between the members of {A} and {{},{A}}. This is not such a bijection.

Again, you fail in your attempt to demonstrate your claim.
Well, your last reply demonstrates again that since you a running in circles within a closed box, you simply unable to understand http://forums.randi.org/showpost.php?p=7006557&postcount=14647 .

So enjoy your run, I am not a participator of it.

[jsfisher]

Well, your last reply demonstrates again that since you a running in circles within a closed box, you simply unable to understand http://forums.randi.org/showpost.php?p=7006557&postcount=14647 .

So enjoy your run, I am not a participator of it.

Excellent. You completely avoided any points raised. Well done!!

[doronshadmi]

Excellent. You completely avoided any points raised. Well done!!
Wrong again, jsfisher, your points are closed within your box, therefore they are not relevant.

Just for example, {} is not a member of N.

Furthermore, we are not talking about finite amount of elements that are mapped with infinitely many elements, so again your reply is a load of nonsense.

[jsfisher]

Wrong again, jsfisher, your points are closed within your box, therefore they are not relevant.

Whatever.

We are all still in awe, though, because you have perfected a method to construct a power set given the power set as a starting point. Well done!

[doronshadmi]

Whatever.

We are all still in awe, though, because you have perfected a method to construct a power set given the power set as a starting point. Well done!
You miss the simple fact, which is:

Definition 1: A systematic construction method of an infinite set does not omit any member of the considered set.

Definition 1 is equivalent to the existence of the considered infinite set (it is a tautology).

Definition 2: By definition 1, for any systematically constructed infinite set, there is a bijection between its members and the members of any arbitrary infinite proper subset of it.

[doronshadmi]

Again that is just your self imposed limitation and once again concentric circles where the center point in not a point on your meets that limitation. Also the first of those circles that intersects your line will do so only at one point (a tangent). The rest will get your pairs. a point of your, well, line. Again please learn some basic geometry.
Even if the set of all circles is defined by consider also the difference of its angle and/or position upon infinitely many dimensional spaces, still this collection does not have the smallest or largest circle, and as a result there is discontinuity between all circles and all points, or between a all circles and and all dimensional spaces.

[epix]

You miss the simple fact, which is:

Definition 1: A systematic construction method of an infinite set does not omit any member of the considered set.

The power set of an infinite set is a set that includes an infinite collection of infinite sets as well as an infinite collection of finite sets. It is something that you've proved to be beyond your imagination. Hence your 1-dim. Definition 1.

[jsfisher]

You miss the simple fact, which is:

Definition 1: A systematic construction method of an infinite set does not omit any member of the considered set.

Definition 1 is equivalent to the existence of the considered infinite set (it is a tautology).

Definition 2: By definition 1, for any systematically constructed infinite set, there is a bijection between its members and the members of any arbitrary infinite proper subset of it.

No, these are not definitions; and the first is a tautology, and the second neither follows from the first nor is it true. I do note you have shifted the goal posts a bit by sliding in the word, infinite, though.

Be that as it may, it is still trivial to produce as output a set that is provided as input. I continue to be impressed by how loudly you tout this particular feat of yours.

And how is that bijection between {A} and {{}, {A}} coming along? Still nothing? Maybe an even easier example would be in order: How about the set {} and its power set {{}}? What would be a bijection between the members of those two sets?

[The Man]

It does not matter.

1) This first circle is omitted from the rest of the circles that get the pairs, so there is a pair of points that does not cover the line, simply becuse there are no two circles with the same curvature in the set of all circles with different curvatures.

"This first circle is omitted from the rest of the circles"? So your set of all circles of different curvatures does not include all circles of different curvatures?


2) Your arbitrary first circle is not the smallest and not the largest circle of the collection of all circles with different curvatures, so nothing was changed by your "tangent trick".

Certainly it is not "the smallest and not the largest circle of the collection of all circles" but as I said it is the first one that intersects your line without being centered on your line. It's no trick but simply a tangent. Again please learn some basic geometry.



In other words, you still have to learn Set theory and Geometry, before you deal with:

http://forums.randi.org/showpost.php?p=6993998&postcount=14607

http://forums.randi.org/showpost.php?p=6998970&postcount=14629

or

http://forums.randi.org/showpost.php?p=6999117&postcount=14631

Nope, that's just you. Please learn some math and geometry then perhaps you can deal with them yourself.

[epix]

Even if the set of all circles is defined by consider also the difference of its angle and/or position upon infinitely many dimensional spaces, still this collection does not have the smallest or largest circle, and as a result there is discontinuity between all circles and all points, or between a all circles and and all dimensional spaces.
Doron, the set of all (in the infinite sense) concentric circles can be defined as A|x in R+ where x is radius. So the set is uncountable. Remember? The diagonal proof? If the set is defined as A|x in Q+, the set is countable. Just define the vision of yours.

[doronshadmi]

"This first circle is omitted from the rest of the circles"? So your set of all circles of different curvatures does not include all circles of different curvatures?
The Man, you still do not know the meaning of: "The set of all circles with different curvatures".

1) Each member of this set has a unique curvature, so if one of the curvatures is used as a tangent, it is not used as one of the centered points, and we get two places along the line, which are not covered by the intersecting points of the circle with the unique curvature that is already used as a tangent.

2) The set of all circles with different curvatures (where pi=circumference/diameter holds) does not have the smallest or the largest circles, so anyway the associated points with the set of all circles with different curvatures, are discontinuous w.r.t the center point (total curvature, where pi=circumference/diameter does not hold) and w.r.t the line (total non-curvature, where pi=circumference/diameter does not hold), or in other words, the set of all these intersecting points, does not completely cover an infinitely long straight line.

[jsfisher]

The Man, you still do not know the meaning of: "The set of all circles with different curvatures".

The fact that it is without meaning might be a factor in that. The rule for set membership must be deterministic. This so-called set would need to contain one and only one circle of unit curvature, but which one? The rule doesn't say, so "the set of all circles with different curvatures" is gibberish.

Then again, it is also clear Doron has concocted his own meaning for the term, curvature, but no one is surprised by this.

[doronshadmi]

Doron, the set of all (in the infinite sense) concentric circles can be defined as A|x in R+ where x is radius. So the set is uncountable. Remember?
No, the whole notion of countable\uncountable sets is a mambo jambo wrong reasoning.

Furthermore, let us change our game.

We take a finitely long straight line.

We can locate a point at each end of it, but these two points do not cover that finitely long straight line.

Actually, any additional point between the extreme endpoints is resulted by more lines with end points, etc. ad infinitum, where the points are actually different than each other exactly because, given any scale level, there is an uncovered line between any closer pair of points, along the finitely long line.

[doronshadmi]

The rule for set membership must be deterministic. This so-called set would need to contain one and only one circle of unit curvature, but which one?
Simply nonsense.

All is needed is the fact that all circles are different than each other by their curvature, that's all.

[jsfisher]

Simply nonsense.

All is needed is the fact that all circles are different than each other by their curvature, that's all.

...except they are not. All unit circles have exactly the same curvature, that being 1.


ETA: How are those bijections coming along? One between {A} and {{},{A}}, and another between {} and {{}}.

[ehcks]

I thought all circles had the same curvature. If you measure the tangent of any number of different sized circles at the same arc, the slopes are all equal. Isn't that what curvature means?

[jsfisher]

I thought all circles had the same curvature. If you measure the tangent of any number of different sized circles at the same arc, the slopes are all equal. Isn't that what curvature means?


The curvature of a circle is the reciprocal of its radius.

[ehcks]

The curvature of a circle is the reciprocal of its radius.

I'm thinking of something else entirely, then. Thanks though.

[jsfisher]

I'm thinking of something else entirely, then. Thanks though.

Perhaps you are not. Curvature can also be expressed in terms of not the slope, but the rate of change of the slope.

[ehcks]

Perhaps you are not. Curvature can also be expressed in terms of not the slope, but the rate of change of the slope.

That's what I'm trying to say. If you read the change of slope of the tangents by moving around the circle by degrees and not distance, the rate of change of the slope is always the same.

I'm not good at describing this.. That sentence is even hard for me to read.

The change in slope per 1 degree rotation is the same, regardless of the radius?

[doronshadmi]

...except they are not. All unit circles have exactly the same curvature, that being 1.

Nonsense again:

http://upload.wikimedia.org/wikipedia/commons/thumb/f/f7/ApollianGasketNested_2-20.svg/360px-ApollianGasketNested_2-20.svg.png


----------------------------------------

By the new game (which is no related to circles) any additional point between the extreme endpoints of a given line segment, is resulted by more lines with end points, etc. ad infinitum, where the points are actually different than each other exactly because given any scale level, there is an uncovered line between any closer pair of points (the closest pair of points does not exist exactly becuse there is always an uncovered line between them) along the finitely long line.

[jsfisher]

By the new game....

You power of non sequitur continues. The adults were talking about circle curvature.

[doronshadmi]

You power of non sequitur continues. The adults were talking about circle curvature.
The adults are wrong about the set of all circles with different curvature degrees because they ignore the difference.

How easy is to be an ignorant adult, isn't its jsfisher?

http://upload.wikimedia.org/wikipedia/commons/thumb/f/f7/ApollianGasketNested_2-20.svg/360px-ApollianGasketNested_2-20.svg.png
_________________________________________________

The curvature of the circles has to be compared with a straight line or with a point, in order to distinguish between all circles' different curvatures, but we already know that the adults reasoning is packed within a box.


--------------------------------------------------------


As for the new game, the adults are ignorant about it too.

[jsfisher]

The adults are wrong....


You continuing to make stuff up doesn't make the adults wrong.

[epix]

No, the whole notion of countable\uncountable sets is a mambo jambo wrong reasoning.

I thought that you liked and accepted Cantor's diagonal proof of unaccountability of real sets. Remember? You enslaved the diagonal proof and raped it repeatedly into the <0,1> argument.

Let me help you . . .

Definition: Any set is countable.

If so, continuity doesn't exist and there are no lines and curves. If there are no curves, there is no curvature. If there is no curvature, there are no circles. So please stop presenting various circles as some arguments of yours -- unless you want to further participate as a subject in the hitherto unknown medical field called "Analytic Psychopathy."

[The Man]

The Man, you still do not know the meaning of: "The set of all circles with different curvatures".

Evidently that is just you as your assertion before was…


1) This first circle is omitted from the rest of the circles that get the pairs, so there is a pair of points that does not cover the line, simply becuse there are no two circles with the same curvature in the set of all circles with different curvatures.

So does your “set of all circles with different curvatures” include “all circles with different curvatures” or just the ones that intersect your line at two points?



1) Each member of this set has a unique curvature, so if one of the curvatures is used as a tangent, it is not used as one of the centered points, and we get two places along the line, which are not covered by the intersecting points of the circle with the unique curvature that is already used as a tangent.

Please indentify these “two places along the line, which are not covered by the intersecting points of the circle with the unique curvature that is already used as a tangent”. How many circles “with the unique curvature” do you have? Have you run out for some reason? You do understand that those “two points” would be intersected and thus covered by some other circle, don’t you? Again please learn some basic geometry.


2) The set of all circles with different curvatures (where pi=circumference/diameter holds) does not have the smallest or the largest circles, so anyway the associated points with the set of all circles with different curvatures, are discontinuous w.r.t the center point (total curvature, where pi=circumference/diameter does not hold) and w.r.t the line (total non-curvature, where pi=circumference/diameter does not hold), or in other words, the set of all these intersecting points, does not completely cover an infinitely long straight line.

However the set of all concentric circles “with different curvatures” that are centered on some point off your line does have a smallest circle that intersects your line as well as a subset of circles “with different curvatures” that do not intersect your line at all. Your line is completely covered by those circles that do intersect your line. Again your assertion was that your line was not completely covered by points yet you could show no point on your line that was not a, well, point. So now you have moved to trying to show that your line can’t be covered by “intersecting points” of “The set of all circles with different curvatures”. Yet again you can identify no point on your line that can not be intersected by a member of some set of concentric circles of different curvatures.


Oh and again…


So by all means please explain to us the difference between changing and unchanging with “no past (before) and no future (after)”?

[epix]

Originally Posted by jsfisher
All unit circles have exactly the same curvature, that being 1.
Nonsense again:
Your love affair with 1+1=3 not even death could part . . . .

http://upload.wikimedia.org/wikipedia/commons/thumb/f/f7/ApollianGasketNested_2-20.svg/360px-ApollianGasketNested_2-20.svg.png

Your inability to understand the term "unit circle" at least fetched a graphic representation of a countably infinite set. Feel free to demonstrate your "systematic mapping" of its power set and bless the board with some more circles.

[!Kaggen]

stop presenting various circles as some arguments of yours -- unless you want to further participate as a subject in the hitherto unknown medical field called "Analytic Psychopathy."

They say it takes one to know one.

[epix]

The change in slope per 1 degree rotation is the same, regardless of the radius?
That's right. The angular change of 1 degree equals 1 degree change in slope for any radius. The rate of change of slope is not the same as the curvature of a circle given by 1/radius. The former measure is used with curves, such as the sine curve, where there is no center point that would connect line m with a given point on the curve resulting in curvature 1/m, like 1/r in the case of the circle.

[doronshadmi]

So does your “set of all circles with different curvatures” include “all circles with different curvatures” or just the ones that intersect your line at two points?

The set of all circles with different curvatures includes those different circles, whether they intersect some infinitely long straight line, or not.

If intersecting, then any intersection is associated to some circle, which its curvature is different than the rest of the circles that are included as unique members (by their different curvatures) of the set of all circles with different curvatures.

Since each circle of the set of all circles with different curvatures is included once and only once as a member of this set, then any unique circle, which does not intersect the considered infinitely long straight line or it is used as a tangent circle with the considered line, its unique curvature is already used, and as a result it does not have a pair of intersecting points, which are associated with it, and this pair does not cover the infinitely long straight line.

As a result the set of all points along an infinitely long striated line, is incomplete, even if one of the unique circles is used as a tangent circle (in this case the other point of the potential pair of the tangent circle, is not on the infinitely long striated line).

[doronshadmi]

You continuing to make stuff up doesn't make the adults wrong.
Wrong, I continuing to make stuff up does make the adults wrong.

[doronshadmi]

I thought that you liked and accepted Cantor's diagonal proof of unaccountability of real sets. Remember? You enslaved the diagonal proof and raped it repeatedly into the <0,1> argument.
Wrong again epix, Cantor's construction method of all P(S) members, is independent of the logical conclusions of his theorem.

[doronshadmi]

I thought that you liked and accepted Cantor's diagonal proof of unaccountability of real sets. Remember? You enslaved the diagonal proof and raped it repeatedly into the <0,1> argument.

epix you still do not understand the independence of Cantor's systematic constriction method of all P(S) members, and the logical conclusions of Cantor's theorem, as explicitly shown in
http://forums.randi.org/showpost.php?p=6986611&postcount=14585


Let me help you . . .

Definition: Any set is countable.

If so, continuity doesn't exist and there are no lines and curves. If there are no curves, there is no curvature. If there is no curvature, there are no circles. So please stop presenting various circles as some arguments of yours -- unless you want to further participate as a subject in the hitherto unknown medical field called "Analytic Psychopathy."
Let me help you, a line segment is not the collection of all the points along it, simply because the smallest pair along the considered line segment, is not satisfied.

In other words the a existence of a line or a circle are independent of the existence of collection of all the points along them.

[laca]

Wrong, I continuing to make stuff up does make the adults wrong.

Sig material.

[epix]

Wrong again epix, Cantor's construction method of all P(S) members, is independent of the logical conclusions of his theorem.
LOL. I thought that any conclusion depends on the examination of the material presented. According to you, Cantor examined x|x in R: 0≤x≤1, went to the bathroom and came up with a proof by looking inside the toilet bowl.

[doronshadmi]

LOL. I thought that any conclusion depends on the examination of the material presented. According to you, Cantor examined x|x in R: 0≤x≤1, went to the bathroom and came up with a proof by looking inside the toilet bowl.
You are right, he did not look also beyond the toilet bowl.

A researcher really does a valuable research only if he\she is both an external observer AND an internal participator of the researched subject.

[doronshadmi]

Your inability to understand the term "unit circle" ...
It is not relevant.

Once again you demonstrate your inability to get http://forums.randi.org/showpost.php?p=7016869&postcount=14679, exactly because you can't get things beyond your (your words) "toilet bowl".

[doronshadmi]

How about the set {} and its power set {{}}? What would be a bijection between the members of those two sets?

Please provide the members of {} to be mapped with.

Since they do not exist we get the member of {{}} which is mapped with nothing:

{} ↔


But if the minimal term must be an existing set then the empty set is mapped with the member of the power set of the empty set, where the power set of the empty set is {{}}.

In this case we get:

{} ↔ {}

[epix]

epix you still do not understand the independence of Cantor's systematic constriction method of all P(S) members, and the logical conclusions of Cantor's theorem, as explicitly shown in
http://forums.randi.org/showpost.php?p=6986611&postcount=14585


Let me help you, a line segment is not the collection of all the points along it, simply because the smallest pair along the considered line segment, is not satisfied.

In other words the a existence of a line or a circle are independent of the existence of collection of all the points along them.
Just unbelievable. :rolleyes:

The existence of objects with dimension d>0 depends on the continuum, which in turn is a result of any 'a' divided by any 'b' apart from zero equals 'c.' As a new point C appears between A and D, another point B appears between A and C and point B' between C and D with the number of new points appearing exponentially. This "dynamic" disables the process of counting defined by the 1-on-1 correspondence also known as bijection, as Cantor proved:
http://wiki.answers.com/Q/How_do_you_prove_that_the_continuum_is_uncountable

In order to prove that any set is countable you need to disprove Cantor's diagonal proof, which is a piece of cake for you -- any random thought of yours wearing the mask of logic will obviously do.

[doronshadmi]

Just unbelievable.
You are right, no belief of any kind is needed, in order to understand the following facts:

1) The members an infinite set are in a 1-to-1 and onto (bijection) with the members of any arbitrary infinite subset of the considered set ( you are still missing http://forums.randi.org/showpost.php?p=6986611&postcount=14585 ).

2) Any additional point between the extreme endpoints of a given line segment, is resulted by more lines with end points, etc. ad infinitum, where the points are actually different than each other exactly because given any scale level, there is an uncovered line between any closer pair of points (the closest pair of points does not exist exactly because there is always an uncovered line between them) along the finitely long line.

[epix]

You are right, no belief of any kind is needed, in order to understand the following facts:

1) The members an infinite set are in a 1-to-1 and onto (bijection) with the members of any arbitrary infinite subset of the considered set ( you are still missing http://forums.randi.org/showpost.php?p=6986611&postcount=14585 ).

Your references reside well outside the realm of reason.

Just focus on disproving Cantor's diagonal argument, otherwise your idea of any set being countable wouldn't free itself from the constrains of pure irrational belief.

[jsfisher]

In this case we get:

{} ↔ {}


Not a bijection between the members of {} and {{}}. Another failure for Doron.

[doronshadmi]

Not a bijection between the members of {} and {{}}. Another failure for Doron.
You are right, it is between {} and itself.

In other words, you still can't comprehend http://forums.randi.org/showpost.php?p=7019615&postcount=14694 .

[doronshadmi]

Your references reside well outside the realm of reason.

Just focus on disproving Cantor's diagonal argument, otherwise your idea of any set being countable wouldn't free itself from the constrains of pure irrational belief.
Once again, no belief is needed in order to understand http://forums.randi.org/showpost.php?p=6986611&postcount=14585 .

[jsfisher]

You are right, it is between {} and itself.

Of course I am right. You presented something completely unrelated to the task at hand. Are you looking to try again, or will you finally admit your claim was bogus?

In case you forgot (or wanted to dodge and weave away from), the claim was that a bijection exists between the members of any set S and its power set.

[doronshadmi]

Of course I am right.
And it is not relevant. You are missing http://forums.randi.org/showpost.php?p=7019883&postcount=14699 .

[doronshadmi]

In case you forgot (or wanted to dodge and weave away from), the claim was that a bijection exists between the members of any set S and its power set.
Forgetting is not the problem here, understanding is the problem here and you indeed do not understand http://forums.randi.org/showpost.php?p=7006557&postcount=14647 .

[jsfisher]

And it is not relevant. You are missing http://forums.randi.org/showpost.php?p=7019883&postcount=14699 .


Nope, didn't miss it. That link is to a post where you link to a post where you tried and failed. Nothing difficult there to understand. You were attempting to present a particular bijection. You got to the end, proudly wrote "Ta da!!", and then sat back smugly. The only problem, of course, was that your result was completely wrong and nothing to be proud of.

So, will you now retract your bogus claim?

[The Man]

The set of all circles with different curvatures includes those different circles, whether they intersect some infinitely long straight line, or not.

So the first circle that intersects your line is not “omitted” as you claimed before.




1) This first circle is omitted from the rest of the circles that get the pairs, so there is a pair of points that does not cover the line, simply becuse there are no two circles with the same curvature in the set of all circles with different curvatures.

Glad we could clear that up.




If intersecting, then any intersection is associated to some circle, which its curvature is different than the rest of the circles that are included as unique members (by their different curvatures) of the set of all circles with different curvatures.

Don’t forget there is a whole subset of “circles with different curvatures” that do not intersect your line when those concentric circles are not centered on your line.



Since each circle of the set of all circles with different curvatures is included once and only once as a member of this set, then any unique circle, which does not intersect the considered infinitely long straight line or it is used as a tangent circle with the considered line, its unique curvature is already used, and as a result it does not have a pair of intersecting points, which are associated with it, and this pair does not cover the infinitely long straight line.

How many “circles with different curvatures” do you have? Have you run out for some reason?

You seem to be thinking as if you had only a finite number of circles to use. So tell us Doron how many circles “with different curvatures” are in your “set of all circles with different curvatures”?

Again as noted above: there is a whole subset of “circles with different curvatures” that do not intersect your line when those concentric circles are not centered on your line. That some of those circles do not intersect your line in no way restricts the rest of those concentric circles from intersecting every point on your line. So you still need to identify what point(s) on your line can not be intersected by some member of that set of concentric circles that is not centered on your line.

You’ve gone a long way around once again only to end up right where you were before, unable to identify any point on your line that is not covered by a point and now even just not intersected by a concentric circle not centered on your line.





As a result the set of all points along an infinitely long striated line, is incomplete, even if one of the unique circles is used as a tangent circle (in this case the other point of the potential pair of the tangent circle, is not on the infinitely long striated line).

Doron the fact that one can define a set that includes some, none or even all but one point of your line in no way makes your line incomplete. In fact we can define any number of sets that include all points along your line, the simplest one being, well, your line.

Again this is just your typical self contradictory nonsense: Please identify what member(s) of “the set of all points along an infinitely long striated line” “is not on the infinitely long striated line”.

[doronshadmi]

Nope, didn't miss it.
Even if you are looking at it you are missing http://forums.randi.org/showpost.php?p=7019615&postcount=14694 , jsfisher.

[jsfisher]

Even if you are looking at it you are missing http://forums.randi.org/showpost.php?p=7019615&postcount=14694 , jsfisher.

Not at all. That's the post where you tried but failed to present a bijection between the elements of {} and {{}}.

Why do you continue to highlight your failures, Doron?

[doronshadmi]

Don’t forget there is a whole subset of “circles with different curvatures” that do not intersect your line when those Don’t forget there is a whole subset of “circles with different curvatures” that do not intersect your line when those concentric circles are not centered on your line. are not centered on your line.

Again you demonstrate your misunderstanding of the expression "the set of all circles with different curvatures".

This set is exactly those concentric circles, and any given circle with a unique curvature is already one of those concentric circles, so your argument does not hold water.

[doronshadmi]

Not at all. That's the post where you tried but failed to present a bijection between the elements of {} and {{}}.

Why do you continue to highlight your failures, Doron?
Wrong, that is a post that your weak reasoning can't comprehend.

Why do you continue to use this weak reasoning?

[The Man]

Again you demonstrate your misunderstanding of the expression "the set of all circles with different curvatures".

This set is exactly those concentric circles, and any given circle with a unique curvature is already one of those concentric circles, so your argument does not hold water.

Again you demonstrate your (apparently deliberate) misunderstanding of the fact that all points along your line are still just points along your line. That can all be intersected by any number of sets of geometrical objects, including concentric circles. Also your, again apparently deliberate, misunderstanding of the fact that simply defining a set that doesn’t include some, all, or just one point along your line does not make that set nor the set of all points along your line incomplete.



ETA: So even though you quoted quoted it twice it twice you’re still deliberately forgetting that it is a subset of your “set of all circles with different curvatures" when those concentric circles are not centered on your line?

[jsfisher]

Wrong, that is a post that your weak reasoning can't comprehend.

Why do you continue to use this weak reasoning?


Doron, stop trying to change the subject. You made a claim. You were called out on it. You have tried repeatedly to support the claim, but every single time you tried to support it with an example you have failed.

You failed every single time.

No amount of your "Oh, but you don't understand this other, equally bogus claim" reverses reality: You failed on every single attempt.

Let's just deal with one of your failures at a time. Once you retract the nonsense claim about bijections between the members of any set and its power set, then we can move on to another of your failures. But let's not jump around.

[doronshadmi]

Again you demonstrate your (apparently deliberate) misunderstanding of the fact that all points along your line are still just points along your line. That can all be intersected by any number of sets of geometrical objects, including concentric circles. Also your, again apparently deliberate, misunderstanding of the fact that simply defining a set that doesn’t include some, all, or just one point along your line does not make that set nor the set of all points along your line incomplete.
Wrong The Man,

The set of all circles with different curvatures is associated with all possible points along the infinitely long straight line, and yet there is discontinuity between these associated points and the center point, and the infinitely long straight line is permanently beyond the set of these associated points, because of a very simple reason:

pi is not a property of the center point (total curvature) or the infinitely long straight line (total non-curvature).

---------------------------------------

Since the Circle's game can't be comprehended by you, I have changed it to the Line segment game, as follows:

Any additional point between the extreme endpoints of a given line segment, is resulted by more lines with end points, etc. ad infinitum, where the points are actually different than each other exactly because given any scale level, there is an uncovered line between any closer pair of points (the closest pair of points does not exist exactly because there is always an uncovered line between them) along the finitely long line.

[doronshadmi]

You made a claim.
But you, jsfisher, can't comprehend it.

[jsfisher]

You made a claim.
But you, jsfisher, can't comprehend it.


Again, you project your own failings onto others. You claimed there was a bijection between the members of any set and its power set. The claim is simple enough to understand, yet you don't. Is there a specific part that is causing you the most trouble?

Are you yet willing to retract it, since it is a bogus claim? It is also key to another of your failings, that for any set S, |S| = |P(S)|, but let's continue on just one failure at a time. Jumping around only confuses you more, Doron.

[doronshadmi]

By using his weak reasoning, jsfisher simply fails to get the following post:

http://forums.randi.org/showpost.php?p=7019615&postcount=14694

He simply can't comprehend that this post is about Set's fundamental foundations.

[doronshadmi]

Again, you project your own failings onto others.
Wrong again jsfisher, you are failings yourself by using your weak reasoning.

[doronshadmi]

jsfisher, you are failing to get:

http://forums.randi.org/showpost.php?p=7006557&postcount=14647

http://forums.randi.org/showpost.php?p=6996924&postcount=14618

http://forums.randi.org/showpost.php?p=6986611&postcount=14585

Your weak reasoning can't help you, but unfortunately this is the only reasoning that you know.

[jsfisher]

By using his weak reasoning, jsfisher simply fails to get the following post:

http://forums.randi.org/showpost.php?p=7019615&postcount=14694

He simply can't comprehend that this post is about Set's fundamental foundations.


Still highlighting your failure, eh? Oh, wait, you aren't trying to compound it, now, are you? Are you perhaps now claiming you did provide a bijection between the members of {} and {{}}??

Your failures only compound, Doron.

Simple question, Doron, that even you should comprehend: Do you still claim there is a bijection between the elements of any set and its power set?

[doronshadmi]

Here is another post that jsfisher can't grasp, about this fine subject:

http://forums.randi.org/showpost.php?p=7003057&postcount=14636

[jsfisher]

Here is another post that jsfisher can't grasp, about this fine subject:

http://forums.randi.org/showpost.php?p=7003057&postcount=14636


Evasion noted.

Simple question, Doron, that even you should comprehend: Do you still claim there is a bijection between the elements of any set and its power set?

[epix]

Once again, no belief is needed in order to understand http://forums.randi.org/showpost.php?p=6986611&postcount=14585 .
The only thing that can be understood from your reference is that involves infantile trivialization of the subject, e.g. after mummy comes home, daddy comes home.

[doronshadmi]

Are you perhaps now claiming you did provide a bijection between the members of {} and {{}}??



http://forums.randi.org/showpost.php?p=7019615&postcount=14694 is not about a bijection, but it is about fundamental notions about the concept of Set.

By Traditional Set theory, Nothing is not considered as a mathematical thing, so in this case we get:

{} ↔

where {} is the member of the powerset of {}.


Since Nothing is not considered as a mathematical thing (by Traditional Set theory), then the minimal existing set is the Empty set and by following the restriction of "Set is an existing mathematical object", we get the following bijection, which is between the Empty set and the member of the powerset of the Empty set (which is actually the Empty set), as follows:

{} ↔ {}


jsfisher's weak reasoning simply can't comprehend what is written above.

[doronshadmi]

The only thing that can be understood from your reference is that involves infantile trivialization of the subject, e.g. after mummy comes home, daddy comes home.
Wrong, it is a simplification of a mambo jambo complication.

[epix]

Doron, stop trying to change the subject. You made a claim. You were called out on it. You have tried repeatedly to support the claim, but every single time you tried to support it with an example you have failed.

You failed every single time.

That reminds me that epix who have been trying to disect the set of all atheists, but epicly failed each time he tried. LOL.

[epix]

Wrong, it is a simplification of a mambo jambo complication.
Your simplification is fully commensurate with the way infants express their feelings toward the hypothesis of the continuum, multi-variable calculus, and other inedible items.
http://4.bp.blogspot.com/_jfwdA0R_fRY/TLaKknbtK9I/AAAAAAAAAnI/PgdH5LDsrC4/s1600/0113101308-01.jpg

[zooterkin]

Since the Circle's game can't be comprehended by you, I have changed it to the Line segment game, as follows:

Any additional point between the extreme endpoints of a given line segment, is resulted by more lines with end points, etc. ad infinitum, where the points are actually different than each other exactly because given any scale level, there is an uncovered line between any closer pair of points (the closest pair of points does not exist exactly because there is always an uncovered line between them) along the finitely long line.

Back to this one again? What uncovered line? Please show where there is a part of the line which has no point on it.

[jsfisher]

http://forums.randi.org/showpost.php?p=7019615&postcount=14694 is not about a bijection, but it is about fundamental notions about the concept of Set.


Yes, that was obvious to everyone including, apparently, even you.

However, the thing you were trying to present has supposed to be a bijection, remember? You were to provide a bijection, any bijection at all, between the members of {} and {{}}.

Try as you may, you keep failing at showing one. You did claim one existed, yet, you cannot point one out. Instead, you keep trying to change the subject.

[The Man]

Wrong The Man,

The set of all circles with different curvatures is associated with all possible points along the infinitely long straight line, and yet there is discontinuity between these associated points and the center point, and the infinitely long straight line is permanently beyond the set of these associated points, because of a very simple reason:

The discontinuity is again only in your head Doron. Your center point is a point on your line as specifically designated by you. You remain with no part of your line that is not covered by points even in your “discontinuity” with yourself.



pi is not a property of the center point (total curvature) or the infinitely long straight line (total non-curvature).

Points still aren’t circles, no matter how much you would like them to be or insist that something is lacking because they aren’t.



---------------------------------------

Since the Circle's game can't be comprehended by you, I have changed it to the Line segment game, as follows:

Any additional point between the extreme endpoints of a given line segment, is resulted by more lines with end points, etc. ad infinitum, where the points are actually different than each other exactly because given any scale level, there is an uncovered line between any closer pair of points (the closest pair of points does not exist exactly because there is always an uncovered line between them) along the finitely long line.

Your game still remains the same; you are just fixated on the smallest and the largest whether they are circles or line segments. You simply insist that something is lacking if they are indefinable in some set, yet fail to understand (again apparently deliberately since it has been explained to you many times) it is exactly that lacking of a definable smallest and/or largest circle and/or line segment which guarantees that your entire line is covered by points (a continuum) as well as line segments and even concentric circles (when not centered on your line).

[doronshadmi]

Back to this one again? What uncovered line? Please show where there is a part of the line which has no point on it.
Please show a pair of points along the line, where there is no line between them.

[doronshadmi]

Points still aren’t circles, no matter how much you would like them to be or insist that something is lacking because they aren’t.
But they are associated with the set of all circles with different curvatures, where these associations can't completely cover an infinitely long straight line, which is intersected by all the circles of this set.


it is exactly that lacking of a definable smallest and/or largest circle and/or line segment which guarantees that your entire line is covered by points
Exactly the opposite.

For example let us focused on the line segment case:

The existence of any arbitrary pair of points along a line segment is guaranteed by the existence of an uncovered line between them.

Without this uncovered line, the pair is merged into a single point.

[epix]

Points still aren’t circles, no matter how much you would like them to be or insist that something is lacking because they aren’t.

Doron sees as the circle is getting smaller and smaller in one instance resembling just a dot. It doesn't take much in his case to switch from 1-dim object to 0-dim object and treat the tiny circle as a point. Doronian equalities such as 1=0 are the banners of his crusades under which he keeps invading the Land of Reason hoping to subdue it. You can see another "equality" as Doron tries to outflank jsfisher's position near the power set. According to Doron, any finite set has the same cardinality as the power set, e.g. n = 2n. These "equalities" are the result of The Super Axiom of Omnipotence, the sole axiom that OM is built upon and which states that 1 + 1 = 3.

Note: In the opposite case, when the circle is getting larger and larger, its circumference gets mercifully out of Doron's sight, and so he can't transform it into some other object like in the case of his jolly circle/point transformation.

[doronshadmi]

Doron sees as the circle is getting smaller and smaller in one instance resembling just a dot. It doesn't take much in his case to switch from 1-dim object to 0-dim object and treat the tiny circle as a point. Doronian equalities such as 1=0 are the banners of his crusades under which he keeps invading the Land of Reason hoping to subdue it.
"Brilliant" conclusion epix. I clearly write that a circle or a line can't be a point (and this is a exactly the reason of why the smallest line or circle do not exist), but you, by using your twisted reasoning, get exactly the opposite (1=0).



You can see another "equality" as Doron tries to outflank jsfisher's position near the power set. According to Doron, any finite set has the same cardinality as the power set,
Wrong, by using Cantor's construction method, one enables to explicitly construct the all possible members of P(S), and because of this fact, one enables to define mapping between these explicit members and the same amount of members, taken to from the set of natural numbers, or any other set with different objects.

Furthermore, the mapping exists between no mapping with P(S) members, and a 1-to-1 and onto with P(S) members, and the degree of mapping is chosen according to one's needs (there is no universality about mapping).


Note: In the opposite case, when the circle is getting larger and larger, its circumference gets mercifully out of Doron's sight, and so he can't transform it into some other object like in the case of his jolly circle/point transformation.
Wrong again epix, since all circles have some curvature > 0 AND < ∞ , then:

1) The largest circle does not exist and as a result, the set of all circles can't completely cover an infinitely long straight line, exactly because such an object has exactly 0 curvature, and as a result it is permanently beyond the range of the set of all circles with different curvatures.

2) The smallest circle does not exist and as a result, the set of all circles can't completely cover an infinitely long straight line, exactly because such a collection do not have an object with ∞ curvature, which is exactly a point (the center point is permanently beyond the range of the set of all circles with different curvatures, which are obviously < ∞ curvature).

[zooterkin]

Please show a pair of points along the line, where there is no line between them.
None at all, but I haven't claimed this to be the case.

[doronshadmi]

None at all, but I haven't claimed this to be the case.
Good.

In that case there is an uncovered line between any arbitrary closer pair of points, along any given line.

[laca]

Good.

In that case there is an uncovered line between any arbitrary closer pair of points, along any given line.

And there it, ladies and gentlemen, the power of Moronetics.

[zooterkin]

Good.

In that case there is an uncovered line between any arbitrary closer pair of points, along any given line.

:rolleyes:

What makes you think that segment is 'uncovered', as you call it? The points exist, they are there, you do not need to place them to bring them into existence. There is nowhere on the line, whether between two other arbitrary points or not, that does not have a point on it.

[jsfisher]

Good.

In that case there is an uncovered line between any arbitrary closer pair of points, along any given line.

In addition to the insertion of the word, uncovered, that makes your statement completely bogus, this is another example of your sloppy powers of expression, Doron.

You didn't not mean "between any arbitrary closer pair of points". You meant that there must exist some pair of points with an "uncovered" line segment between them.

The statement is still wrong, as zooterkin points out. It is just less wrong this way.

[doronshadmi]

:rolleyes:

What makes you think that segment is 'uncovered', as you call it? The points exist, they are there, you do not need to place them to bring them into existence. There is nowhere on the line, whether between two other arbitrary points or not, that does not have a point on it.
Without the uncovered line, no arbitrary closer pair of points exists.

[doronshadmi]

You didn't not mean "between any arbitrary closer pair of points".
Now you, jsfisher, tell me what I mean, what a guy you are !

Anyway, without the uncovered line, no arbitrary closer pair of points exists.

[jsfisher]

Now you, jsfisher, tell me what I mean, what a guy you are !

You're welcome.

Anyway, without the uncovered line, no arbitrary closer pair of points exists.

Mathematics doesn't support this assertion, Doron. In fact, mathematics contradicts it.

[doronshadmi]

There is nowhere on the line, whether between two other arbitrary points or not, that does not have a point on it.
Given any arbitrary point along a line, it is distinguished from the rest of the points along that line only if there is an arbitrary smaller AND uncovered line between that point and some arbitrary closer point, which is not that point.

EDIT:

Without this simple fact, there is no more than one and only point along the considered line.

This fine reasoning distinguishes between "permanently closer" (not the considered point) and "closest" (the considered point).

[doronshadmi]

Mathematics doesn't support this assertion, Doron. In fact, mathematics contradicts it.
Your reasoning (which you like to call Mathematics) is too weak in order to comprehend http://forums.randi.org/showpost.php?p=7022911&postcount=14741 .

EDIT:

In other words, your reasoning is not fine enough in order to distinguish between "permanently closer" (not the considered point) and "closest" (the considered point).

[epix]

"Brilliant" conclusion epix. I clearly write that a circle or a line can't be a point (and this is a exactly the reason of why the smallest line or circle do not exist), but you, by using your twisted reasoning, get exactly the opposite (1=0).
I don't think your writing on the subject has been crystal clear, otherwise The Man, whom I replied to, wouldn't write

Points still aren’t circles, no matter how much you would like them to be or insist that something is lacking because they aren’t.
Doron:

Wrong, by using Cantor's construction method, one enables to explicitly construct the all possible members of P(S), and because of this fact, one enables to define mapping between these explicit members and the same amount of members, taken to from the set of natural numbers, or any other set with different objects.

Furthermore, the mapping exists between no mapping with P(S) members, and a 1-to-1 and onto with P(S) members, and the degree of mapping is chosen according to one's needs (there is no universality about mapping).

Calm down, Doron. Phrases such as "the mapping between no mapping" may suggest that you write these things under duress.
Surely, you can match the members of any finite power set with natural numbers, but the relationship in question is |S| <--> |P(S)| and not N <--> |P(S)|.


Wrong again epix, since all circles have some curvature > 0 AND < ∞ , then:
...then you should realize that the curvature is redundant to the conclusion you list below. If radius is in R and radius is in (0,∞) and radius is an orthogonal vector, you don't need any curvature to obscure the issue.

1) The largest circle does not exist and as a result, the set of all circles can't completely cover an infinitely long straight line, exactly because such an object has exactly 0 curvature, and as a result it is permanently beyond the range of the set of all circles with different curvatures.

If the magnitude of the radius grows unbound in R, then there obviously can't be the largest circle. DUH.
Your conclusion in (1) doesn't make sense, coz for any point X and r: X(-∞,∞)) = |r|(-∞,∞)


2) The smallest circle does not exist and as a result, the set of all circles can't completely cover an infinitely long straight line, exactly because such a collection do not have an object with ∞ curvature, which is exactly a point (the center point is permanently beyond the range of the set of all circles with different curvatures, which are obviously < ∞ curvature).
The center point of those concentric circles is not in the set. Despite your poor definition, the members of the set are points created by the intersection of circumferences and the straight line on which the center point lies.

[zooterkin]

Given any arbitrary point along a line, it is distinguished from the rest of the points along that line only if there is an arbitrary smaller AND uncovered line between that point and some arbitrary closer point, which is not that point.


Says who?

[doronshadmi]

I don't think your writing on the subject has been crystal clear, otherwise The Man, whom I replied to, wouldn't write

Doron:

Calm down, Doron. Phrases such as "the mapping between no mapping" may suggest that you write these things under duress.
Surely, you can match the members of any finite power set with natural numbers, but the relationship in question is |S| <--> |P(S)| and not N <--> |P(S)|.
You have missed this part:

... , or any other set with different objects.


...then you should realize that the curvature is redundant to the conclusion you list below. If radius is in R and radius is in (0,∞) and radius is an orthogonal vector, you don't need any curvature to obscure the issue.

If the magnitude of the radius grows unbound in R, then there obviously can't be the largest circle. DUH.
Your conclusion in (1) doesn't make sense, coz for any point X and r: X(-∞,∞)) = |r|(-∞,∞)


The center point of those concentric circles is not in the set. Despite your poor definition, the members of the set are points created by the intersection of circumferences and the straight line on which the center point lies.
epix, again your reasoning ignores the fact that there is no pi at diameter=0 or diameter=∞ states (since you like to use radius instead of diameter, then there is no 2pi at diameter=0 or diameter=∞ states).

Once again you are failing to get my argument about the set of all circles (pi or 2pi must exist) with different curvatures.

It is quite fascinating how your friends here and you can't comprehend such simple facts.

EDIT:

Also you ignore the points\line game (circles are not involved), so please look at http://forums.randi.org/showpost.php?p=7022911&postcount=14741 .

[doronshadmi]

Says who?
Do you really need somebody else, but you, in order to say it?

[epix]

Good.

In that case there is an uncovered line between any arbitrary closer pair of points, along any given line.
No, it's not good. In this type of contention, you need to define the term "line," coz

The first class of definitions follows Euclid's approach and consists in defining a line as an abstract primitive object whose properties are defined by a set of axioms. Such systems of axioms have been given by Karl von Staudt, David Hilbert, Giuseppe Peano, Mario Pieri and Alessandro Padoa.

The second definition is presently the most commonly used and relies to coordinate geometry introduced by René Descartes. It consists in defining a line in the Euclidean plane as the set of the points whose coordinates satisfy a given linear equation. More generally, a line in a Euclidean space of dimension n is the set of the points whose coordinates satisfy a given set of n−1 independent linear equations.

According to the second definition, there is no such a thing as an uncovered line between two points p and q, where p≠q.

So define the line to your likeness, to your image, so it shall become uncovered.

[jsfisher]

Your reasoning (which you like to call Mathematics) is too weak in order to comprehend http://forums.randi.org/showpost.php?p=7022911&postcount=14741 .

Nonsense. Your post is easily comprehended. It is nonsense.

EDIT:

In other words, your reasoning is not fine enough in order to distinguish between "permanently closer" (not the considered point) and "closest" (the considered point).

Still making up terminology to cover your ignorance, I see.


How's that bijection coming? Ready to retract your bogus claims about power sets, yet?

[zooterkin]

Do you really need somebody else, but you, in order to say it?

:rolleyes:

To remind you, you stated:
Given any arbitrary point along a line, it is distinguished from the rest of the points along that line only if there is an arbitrary smaller AND uncovered line between that point and some arbitrary closer point, which is not that point.

I was asking if that was something you made up yourself, or if you pinched that nonsense from somewhere else.

ETA: How big do you think a point is?

[epix]

epix, again your reasoning ignores the fact that there is no pi at diameter=0 or diameter=∞ states (since you like to use radius instead of diameter, then there is no 2pi at diameter=0 or diameter=∞ states).

I don't see how division by zero could affect the fact that you misdefined the set which is particular to your argument. Btw, "diameter = ∞" is a clear sign of you regarding infinity as a number. You should make an attempt to land forward in the 19th century and call a taxi for the trip to the 21st century.