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calcmandan
1st October 2011, 11:54 PM
A notion crossed my mind the other night about odd and even integers, during dinner actually. It got me curious enough to do some searching on google but came up empty handed. And I spent plenty of time reading through a bunch of existing proofs and theorems before I got tired.

Unfortunately, it's been ages since my proof writing days in college, so here goes.

Proof: In the universe of integers, there are an equal number of odd and even integers.

I'm just curious.

Daniel

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Vorpal
1st October 2011, 11:59 PM
"Equal number" just means that you can put them into bijection (one-to-one correspondence, so that every odd number is mapped to exactly one even number and every even number is mapped to by some odd number).
f(n) = n+1 works.

Actually, every infinite subset of the integers has an equal number of members as the integers.

DrDave
2nd October 2011, 12:01 AM
Yes it's true. See this thread for a discussion on similar proofs http://forums.randi.org/showthread.php?t=219795

In brief, if you can show that for each natural number there is one odd number (or even) you show there are necessarily the same number.

For even numbers the formula is f(x)=2x
For odd numbers the formula is f(x)=2x+1

So we have shown there are infinite of each. To quickly show there are the same number of each.

Drachasor
2nd October 2011, 12:03 AM
There's an equal number of odd and even integers. this number is equal to the total number of integers as well as the number of primes, perfect squares, etc, etc, etc.

The basic idea is you make a 1-1 ratio with the positive integers.

1 2 3 4 5 6 7 8 9 10...
1 -1 3 -3 5 -5 7 -7 9 -9...
0 2 -2 4 -4 6 -6 8 -8 10...
2 3 5 7 11 13 17 19 23 29...
1 4 9 16 25 36 49 64 81...

And so forth. It's part of Set Theory.

Tim Thompson
2nd October 2011, 12:05 AM
I can't give you the formal proof, but I think you will find that the paradoxical answer is this: Yes, there are just as many odd integers as there are even integers, and there are just as many odd or even integers as there are integers total. The reason for this is that the set of all even or odd integers is a countable infinity. In other words, you can put either set of odd or even integers into a one-to-one correspondence with the set of all integers (which is really what you do whenever you count anything).

If you stop short at a finite number, then of course there will always be more integers in your finite set than there are even or odd integers. But if you include the entire infinite sets, then they are the same order of infinity.

I remember learning that when I was a kid in the book One, Two, Three ... Infinity by George Gamow (http://en.wikipedia.org/wiki/George_Gamow).

calcmandan
2nd October 2011, 12:10 AM
There's an equal number of odd and even integers. this number is equal to the total number of integers as well as the number of primes, perfect squares, etc, etc, etc.

The basic idea is you make a 1-1 ratio with the positive integers.

1 2 3 4 5 6 7 8 9 10...
1 -1 3 -3 5 -5 7 -7 9 -9...
0 2 -2 4 -4 6 -6 8 -8 10...
2 3 5 7 11 13 17 19 23 29...
1 4 9 16 25 36 49 64 81...

And so forth. It's part of Set Theory.

My initial impression was set theory, but I kept second guessing myself. Yall are so FAST at responding!

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calcmandan
2nd October 2011, 12:15 AM
Yes it's true. See this thread for a discussion on similar proofs http://forums.randi.org/showthread.php?t=219795

In brief, if you can show that for each natural number there is one odd number (or even) you show there are necessarily the same number.

For even numbers the formula is f(x)=2x
For odd numbers the formula is f(x)=2x+1

So we have shown there are infinite of each. To quickly show there are the same number of each.

One to one correspondence, phew takes me way back to math 108.

So basically, it's more or less a tautology?

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sol invictus
2nd October 2011, 01:03 AM
One to one correspondence, phew takes me way back to math 108.

So basically, it's more or less a tautology?

Sent from my HP TouchPad using Communities

I don't think it's a tautology, no. As Tim mentioned, the same techniques can be used to prove that there are as many positive even integers as there are integers, that there are as many primes as integers, that there are as many integers as rational numbers, as many reals between 0 and 1 as all reals, etc. Most people do not find those facts very obvious.

calcmandan
2nd October 2011, 01:20 AM
I don't think it's a tautology, no. As Tim mentioned, the same techniques can be used to prove that there are as many positive even integers as there are integers, that there are as many primes as integers, that there are as many integers as rational numbers, as many reals between 0 and 1 as all reals, etc. Most people do not find those facts very obvious.

Some of those facts are starting to come back to me. I studied math in college but it's been a while, so a lot of it is misfiled in the head. Thanks for your input.

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Acleron
2nd October 2011, 11:32 PM
Actually, every infinite subset of the integers has an equal number of members as the integers.

Yes, I particularly like that it can be proved that 100% of integers contain the digit 3.

WhatRoughBeast
3rd October 2011, 07:13 AM
Yes, I particularly like that it can be proved that 100% of integers contain the digit 3.

No, that's not what he said. Please pay attention.

The number of integers which end in the number 3 is equal to the number of integers. That two sets have the same number of elements does not mean that the two sets are equal. Even if one set is a subset of the other. As long as both sets are infinite.

Acleron
3rd October 2011, 09:26 AM
No, that's not what he said. Please pay attention.

The number of integers which end in the number 3 is equal to the number of integers. That two sets have the same number of elements does not mean that the two sets are equal. Even if one set is a subset of the other. As long as both sets are infinite.

Vorpal said
Actually, every infinite subset of the integers has an equal number of members as the integers.

Please tell me where my statement conflicts with either Vorpal's or yours. Except of course where you talk about integers ending in 3, closer reading perhaps.

psionl0
3rd October 2011, 09:44 AM
It is pretty pointless to compare the number of elements in the set of even numbers to that in the set of odd numbers because both sets are infinitely large.

Trying to compare one infinitely large number with another is meaningless because infinity is not a number.

The folly of treating infinity like it was a number can be demonstrated in the following example:

What is 1 - 1 + 1 - 1 + 1 - 1 + . . .? It turns out that there are three possible answers depending on how you do the calculation.

The expression could be written this way: (1 - 1) + (1 - 1) + (1 - 1) + . . . = 0
Alternatively, we could write 1 - (1 + 1 - 1 + 1 - 1 + 1 - 1 + . . .) = 1 - 0 = 1
Finally, if the total is X then we could write an algebraic expression: X = 1 - X for which the solution is X = 0.5

We will never know the actual sum because it depends on the last digit in the series and we will never get to that number.

lomiller
3rd October 2011, 10:43 AM
Yes it's true. See this thread for a discussion on similar proofs http://forums.randi.org/showthread.php?t=219795

In brief, if you can show that for each natural number there is one odd number (or even) you show there are necessarily the same number.

For even numbers the formula is f(x)=2x
For odd numbers the formula is f(x)=2x+1

So we have shown there are infinite of each. To quickly show there are the same number of each.

I don’t like the highlighted part. Just because 2 sets are both infinite doesn’t mean they are equal.

If you go back a step and say that you have shown that for every odd number there exists an even number that can be found by adding 1(and that for every odd number there exists an even number that can be found by subtracting 1). Since even and odd numbers always occur in pairs there must be an equal number of them.

On the other hand you wanted to compare numbers divisible by 2 to numbers dividable by 3 again both sets are infinite but for every 3 numbers divisible by 2 there are only 2 that are divisible by 3 so numbers divisible by 2 will always be a larger value as you move towards infinity.

Ferguson
3rd October 2011, 01:13 PM
I don’t like the highlighted part. Just because 2 sets are both infinite doesn’t mean they are equal.

Sure does.

On the other hand you wanted to compare numbers divisible by 2 to numbers dividable by 3 again both sets are infinite but for every 3 numbers divisible by 2 there are only 2 that are divisible by 3 so numbers divisible by 2 will always be a larger value as you move towards infinity.

For each finite set you pick, yes. But once we're talking infinity, there are as many. You will never "run out" of numbers divisible by 3.

Modified
3rd October 2011, 01:40 PM
Vorpal said
Please tell me where my statement conflicts with either Vorpal's or yours. Except of course where you talk about integers ending in 3, closer reading perhaps.

You seem to have assumed that "sets A and B have the same number of members and B is a subset of A" implies that every property shared by all members of B is also shared by all members of A. That is only true if A and B are of finite length, in which case they must be identical.

lomiller
3rd October 2011, 02:42 PM
Sure does.

For each finite set you pick, yes. But once we're talking infinity, there are as many. You will never "run out" of numbers divisible by 3.

No. Infinity isn’t a number. This means you can and do get cases where two things that both equal infinity are not equal to each other.
x=z
y=2*z
y/x will still equal 2 even if you allow z to go to infinity (making both x and y infinity)

Acleron
3rd October 2011, 03:00 PM
You seem to have assumed that "sets A and B have the same number of members and B is a subset of A" implies that every property shared by all members of B is also shared by all members of A. That is only true if A and B are of finite length, in which case they must be identical.

If I did, I didn't mean to, but I don't think I did. I was more amused that the calculation of the percentage of the numbers containing a 3 is 100% of all numbers. The amusement is because every property shared by A ie containing the digit 3 is not shared by the set of all numbers.

Drachasor
3rd October 2011, 03:04 PM
No. Infinity isn’t a number. This means you can and do get cases where two things that both equal infinity are not equal to each other.
x=z
y=2*z
y/x will still equal 2 even if you allow z to go to infinity (making both x and y infinity)

Infinity is like a number in several ways, but of course there are a lot of different types of infinity. How it might show up in a given problem can vary a lot.

There's a significant difference between limits (what you have above) and an infinity like "the number of integers". X = Z could be "the number of positive integers equal or less than Z". y = 2z could be "the number of positive even integers equal or less than 4z." You have, by your argument, shown that the latter set of numbers is twice as big as the former, which is decidedly not the case.

It's much more useful to consider the fact you can match up any countable and infinity set as a 1-1 ratio with the positive integers. You can also do it at a 2-1, 1-2, 4-1, or any other sort of ratio that you like. You'll never reach a point where you can't find numbers to fill in for this matching process. It's on this basis they are said to be of equal size, because we are comparing elements in a set.

Of course, there are infinities much bigger than this, such as the set of Real numbers (this includes fractions and irrational numbers...the set of Rational numbers, which has fractions composed of integers, is countable). This set can't be set up in a 1-1 correspondence with the natural numbers, because they simply can't be placed into an ordered list. There are then infinities bigger still than this, which can't be placed into a 1-1 correspondence with the Real numbers. In fact, there are an infinite number of infinity sizes.

W.D.Clinger
3rd October 2011, 03:17 PM
No. Infinity isn’t a number. This means you can and do get cases where two things that both equal infinity are not equal to each other.
:confused:
Since infinity isn't a real number, no real numbers equal infinity.

Besides, this thread is about bijections between sets and the notion of cardinal numbers, which don't have much in common with the real numbers you seem to have in mind.

x=z
y=2*z
y/x will still equal 2 even if you allow z to go to infinity (making both x and y infinity)
You appear to be thinking about the limit of a function as its argument increases without bound ("goes to infinity" in the vernacular). The use of the infinity symbol in

\begin{align}
x(z) &= z \\
y(z) &= 2 z \\
f(z) &= \frac{y(z)}{x(z)} \\
\lim_{z \rightarrow \infty} f(z) &= 2
\end{align}

doesn't mean there's any such thing as a real infinity. Line (4) above actually means

$(\forall \epsilon > 0)(\exists z_0)(\forall z > z_0) \left| {f(z) - 2} \right| < \epsilon$

where the variables in that last line range over real numbers (so there's no actual infinity anywhere).

As I said, that doesn't have anything to do with the topic of this thread.

Modified
3rd October 2011, 03:24 PM
If I did, I didn't mean to, but I don't think I did. I was more amused that the calculation of the percentage of the numbers containing a 3 is 100% of all numbers. The amusement is because every property shared by A ie containing the digit 3 is not shared by the set of all numbers.

If you have a proof of that, then there is a mistake somewhere, since the counterargument is obvious. I'd guess the percentage of numbers less than n containing the digit 3 approaches 100 as n approaches infinity, but that is not the same thing unless your definition of "100% of all numbers" is different from the norm.

lomiller
4th October 2011, 08:07 AM
Infinity is like a number in several ways, but of course there are a lot of different types of infinity. How it might show up in a given problem can vary a lot.

There's a significant difference between limits (what you have above) and an infinity like "the number of integers". X = Z could be "the number of positive integers equal or less than Z". y = 2z could be "the number of positive even integers equal or less than 4z." You have, by your argument, shown that the latter set of numbers is twice as big as the former, which is decidedly not the case.

It's much more useful to consider the fact you can match up any countable and infinity set as a 1-1 ratio with the positive integers. You can also do it at a 2-1, 1-2, 4-1, or any other sort of ratio that you like. You'll never reach a point where you can't find numbers to fill in for this matching process. It's on this basis they are said to be of equal size, because we are comparing elements in a set.

Of course, there are infinities much bigger than this, such as the set of Real numbers (this includes fractions and irrational numbers...the set of Rational numbers, which has fractions composed of integers, is countable). This set can't be set up in a 1-1 correspondence with the natural numbers, because they simply can't be placed into an ordered list. There are then infinities bigger still than this, which can't be placed into a 1-1 correspondence with the Real numbers. In fact, there are an infinite number of infinity sizes.

You are pretty much saying the same thing as I did above to explain why the number if even and odd numbers must be equal. I suspect you missed part of the discussion leading up to my post so I’ll backtrack a bit

The question being addressed in the post you were quoting is a little different, however. The argument is being made that the number of even and odd numbers are equal because they are both infinity, while the number of even and odd numbers are equal it is not because both sets are infinite so I went on to give an example of two infinite sets where the number of members is not equal
The first set is the numbers evenly divisible by 2 (even numbers)
The second set is the numbers evenly divisible by 3

In this case every time we count to a multiple of 6 w get 3 numbers divisible by 2 and only 2 that are divisible by 3. No matter how high we count this relationship does not change so even though both sets are infinite the set of numbers divisible by 2 is larger.

Ferguson tried to argue that this isn’t the case by saying the number of integers in each set were both equal to infinity and therefore equal. While this would be true for equality to a real number infinity is not a real number and you can apply equality in this way.

lomiller
4th October 2011, 08:14 AM
As I said, that doesn't have anything to do with the topic of this thread.

Several people in this thread are attempting to argue that the number of even and odd numbers are equal because both equal infinity. This is not a valid proof because infinity is not a number and you can’t apply equality in that way with it.

Even though they get lucky and get the correct answer in this case, there are other cases where their logic fails.

Acleron
4th October 2011, 08:54 AM
If you have a proof of that, then there is a mistake somewhere, since the counterargument is obvious. I'd guess the percentage of numbers less than n containing the digit 3 approaches 100 as n approaches infinity, but that is not the same thing unless your definition of "100% of all numbers" is different from the norm.

Hmm. what I'm saying is that it is 100% when you have reached infinity.

Look here for the proof. (http://trickofmind.com/?p=970)

sol invictus
4th October 2011, 08:55 AM
I went on to give an example of two infinite sets where the number of members is not equal
The first set is the numbers evenly divisible by 2 (even numbers)
The second set is the numbers evenly divisible by 3

Your "example" is not an example. Those two sets have exactly the same number of elements. The proof is identical in structure to the proof for evens and odds: construct a bijection between the two sets. In the example you gave, that is trivial: map each even number 2n to the divisible-by-3 number 3n, and vice versa (where n is any integer). That's a 1-to-1 map, so it proves that the number of elements in those two sets are equal.

An example of two infinite sets with different numbers of elements are the set of all real numbers between 0 and 1, and the integers. The set of reals is larger.

W.D.Clinger
4th October 2011, 09:09 AM
As I said, that doesn't have anything to do with the topic of this thread.

Several people in this thread are attempting to argue that the number of even and odd numbers are equal because both equal infinity. This is not a valid proof because infinity is not a number and you can’t apply equality in that way with it.
No, their proofs are valid because

The integers are countable (obviously).
The set of even numbers is a subset of the set of integers (obviously).
The set of odd numbers is a subset of the set of integers (obviously).
There are infinitely many even numbers (obviously).
There are infinitely many odd numbers (obviously).
So there's a countable infinity of even numbers (from facts 1, 2, 3), and there's a countable infinity of odd numbers (from facts 1, 3, 5).
All countably infinite sets have the same size; to put it more precisely, given any two countably infinite sets, there's a bijection between them (this is trivial, as it follows immediately from the definition of countable infinity).
The set of even numbers has the same size as the set of odd numbers (from facts 6 and 7).

It would be easier just to describe a one-to-one correspondence (bijection) between the even and odd integers, but it's important to recognize the validity of the above proof because that kind of reasoning is easier to apply in some less obvious cases, and generalizes to more powerful proof techniques for truly difficult cases.

lomiller
4th October 2011, 09:26 AM
Your "example" is not an example. Those two sets have exactly the same number of elements. The proof is identical in structure to the proof for evens and odds: construct a bijection between the two sets. In the example you gave, that is trivial: map each even number 2n to the divisible-by-3 number 3n, and vice versa (where n is any integer). That's a 1-to-1 map, so it proves that the number of elements in those two sets are equal.r.

You haven’t correctly characterised the example.

Take x as the set of integers from 4 to n
Count the number of times x is evenly divisible by 2
Count the number of times x is evenly divisible by 3
If you allow n to go to infinity both counts are infinite but for any value of n the first will always be greater.

WhatRoughBeast
4th October 2011, 09:28 AM
If I did, I didn't mean to, but I don't think I did. I was more amused that the calculation of the percentage of the numbers containing a 3 is 100% of all numbers. The amusement is because every property shared by A ie containing the digit 3 is not shared by the set of all numbers.

That's what I thought what was going on. And yes, it's sufficiently counter-intuitive (one might even say absurd) to be amusing.

But that's infinity for you.

Drachasor
4th October 2011, 10:11 AM
You are pretty much saying the same thing as I did above to explain why the number if even and odd numbers must be equal. I suspect you missed part of the discussion leading up to my post so I’ll backtrack a bit

No, I'm saying you are wrong. I suspect you missed my first post on this thread, because I explained how all countable sets are the same size.

The question being addressed in the post you were quoting is a little different, however. The argument is being made that the number of even and odd numbers are equal because they are both infinity, while the number of even and odd numbers are equal it is not because both sets are infinite so I went on to give an example of two infinite sets where the number of members is not equal
The first set is the numbers evenly divisible by 2 (even numbers)
The second set is the numbers evenly divisible by 3

Those two sets are NOT different in size. This is part of Set Theory. (http://en.wikipedia.org/wiki/Cardinality#Infinite_sets)

In this case every time we count to a multiple of 6 w get 3 numbers divisible by 2 and only 2 that are divisible by 3. No matter how high we count this relationship does not change so even though both sets are infinite the set of numbers divisible by 2 is larger.

You can line up the members of these sets so that they are in a 1-1 relationship with each other. For any member of one set, you can produce a member of the other in that manner. They are the same size.

This is how you count things. You line up the members of one and compare them to the members of another. You don't go "ok, how many members are there less than 5? How about 6? How about 7?"

Ferguson tried to argue that this isn’t the case by saying the number of integers in each set were both equal to infinity and therefore equal. While this would be true for equality to a real number infinity is not a real number and you can apply equality in this way.

Sometimes I wonder if anyone reads my posts. I mean, I'm the third reply, but you act like I haven't said anything in this thread.

As you say, infinity isn't quite a number. Certainly not an ordinary number. Yet you are treating it like one. You are taking a limit as two numbers go to infinity and divide each other and acting like this is a solid comparison of the size of the two infinities. It isn't. As I demonstrated you can have that ratio come out to be ANY number you wish by adjusting how you count and still counting ALL members of both sets. That's how we know your counting system is messed up.

Instead, we say that when things are like that, the two sets are equal in size. If the ratio must be infinity (e.g. you can't make a 1-1 correspondence between two sets), then we say one is bigger than the other.

Solitaire
4th October 2011, 08:49 PM
It is pretty pointless to compare the number of elements in the set of even numbers to that in the set of odd numbers because both sets are infinitely large.

Trying to compare one infinitely large number with another is meaningless because infinity is not a number.

I must disagree.

Ordinality And Cardinality

An ordinal is number you can reach by counting the elements finite set. For example, the set {0, 1, 2, 3}, has an ordinality of four – the same ordinality as the set {-1, 1, -2, 2}.

A cardinal is number you cannot simply count them by you can compare the size of them by lining up the members – see illustration. (http://scidiv.bellevuecollege.edu/math/infinity.html)

Thus when I say infinity is a number I mean it is a cardinal number not an ordinal number.

The Function Trap

Functions with an infinite number terms in series often get confused with the issue of the size of infinity. Infinity has a definite size, not an arbitrary one. Hence when you create a term series such as the 1 - 1 + 1 - 1 summing series depends upon how you evaluate it.

(1 - 1) + (1 - 1) + (1 - 1) + (1 - 1) + (1 - 1) + (1 - 1) . . . = 0

Move the parenthesis and extract the first term.

1 - (1 + 1) - (1 + 1) - (1 + 1) - (1 + 1) - (1 + 1) - (1 + 1) . . . = 1

Rotate the terms so that the -1 is in front.

(-1 + 1) + (-1 + 1) + (-1 + 1) + (-1 + 1) + (-1 + 1) + (-1 + 1) . . . = 0

Move the parenthesis and extract the first term.

-1 + (1 - 1) + (1 - 1) + (1 - 1) + (1 - 1) + (1 - 1) + (1 - 1) + (1 - 1) . . . = -1

None of these series expresses the true size of infinity. They only show the possible values of a function with an infinite number of terms.

Infinity As A Constant

Infinity acts like a constant similar to Pi, e, or phi. You can add, divide, multiply, and subtract a number from it. Such operations do not change its size, for example: ∞ + 1 = ∞ - 1. But, if you subtract two functions with these values you won't find ∞ + 1 ≡ ∞ - 1 because the two functions subtracted will give you ∞ + 1 - (∞ - 1) = 2.

This subtraction of two infinite functions that differ shows up in Quantum Mechanics as the Casimir Force between two plates. The infinite wave modes outside the plates generates an infinite force inward. The infinite wave modes inside the plages generate an infinite force outward. Yet when you do the math and the experiment in the lab you find the two don't cancel out perfectly at zero, but produce a predictable inward force on the plates depending upon the distance.

Does this clear things up?
Or did I make the subject more confusing?

http://forums.randi.org/imagehosting/154bf82d24398b8.gif (http://forums.randi.org/vbimghost.php?do=displayimg&imgid=20000)

psionl0
4th October 2011, 09:24 PM
No. Infinity isn’t a number. This means you can and do get cases where two things that both equal infinity are not equal to each other.
x=z
y=2*z
y/x will still equal 2 even if you allow z to go to infinity (making both x and y infinity)You accidentally made the mistake of treating infinity as a number in this example after all. This becomes clear if we add a couple of words to the example:

For any finite number z NO MATTER HOW LARGE
if x = z
and y = 2 * z
then y/x = 2

Although this is described in calculus as a "limit to infinity", it is still dealing with finite numbers. The expression "∞ / ∞" is known in calculus as "indeterminate".

psionl0
4th October 2011, 09:45 PM
Does this clear things up?
Or did I make the subject more confusing?Definitely more confusing.

It is nonsense to talk of such things as ∞ + 1 or ∞ - 1 but let us assume that your claim that ∞ + 1 = ∞ - 1 is true.

You also claim that (∞ + 1) - (∞ - 1) = 2 but because of your first postulate we can write (∞ + 1) - (∞ - 1) = (∞ - 1) - (∞ - 1) = 0.

So, does 0 = 2?

Vorpal
4th October 2011, 10:28 PM
There, Solitaire is probably talking about the extended reals. Affinely extended reals (http://mathworld.wolfram.com/AffinelyExtendedRealNumbers.html) (with +∞ and -∞) are standard in real analysis, while projectively extended reals (http://mathworld.wolfram.com/ProjectivelyExtendedRealNumbers.html) (with just ∞) are the analogue of the Riemann sphere, the one-point compatification of the complex numbers, commonly used in complex analysis. For neither one are the expression ∞ - ∞ (or +∞ - +∞) defined, but many artihmetic operations, such as ∞+1 = ∞-1 = ∞, are meaningful and useful.

What happens in Casimir force or physics in general is a bit more subtle, though. More relevant are the Cauchy principal value (http://mathworld.wolfram.com/CauchyPrincipalValue.html) of integrals and Euler's zeta function trick for series (or exponential regulator for physics-folk), or dark sorcery.

Drachasor
4th October 2011, 11:08 PM
Definitely more confusing.

It is nonsense to talk of such things as ∞ + 1 or ∞ - 1 but let us assume that your claim that ∞ + 1 = ∞ - 1 is true.

You also claim that (∞ + 1) - (∞ - 1) = 2 but because of your first postulate we can write (∞ + 1) - (∞ - 1) = (∞ - 1) - (∞ - 1) = 0.

So, does 0 = 2?

Quite. It becomes unclear what you get when you subtract one infinity from another. When it happens in mathematics dealing with reality, you can take some sort of limit to figure out the answer (and indeed it can be pretty much anything, and depends on the physical constraints).

sol invictus
5th October 2011, 03:04 PM
You haven’t correctly characterised the example.

I think I have.

Take x as the set of integers from 4 to n
Count the number of times x is evenly divisible by 2
Count the number of times x is evenly divisible by 3
If you allow n to go to infinity both counts are infinite but for any value of n the first will always be greater.

That's true. Nevertheless, in the limit, those two sets are the same size.

psionl0
5th October 2011, 05:15 PM
That's true. Nevertheless, in the limit, those two sets are the same size.Wrong!

If the number of elements in the first set is n/2 and the number of elements in the second set is n/3 then the ratio between the two is 3:2 and in the limit the ratio tends to 3:2.

Again this shows the absurdity of treating infinity as a number. You can't count up to infinity nor can you add up finite numbers and get an infinite result.

Modified
5th October 2011, 05:19 PM
Hmm. what I'm saying is that it is 100% when you have reached infinity.

What does "reached infinity" mean?

Solitaire
5th October 2011, 05:31 PM
Definitely more confusing.

It is nonsense to talk of such things as ∞ + 1 or ∞ - 1 but let us assume that your claim that ∞ + 1 = ∞ - 1 is true.

You also claim that (∞ + 1) - (∞ - 1) = 2 but because of your first postulate we can write (∞ + 1) - (∞ - 1) = (∞ - 1) - (∞ - 1) = 0.

So, does 0 = 2?

Clearly not. My approach isn't working, try this.

You can have a function f(x) whose series equals infinity and a function g(x) whose series equals infinity, but when you subtract f(x) from g(x), term by term, you get a finite nonzero number. The size of f(x) and g(x) remain the same, but something remains. Hence the desire to include this in some sort of notation.

Perhaps even the equation, ∞g = ∞f + 3, is overly ambitious in trying to express this.

There, Solitaire is probably talking about the extended reals. Affinely extended reals (http://mathworld.wolfram.com/AffinelyExtendedRealNumbers.html) (with +∞ and -∞) are standard in real analysis, while projectively extended reals (http://mathworld.wolfram.com/ProjectivelyExtendedRealNumbers.html) (with just ∞) are the analogue of the Riemann sphere, the one-point compatification of the complex numbers, commonly used in complex analysis. For neither one are the expression ∞ - ∞ (or +∞ - +∞) defined, but many artihmetic operations, such as ∞+1 = ∞-1 = ∞, are meaningful and useful.

What happens in Casimir force or physics in general is a bit more subtle, though. More relevant are the Cauchy principal value (http://mathworld.wolfram.com/CauchyPrincipalValue.html) of integrals and Euler's zeta function trick for series (or exponential regulator for physics-folk), or dark sorcery.

Yes. I couldn't think of Cauchy at the moment.

Acleron
5th October 2011, 05:45 PM
What does "reached infinity" mean?

Look at the link I gave

sol invictus
5th October 2011, 05:47 PM
Wrong!

No, it really isn't.

If the number of elements in the first set is n/2 and the number of elements in the second set is n/3 then the ratio between the two is 3:2 and in the limit the ratio tends to 3:2.

You're assuming the value in the limit is equal to the value the ratio tends to under your chosen procedure. That assumption is unwarranted, and false in this instance.

Again this shows the absurdity of treating infinity as a number. You can't count up to infinity nor can you add up finite numbers and get an infinite result.

Not at all - it simply shows the absurdity of making incorrect assumptions about infinite sets.

There's a very simple proof that shows that those two sets have exactly the same number of elements. It was explained at length in the thread on primes; I recommend you read that.

W.D.Clinger
5th October 2011, 06:00 PM
There, Solitaire is probably talking about the extended reals. Affinely extended reals (http://mathworld.wolfram.com/AffinelyExtendedRealNumbers.html) (with +∞ and -∞) are standard in real analysis, while projectively extended reals (http://mathworld.wolfram.com/ProjectivelyExtendedRealNumbers.html) (with just ∞) are the analogue of the Riemann sphere, the one-point compatification of the complex numbers, commonly used in complex analysis. For neither one are the expression ∞ - ∞ (or +∞ - +∞) defined, but many artihmetic operations, such as ∞+1 = ∞-1 = ∞, are meaningful and useful.
There might be a good reason why we use different notations for these distinct notions of infinity:

+∞ and -∞ for the infinities of projectively affinely extended reals
∞ for the infinity of projectively extended reals
ω0 for the least infinite ordinal
$\aleph_0$ for the least infinite cardinal, which is the only kind of infinity that's relevant to the OP

(Many of us write ω0 instead of $\aleph_0$, but that's less likely to cause confusion. Most of the people who recognize ω0 as a notation for infinity are aware that the cardinal numbers can be identified with a certain subclass of the ordinal numbers. They will also know why I wrote "subclass" instead of "subset".)

psionl0
5th October 2011, 08:33 PM
There's a very simple proof that shows that those two sets have exactly the same number of elements. It was explained at length in the thread on primes; I recommend you read that.We have seen two proofs here: one that shows that the number of elements in each set is equal and the other showing that the ratio of elements in each set is 3:2.

Since each proof started with sets with a finite number of elements and extended that to the case of sets with and infinite "number" of elements, both proofs are equally valid - or invalid. Unless you can show otherwise, there is no justification for considering infinity as anything other than a limit (ie for any real number n, ∞ > n).

epix
5th October 2011, 08:51 PM
I don't think it's a tautology, no. As Tim mentioned, the same techniques can be used to prove that there are as many positive even integers as there are integers, that there are as many primes as integers, that there are as many integers as rational numbers, as many reals between 0 and 1 as all reals, etc. Most people do not find those facts very obvious.
The reason is that folks regard primes, for example, as a part of natural numbers and therefore a subject to Euclid's Common Notion No. 5. The reinforcement comes from the fact that the Fifth Common Notion is a destination of many math inquiries regarding science and technology. But Cantor's approach regards primes and natural numbers as a collection called "sets"; that means primes are NOT A PART of natural numbers. That's why Cantor met a substantial opposition in his time from those mathematicians who couldn't reconcile the difference.

Vorpal
5th October 2011, 09:50 PM
We have seen two proofs here: one that shows that the number of elements in each set is equal and the other showing that the ratio of elements in each set is 3:2.
1) Those proofs are not contradictory at all.
2) Actually, the latter statement is a bit ambiguous, because it relies on a pretty arbitrary choice of how you're taking the limit, though that's not a huge deal.

Since each proof started with sets with a finite number of elements and extended that to the case of sets with and infinite "number" of elements, both proofs are equally valid - or invalid.
Well, for one thing, your definition can only talk about sets of numbers. This completely removes the point of a set-theoretic definition in which we should be able to count the number of objects in it regardless of their nature or any other structure on that set.

Unless you can show otherwise, there is no justification for considering infinity as anything other than a limit (ie for any real number n, ∞ > n).
Is it being a straightforward part of a very rich mathematical structure with lots a good justification for adopting the standard set-theoretic definition of sizes of sets? We're talking about infinite cardinalities in particular here, not "infinity in general" (there's no such thing). There's nothing preventing you from using limits when they are relevant, but for sizes of sets, there is a much better, more powerful, and more useful definition that you're ignoring.

In standard set theory, the following can be taken as a definition: a set X is infinite if, and only if, it is equinumerous with some proper subset Y⊊X, i.e., it has a part that's as large as the whole thing.

psionl0
5th October 2011, 11:32 PM
1) Those proofs are not contradictory at all.Apart from their conclusions.

2) Actually, the latter statement is a bit ambiguous, because it relies on a pretty arbitrary choice of how you're taking the limit, though that's not a huge deal.A limit exists only if it is independent of the path taken. Of course, path dependent limits can exist. In a one-dimensional domain we talk of "one-sided" limits. In this instance, there is only one path and that is to increase n.

Well, for one thing, your definition can only talk about sets of numbers. This completely removes the point of a set-theoretic definition in which we should be able to count the number of objects in it regardless of their nature or any other structure on that set.I was not aware that "element" means "number". In any case, we are dealing with sets of discrete numbers here. If what you mean is that I am dealing with sets whose size can be determined by counting the number of elements then you are correct. Sets that have a finite size but an infinite number of elements (eg the set of real numbers between 0 and 1) do not need to be considered for the purpose of this thread.

Is it being a straightforward part of a very rich mathematical structure with lots a good justification for adopting the standard set-theoretic definition of sizes of sets? We're talking about infinite cardinalities in particular here, not "infinity in general" (there's no such thing).Huh???

There's nothing preventing you from using limits when they are relevant, but for sizes of sets, there is a much better, more powerful, and more useful definition that you're ignoring.There may well be some abstract branches of mathematics where the "size of infinity" as a meaning and there may well be practical applications of this branch of mathematics. However, unless you know exactly what you are doing, you will get tied up in knots and come up with inconsistent conclusions as was done in this case.

In standard set theory, the following can be taken as a definition: a set X is infinite if, and only if, it is equinumerous with some proper subset Y⊊X, i.e., it has a part that's as large as the whole thing.So the set of all even numbers has the same number of elements as the set of all whole numbers because they are "equinumerous"? To me that sounds more like "begging the question" than a definition.

Vorpal
6th October 2011, 12:19 AM
Apart from their conclusions.
You have two notions of size giving distinct answers. There is no real contradiction because you're they are different notions. If you make up some nonstandard procedure of comparing sizes of sets, as you have done, why do you think it's a problem that it gives different answers than the standard one?

A limit exists only if it is independent of the path taken. Of course, path dependent limits can exist. In a one-dimensional domain we talk of "one-sided" limits. In this instance, there is only one path and that is to increase n.
Yet there are many ways to pick what you're taking the limit of. For example, you're assuming a particular order to the set you're given.

I was not aware that "element" means "number".
It doesn't. Why would you think that? My point is that your notion of "size" only applies to certain kinds of sets of numbers. It's simply meaningless for sets in general.

If ∅ is the empty set, how does the set {∅, {∅}, {∅,{∅}}, {∅,{∅},{∅,{∅}}}, ... }, where the next element is formed from the previous through X U {X}, compare in size to the set of even integers? All the integers? How does your procedure deal with, say, the set of stars in an infinite universe? What about {bob, snake, hat, Sgr A*, ...}?

It's not even well-defined for all subsets of natural numbers! Say X = { n: 22k ≤ n ≤ 22k+1, k natural }. Your procedure of comparing it to the set of all natural numbers fails. Edit: Fixed criterion.

In any case, we are dealing with sets of discrete numbers here. If what you mean is that I am dealing with sets whose size can be determined by counting the number of elements then you are correct. Sets that have a finite size but an infinite number of elements (eg the set of real numbers between 0 and 1) do not need to be considered for the purpose of this thread.
In other words, you're just making a completely ad hoc definition of the sizes of sets tailored to "prove" your preferred conclusion. And you're admonishing others of begging the question?

Huh???
See: set theory.

There may well be some abstract branches of mathematics where the "size of infinity" as a meaning and there may well be practical applications of this branch of mathematics. However, unless you know exactly what you are doing, you will get tied up in knots and come up with inconsistent conclusions as was done in this case.
The issue is much more clear and straightforward than you think. The 'inconsistent conclusions' are only there to those unfamiliar with these areas of mathematics.

So the set of all even numbers has the same number of elements as the set of all whole numbers because they are "equinumerous"? To me that sounds more like "begging the question" than a definition.
The irony is strong here.

Look, at the end of the day there's a very straightforward, very intuitive way of comparing sizes of sets in set theory that's completely independent of what kind of things you're dealing with in that set and what kind of order is defined on them (if any at all!). According to it, the set of even integers is equinumerous with the set of all integers. And the property I've stated is equivalent to the set being infinite (defined as not equinumerous with a set of a cardinality n for natural n), so it can itself serve as a definition of 'infinite set' if one is so inclined. It works for any sets whatsoever.

psionl0
6th October 2011, 02:17 AM
You have two notions of size giving distinct answers. There is no real contradiction because you're they are different notions. If you make up some nonstandard procedure of comparing sizes of sets, as you have done, why do you think it's a problem that it gives different answers than the standard one?In any set with a finite number of discrete elements, the "size" of the set has always been the number of elements in the set. What is the other notion of size that I have? Is the number of elements irrelevant in the "standard" definition of set size?

For example, you're assuming a particular order to the set you're given.If the order that the elements of a set are listed is unimportant then the results should similarly be independent of the order that the elements are listed. Are you suggesting that if we list the elements in the two sets given by the OP in a different order that we can conclude something else?

If ∅ is the empty set, how does the set {∅, {∅}, {∅,{∅}}, {∅,{∅},{∅,{∅}}}, ... }, where the next element is formed from the previous through X U {X}, compare in size to the set of even integers? All the integers? How does your procedure deal with, say, the set of stars in an infinite universe? What about {bob, snake, hat, Sgr A*, ...}?They are all sets with an infinite number of discrete elements. The notion of "size" is meaningless because infinity is not a number. All you can say is that these sets are "bigger" than any set with a finite number of discrete elements.

It's not even well-defined for all subsets of natural numbers! Say X = { n: 22k ≤ n ≤ 22k+1, k natural }. Your procedure of comparing it to the set of all natural numbers fails. Edit: Fixed criterion.That isn't even a set! It is a rule for generating a set. For example, if k = 1 then according to this rule, we get the set {4, 5, 6, 7, 8}. The "size" of the set generated by this rule can be calculated from the following formula: n(X) = 22k+1 - 22k + 1. I see nothing inconsistent here.

EDIT: I just realized that you haven't necessarily restricted n to a subset of integers. Clearly, if n can be any real number then "size" and element count are not necessarily the same thing in this example but we are dealing with subsets of integers here so talking about something else doesn't clarify the issue.

In other words, you're just making a completely ad hoc definition of the sizes of sets tailored to "prove" your preferred conclusion. And you're admonishing others of begging the question?Counting the number of elements in a set with a finite number of discrete elements is an "ad hoc rule"?

Look, at the end of the day there's a very straightforward, very intuitive way of comparing sizes of sets in set theory that's completely independent of what kind of things you're dealing with in that set and what kind of order is defined on them (if any at all!).And if I ask you what this "very straightforward, very intuitive way" is you would probably tell me to go study set theory.

According to it, the set of even integers is equinumerous with the set of all integers.You don't prove this. You just say this is "standard set theory". That is an appeal to authority.

At the end of the day, all you are arguing is that the method used by sol invictus (http://forums.randi.org/showthread.php?postid=7641289#post7641289) is valid and the method used by Iomiller (http://forums.randi.org/showthread.php?postid=7641389#post7641389) isn't.

sol invictus
6th October 2011, 02:18 AM
We have seen two proofs here: one that shows that the number of elements in each set is equal and the other showing that the ratio of elements in each set is 3:2.

Since each proof started with sets with a finite number of elements and extended that to the case of sets with and infinite "number" of elements, both proofs are equally valid - or invalid. Unless you can show otherwise, there is no justification for considering infinity as anything other than a limit (ie for any real number n, ∞ > n).

Nope. The (correct) proof that the sets have equal size does not "start" with a finite set, nor does it involve any kind of limit.

Look: take the set of all natural numbers {1, 2, 3, 4, ...}. Now multiply every element in that set by 3: {3, 6, 9 ,12, ...}. It should be intuitively obvious that you haven't changed the number of elements - all you've done is multiply each element by 3, and how can that possibly change the number of them?

For each element in the first set, there's exactly one element in the second and vice versa - and that proves they are equinumerous.

sol invictus
6th October 2011, 02:25 AM
If the order that the elements of a set are listed is unimportant then the results should similarly be independent of the order that the elements are listed. Are you suggesting that if we list the elements in the two sets given by the OP in a different order that we can conclude something else?

For the procedure you used, yes. The answers you'd get - including which set is bigger - depend on the (entirely arbitrary) order you choose to list the elements in.

That's one reason why your definition of "size" is not a useful one - it depends on order.

They are all sets with an infinite number of discrete elements. The notion of "size" is meaningless because infinity is not a number. All you can say is that these sets are "bigger" than any set with a finite number of discrete elements.

You keep saying that. And yet, mathematicians have a perfectly good notion of "size" that applies to infinite sets, and that shows that some infinite sets are bigger than others (I have an example earlier).

psionl0
6th October 2011, 02:43 AM
Look: take the set of all natural numbers {1, 2, 3, 4, ...}. Now multiply every element in that set by 3: {3, 6, 9 ,12, ...}. It should be intuitively obvious that you haven't changed the number of elements - all you've done is multiply each element by 3, and how can that possibly change the number of them? Look: Take the set of positive integers {1, 2, 3, 4, 5, ...}. Now take every third element and make a subset of them: {3, 6, 9, 12, ...}. It should be intuitively obvious that for every element in the second set, there are three elements in the first set. How can you say they have the same number of elements?

The Don
6th October 2011, 02:47 AM
Look: Take the set of positive integers {1, 2, 3, 4, 5, ...}. Now take every third element and make a subset of them: {3, 6, 9, 12, ...}. It should be intuitively obvious that for every element in the second set, there are three elements in the first set. How can you say they have the same number of elements?

There are an infinite number of positive integers. One third of these are divisible by three.

What's one third of infinity ?

psionl0
6th October 2011, 02:50 AM
What's one third of infinity ?Precisely what I am getting at. What does it mean to compare the "size" of one set to that of another if both have an infinite number of discrete elements?

sol invictus
6th October 2011, 02:54 AM
Look: Take the set of positive integers {1, 2, 3, 4, 5, ...}. Now take every third element and make a subset of them: {3, 6, 9, 12, ...}. It should be intuitively obvious that for every element in the second set, there are three elements in the first set. How can you say they have the same number of elements?

I'm trying to show you that, at the very least, there are two intuitive arguments that give contradictory answers. Once you see that, it should make you pause and think: "I've got two conflicting intuitions, so clearly I can't rely only on intuition. I need a more rigorous criterion. Gee, what could that be?"

"My" definition (that two sets are the same size if and only if there exists a 1-to-1 map between them) gives consistent answers and can be applied to any pair pf sets, while yours (to the extent that I can even understand what it is) does not. For instance, yours gives answers that depend on order. That alone is enough to prefer "mine". But there are many other reasons as well.

Precisely what I am getting at. What does it mean to compare the "size" of one set to that of another if both have an infinite number of discrete elements?

Mashuna
6th October 2011, 02:56 AM
Look: Take the set of positive integers {1, 2, 3, 4, 5, ...}. Now take every third element and make a subset of them: {3, 6, 9, 12, ...}. It should be intuitively obvious that for every element in the second set, there are three elements in the first set. How can you say they have the same number of elements?

It looks like you're trying to match the identical numbers rather than the elements.

Vorpal
6th October 2011, 03:13 AM
In any set with a finite number of discrete elements, the "size" of the set has always been the number of elements in the set. What is the other notion of size that I have? Is the number of elements irrelevant in the "standard" definition of set size?
Can you tell the difference between the number of elements in a set and the numbers the set happens to contain?
The notion of size that you have is the limit procedure.

If the order that the elements of a set are listed is unimportant then the results should similarly be independent of the order that the elements are listed. Are you suggesting that if we list the elements in the two sets given by the OP in a different order that we can conclude something else?
By the standard definition, order is irrelevant. By your definition, order is critical.

They are all sets with an infinite number of discrete elements. The notion of "size" is meaningless because infinity is not a number. All you can say is that these sets are "bigger" than any set with a finite number of discrete elements.
The notion of "size" is very meaningful even for infinite sets.
Just because it doesn't jive with your personal intuition doesn't mean it's not meaningful.

That isn't even a set! It is a rule for generating a set. For example, if k = 1 then according to this rule, we get the set {4, 5, 6, 7, 8}.
Nonsense. It's a set. If you want more formal notation,
$X = \{n\in{\mathbb{N}}: (\exists k\in{\mathbb{N}})(2^{2k}\leq n\leq 2^{2k+1}\}$
This specifies a subset of the natural numbers that satisfy a certain property. It's a set.

The "size" of the set generated by this rule can be calculated from the following formula: n(X) = 22k+1 - 22k + 1. I see nothing inconsistent here.
Did you actually try comparing it to the naturals? The limit of the ratio
$\lim_{n\to\infty}\frac{|\{m\in X: m\leq n\}|}{|\{m\in{\mathbb{N}}: m\leq n\}|}$
does not exist, where |{...}| denotes the size of the set. This is is exactly what you tried to do with your 3:2 ratio argument from earlier.

Counting the number of elements in a set with a finite number of discrete elements is an "ad hoc rule"?
Taking ratios of these quantites as size comparison? Absolutely ad hoc. It's not even well-defined for all subsets of the natural numbers, much less sets in general.

Just counting a limit without a ratio? Well, that works, but the only thing that tells you is that set is infinite, on which everyone agrees. It does not tell you how it compared to any other infinite set.

And if I ask you what this "very straightforward, very intuitive way" is you would probably tell me to go study set theory.
Considering it's been stated by many posters in this very thread (incl. the very first reply), I'd suggest just reading.
But studying set theory is not a bad idea either.

You don't prove this. You just say this is "standard set theory". That is an appeal to authority.
Nonsense. It's not because authority that you're wrong. You're wrong because there's an intuitive, powerful, and useful notion of comparing "size" that works for any set whatsoever, you've presented precisely zero reasons why it shouldn't be used, and your alternative notion of ratios of limits is weak, inapplicable for arbitrary sets, and actually inapplicable even to the the simplest case of a subsets of the natural numbers.

At the end of the day, all you are arguing is that the method used by sol invictus (http://forums.randi.org/showthread.php?postid=7641289#post7641289) is valid and the method used by Iomiller (http://forums.randi.org/showthread.php?postid=7641389#post7641389) isn't.
At the end of the day, there is a meaningful notion of "size" that's able to distinguish many infinite sets from one another and is applicable to every set. All you have is a broken little notion that doesn't work most of the time. You can keep it, if you want~let's call it the iosize of a set or something.

psionl0
6th October 2011, 05:37 AM
Can you tell the difference between the number of elements in a set and the numbers the set happens to contain?Seriously? You can't tell me what the number of elements is in a set like {dog, cat, horse, fish}? (hint: try counting)
And guess what? It is the same as the number of elements in {cat, fish, dog, horse}. Why you think that changing the order that the elements are listed would change the number of elements (under my "unique" definition of size) is beyond me.

The notion that two sets are the same size iff there is a 1 - 1 correspondence between the elements might be a more useful generalized definition of size but in the case of sets with a finite number of discrete elements it still comes down to counting.

Unfortunately, this "official"(?) definition of "size" still doesn't prove your case because if two sets have an infinite "number" of elements then you can't prove a 1 - 1 correspondence.

If A = {1, 2, 3, 4, ...} and B = {3, 6, 9, 12, ...} then there are (at least) two ways of generating set B. The first method is to multiply each element in set A by 3 in which case there is a 1 - 1 correspondence. The other method is to copy every multiple of 3 from set A to set B. This would imply a 1 - 3 correspondence.

Nonsense. It's a set. If you want more formal notation,
$X = \{n\in{\mathbb{N}}: (\exists k\in{\mathbb{N}})(2^{2k}\leq n\leq 2^{2k+1}\}$
This specifies a subset of the natural numbers that satisfy a certain property. It's a set.That notation is more impressive but it only seems to change the independent variable from k to n. ie Pick a value for n, work out the value of k then list each other n that falls within the required range. (At least, that's how I interpret the rule).

Did you actually try comparing it to the naturals? The limit of the ratio
$\lim_{n\to\infty}\frac{|\{m\in X: m\leq n\}|}{|\{m\in{\mathbb{N}}: m\leq n\}|}$
does not exist, where |{...}| denotes the size of the set. This is is exactly what you tried to do with your 3:2 ratio argument from earlier.This is the first time you have shown this particular formula (as well as your modified X) so, no.

The denominator is clearly n but the numerator is harder to calculate. In fact, X doesn't even exist if n = 3. For k = 0 the required range is [1, 2] while for k = 1 the required range is [4, 8]. No k exists that will satisfy the equation if n = 3.

Nonsense. It's not because authority that you're wrong. You're wrong because there's an intuitive, powerful, and useful notion of comparing "size" that works for any set whatsoever, you've presented precisely zero reasons why it shouldn't be used, and your alternative notion of ratios of limits is weak, inapplicable for arbitrary sets, and actually inapplicable even to the the simplest case of a subsets of the natural numbers.That's not a mathematical argument. That is just a vague assertion coupled with some incorrect statements about my position based on the fact that you didn't have a clue what "number of elements" means.

I'm not saying that you are wrong about all of this - just that you haven't proven your case. In fact, you aren't even making a case anymore. You seem to think you can win this argument by simply implying that you know more than me.

sol invictus
6th October 2011, 05:50 AM
Seriously? You can't tell me what the number of elements is in a set like {dog, cat, horse, fish}? (hint: try counting)
And guess what? It is the same as the number of elements in {cat, fish, dog, horse}.

You've mis-read what Vorpal wrote. That seems to be a pattern.

Why you think that changing the order that the elements are listed would change the number of elements (under my "unique" definition of size) is beyond me.

Because it does. I'm in a hurry, but I'm sure someone will show you why if you can't figure it out yourself.

The notion that two sets are the same size iff there is a 1 - 1 correspondence between the elements might be a more useful generalized definition of size but in the case of sets with a finite number of discrete elements it still comes down to counting.

Unfortunately, this "official"(?) definition of "size" still doesn't prove your case because if two sets have an infinite "number" of elements then you can't prove a 1 - 1 correspondence.

Nonsense. I already gave you an explicit 1-to-1 map for the case at hand.

If A = {1, 2, 3, 4, ...} and B = {3, 6, 9, 12, ...} then there are (at least) two ways of generating set B. The first method is to multiply each element in set A by 3 in which case there is a 1 - 1 correspondence. The other method is to copy every multiple of 3 from set A to set B. This would imply a 1 - 3 correspondence.

Again, you need to read more carefully: "two sets are the same size if and only if there exists a 1-to-1 map between them".

If there's at least one 1-to-1 map, then the existence of other maps, 1-to-1 or not, is not relevant.

psionl0
6th October 2011, 06:05 AM
You've mis-read what Vorpal wrote. That seems to be a pattern.You mean that Vorpal didn't ask me to "tell the difference between the number of elements in a set and the numbers the set happens to contain"?

Because it does. I'm in a hurry, but I'm sure someone will show you why if you can't figure it out yourself.:dl:
What a convincing argument!

Nonsense. I already gave you an explicit 1-to-1 map for the case at hand.And I gave you an explicit 1-3 map.

Again, you need to read more carefully: "two sets are the same size if and only if there exists a 1-to-1 map between them".

If there's at least one 1-to-1 map, then the existence of other maps, 1-to-1 or not, is not relevant.Why? Because it is.

W.D.Clinger
6th October 2011, 06:21 AM
The notion that two sets are the same size iff there is a 1 - 1 correspondence between the elements might be a more useful generalized definition of size but in the case of sets with a finite number of discrete elements it still comes down to counting.
There's no "might be" about it. It is a useful generalization of size.

Unfortunately, this "official"(?) definition of "size" still doesn't prove your case because if two sets have an infinite "number" of elements then you can't prove a 1 - 1 correspondence.
False. Such proofs are given all the time. Several such proofs have been given in this thread, and several such proofs have been given in the "Prime Numbers" thread that's been cited within this thread. Several of those proofs include explicit one-to-one correspondences.

If A = {1, 2, 3, 4, ...} and B = {3, 6, 9, 12, ...} then there are (at least) two ways of generating set B. The first method is to multiply each element in set A by 3 in which case there is a 1 - 1 correspondence. The other method is to copy every multiple of 3 from set A to set B. This would imply a 1 - 3 correspondence.
That last sentence is false. I suspect you just guessed wrong about the definition of a 1 - 1 correspondence. Here's the definition:
Definition. A function f:A→B is a 1 - 1 correspondence if and only if f is both
one-to-one (aka injective): For all x and y in A, if f(x)=f(y) then x=y.
and
onto (aka surjective): For all b in B, there exists an a in A such that f(a)=b.

One-to-one correspondences are also known as bijections.

Nonsense. It's not because authority that you're wrong. You're wrong because there's an intuitive, powerful, and useful notion of comparing "size" that works for any set whatsoever, you've presented precisely zero reasons why it shouldn't be used, and your alternative notion of ratios of limits is weak, inapplicable for arbitrary sets, and actually inapplicable even to the the simplest case of a subsets of the natural numbers.That's not a mathematical argument. That is just a vague assertion coupled with some incorrect statements about my position based on the fact that you didn't have a clue what "number of elements" means.
It seemed like a mathematical argument to me. It wasn't vague at all: Vorpal has supplied specific examples to back up all of his criticisms.

I'm not saying that you are wrong about all of this - just that you haven't proven your case. In fact, you aren't even making a case anymore. You seem to think you can win this argument by simply implying that you know more than me.
Vorpal is trying hard not to use any arguments from authority, but he is using arguments that may be too subtle for you. You might benefit from going back and reading the first four replies in this thread, which were all quite good.

sol invictus
6th October 2011, 06:26 AM
You mean that Vorpal didn't ask me to "tell the difference between the number of elements in a set and the numbers the set happens to contain"?

That's what he asked. If in fact you read that carefully, apparently you actually cannot tell the difference. The set {cat, fish, dog, horse} does not contain any numbers.

:dl:
What a convincing argument!

That just makes you look like a troll.

I can think of many different definitions of the ratio of set size you might have meant. All of them are flawed. Rather than me wasting my time (I am now late) guessing what you mean, why don't you tell us your procedure for determining which of two infinite sets A and B is larger?

Note: your definition should apply to any sets, including those with no numbers in them, it should apply to infinite sets like the number of reals or rationals in an interval, it should apply to sets with negative or complex or quanternionic numbers in them, etc. If you can't accomplish that, then at least give us one that applies to subsets of the natural numbers.

And I gave you an explicit 1-3 map.

Again, you've failed to understand what you read. According to the definition I gave you, the existence or non-existence of 1-3 maps is totally irrelevant.

AvalonXQ
6th October 2011, 07:12 AM
Seriously? You can't tell me what the number of elements is in a set like {dog, cat, horse, fish}? (hint: try counting)

Yes, exactly - try counting! That's what you seem unwilling to do.

The way we know that the above set has four elements is by creating a bijection between that set, and the set {1,2,3,4}. If we can do that, then the set has four elements -- and every other set that has this property has the same number of elements. This is how counting, and cardinality, is defined.

We can extend this to infinite sets. Again, if a bijection exists, then the sets are the same size.

Furthermore, if a bijection exists between one full set and a subset of the other set (such as in the 1-to-3 example), then we know the one set is at least as big as the other, but not necessarily bigger. That's what you're missing -- a subset bijection proves less than or equal, not strictly less than.

psionl0
6th October 2011, 07:13 AM
You might benefit from going back and reading the first four replies in this thread, which were all quite good.Looking at Drachasor's argument in post 4 and your own post, let me see if I can put it more clearly:

If N = {1, 2, 3, 4, ...} and X = {x1, x2, x3, x4, ...} and there exists a function f such that f(i) = xi
Then by definition the sets {1, 2, 3, 4, ...} and {f(1), f(2), f(3), f(4), ...} are said to be the same size.

I say "by definition" because it is neither a proof nor a theorem. Its truth depends upon what we mean by the "size" of an infinite set. Unfortunately, we can't say they have the same number of elements because we can't count them.

The problem is that put this way, we are basically saying that there are as many even numbers as there are natural numbers because the sets of these numbers are defined to be the same size.

psionl0
6th October 2011, 07:23 AM
The set {cat, fish, dog, horse} does not contain any numbers.Well, at least we agree on that!

Since you are so determined to twist my posts, let's put it this way:
What is the number of elements in {3, -1, 10, 17}?
What is the number of elements in (10, 3, 17, -1}?
4 and 4

W.D.Clinger
6th October 2011, 08:36 AM
I say "by definition" because it is neither a proof nor a theorem. Its truth depends upon what we mean by the "size" of an infinite set.
The existence of the one-to-one correspondence is a theorem, and we can prove that theorem. That theorem depends upon the definition of a one-to-one correspondence, but that definition is standard; disputing that definition would be like disputing the meaning of the words "theorem" or "proof".

The "size" of an infinite set is not quite so well defined. For one thing, mathematicians are more likely to write about cardinality than size; the word "size" is informal, hence more likely to be used informally than in journal articles.

More importantly, you can't prove that every set has a well-defined cardinality unless you assume the axiom of choice or its equivalent, and that axiom is somewhat controversial. What's more, you can't use the usual axioms of mathematics or set theory to prove that the set of real numbers has any specific cardinality: You can assume that its cardinality is ω1, or ω17, or whatever, but you can't prove anything like that until you've assumed your conclusion. That's why some of us prefer to talk about the existence of one-to-one correspondences between sets than about the cardinalities or sizes of arbitrary sets.

For the countable sets that have been mentioned within this thread, however, there is no problem whatsoever. It's easy to prove that every one of those countably infinite sets can be placed in one-to-one correspondence with any other countably infinite set.

Among those who are familiar with logic and mathematics, there is no more controversy about the definition of cardinality in terms of one-to-one correspondences than about the definitions of "theorem" or "proof", so all of those countably infinite sets have cardinal ω0 (or $\aleph_0$, if you want to get picky).

Unfortunately, we can't say they have the same number of elements because we can't count them.

The problem is that put this way, we are basically saying that there are as many even numbers as there are natural numbers because the sets of these numbers are defined to be the same size.
That's something of an exaggeration. Here's a more careful statement of what you're saying:

We can prove the existence of a one-to-one correspondence between the even numbers and the natural numbers (or integers).
We define "have the same size" for sets to mean there's a one-to-one correspondence between them.
Therefore, by fact 1 and definition 2, the set of even numbers has the same size as the set of natural numbers (or integers).

psionl0
6th October 2011, 09:33 AM
That's something of an exaggeration. Here's a more careful statement of what you're saying:

We can prove the existence of a one-to-one correspondence between the even numbers and the natural numbers (or integers).
We define "have the same size" for sets to mean there's a one-to-one correspondence between them.
Therefore, by fact 1 and definition 2, the set of even numbers has the same size as the set of natural numbers (or integers).
I agree.

Now that we have an unambiguous definition for the size of an infinite set that allows us to compare the sizes of two infinite sets, there is absolutely no controversy.

Of course this is not the same as saying, "every infinite subset of the integers has an equal number of members as the integers" which was controversial because it wasn't clear what was meant by "number of members".

Vorticity
6th October 2011, 09:35 AM
Edited to add: W. D. Clinger covered a number of my points while I was putting this post together.

psionl0:

Vorpal and sol invictus are unambiguously right, and you are unambiguously wrong.

They have been shying away from making an argument from authority, but I'm willing to briefly dip into that territory. You should know that the 1-1 correspondence definition of set size is not some new thing that people have cooked up in this thread. It really is the standard definition, and is based on more than 100 years of careful thought. The entire reason it was created was because of the inadequacies and inconsistencies of the intuitive notion of counting the number of elements in a set.

You may wish to keep that in mind as this discussion continues.

Now:

The notion that two sets are the same size iff there is a 1 - 1 correspondence between the elements might be a more useful generalized definition of size but in the case of sets with a finite number of discrete elements it still comes down to counting.

I think you have that backwards: The notion that two sets are the same size iff there is a 1 - 1 correspondence between the elements is a useful generalization of counting. Ordinary counting is a special case which becomes available to us when we are dealing with finite sets. But we don't really need it. If we can prove a 1-1 correspondence between the elements of two finite sets, we have just as effectively proven that they have the same size.

Counting the elements in set A and then counting the elements in set B, and then noting that the counts are equal is the same as finding a 1-1 correspondence between A's elements and a subset of the naturals, and then finding a 1-1 correspondence between B's elements and the very same subset of the naturals. Why not cut out the middle man and just find a 1-1 correspondence between the elements of A and the elements of B? The advantage of this method is that it works for both finite and infinite sets.

Unfortunately, this "official"(?) definition of "size" still doesn't prove your case because if two sets have an infinite "number" of elements then you can't prove a 1 - 1 correspondence.

Sure you can. I think there may be some confusion here about what it means to "prove a 1 - 1 correspondence" between the elements of two sets. It doesn't mean that I have to prove that there is a mystical partnership between particular pairs of elements in set A and set B, and that they're connected... just 'cause. I just need to find some function - any function - that maps each element of A to one unique element of B, and vice-versa, so that none are left out on either end. It doesn't matter if there is only one such function or 1000 such functions, I just need to find one of them.

Example:

A = {1,2,3,4,5...}
B = {2,4,6,8,10...}

Let f:A->B be f(n) = 2n. And the inverse g:B->A, which goes the other way, is g(n) = n/2. For any number you care to name in A, I can use f to tell you which number in B it corresponds to, and g for vice versa.

Nonsense. It's a set. If you want more formal notation,
$X = \{n\in{\mathbb{N}}: (\exists k\in{\mathbb{N}})(2^{2k}\leq n\leq 2^{2k+1}\}$
This specifies a subset of the natural numbers that satisfy a certain property. It's a set.

That notation is more impressive but it only seems to change the independent variable from k to n. ie Pick a value for n, work out the value of k then list each other n that falls within the required range. (At least, that's how I interpret the rule).

It's a perfectly reasonable and standard notation for defining a particular infinite subset of the natural numbers. I don't see what the problem is.

Just for reference, X starts out as:

X = {1, 2, 4, 5, 6, 7, 8, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 64, 65, 66, 67, 68, 69, 70, 71, 72, 73, 74, 75, 76, 77, 78, 79, 80, 81, 82, 83, 84, 85, 86, 87, 88, 89, 90, 91, 92, 93, 94, 95, 96, 97, 98, 99, 100, 101, 102, 103, 104, 105, 106, 107, 108, 109, 110, 111, 112, 113, 114, 115, 116, 117, 118, 119, 120, 121, 122, 123, 124, 125, 126, 127, 128, 256, 257, 258, 259, 260, 261, 262, 263, 264, 265, 266, 267, 268, 269, 270, 271, 272, 273, 274, 275, 276, 277, 278, 279, 280, 281, 282, 283, 284, 285, 286, 287, 288, 289, 290, 291, 292, 293, 294, 295, 296, 297, 298, 299, 300, 301, 302, 303, 304, 305, 306, 307, 308, 309, 310, 311, 312, 313, 314, 315, 316, 317, 318, 319, 320, 321, 322, 323, 324, 325, 326, 327, 328, 329, 330, 331, 332, 333, 334, 335, 336, 337, 338, 339, 340, 341, 342, 343, 344, 345, 346, 347, 348, 349, 350, 351, 352, 353, 354, 355, 356, 357, 358, 359, 360, 361, 362, 363, 364, 365, 366, 367, 368, 369, 370, 371, 372, 373, 374, 375, 376, 377, 378, 379, 380, 381, 382, 383, 384, 385, 386, 387, 388, 389, 390, 391, 392, 393, 394, 395, 396, 397, 398, 399, 400, 401, 402, 403, 404, 405, 406, 407, 408, 409, 410, 411, 412, 413, 414, 415, 416, 417, 418, 419, 420, 421, 422, 423, 424, 425, 426, 427, 428, 429, 430, 431, 432, 433, 434, 435, 436, 437, 438, 439, 440, 441, 442, 443, 444, 445, 446, 447, 448, 449, 450, 451, 452, 453, 454, 455, 456, 457, 458, 459, 460, 461, 462, 463, 464, 465, 466, 467, 468, 469, 470, 471, 472, 473, 474, 475, 476, 477, 478, 479, 480, 481, 482, 483, 484, 485, 486, 487, 488, 489, 490, 491, 492, 493, 494, 495, 496, 497, 498, 499, 500, 501, 502, 503, 504, 505, 506, 507, 508, 509, 510, 511, 512, 1024, 1025, 1026, ...}

(There is sometimes some ambiguity about whether N, the set of natural numbers, contains 0 or not. Here I've assumed it does, although that assumptions affects the argument not at all.)

Did you actually try comparing it to the naturals? The limit of the ratio
$\lim_{n\to\infty}\frac{|\{m\in X: m\leq n\}|}{|\{m\in{\mathbb{N}}: m\leq n\}|}$
does not exist, where |{...}| denotes the size of the set. This is is exactly what you tried to do with your 3:2 ratio argument from earlier.

This is the first time you have shown this particular formula (as well as your modified X) so, no.

The denominator is clearly n but the numerator is harder to calculate. In fact, X doesn't even exist if n = 3. For k = 0 the required range is [1, 2] while for k = 1 the required range is [4, 8]. No k exists that will satisfy the equation if n = 3.

The formula Vorpal shows here is your formula. It's the formula you have implicitly been using to compare the sizes of infinite sets. It compares the size of X to the size of N by looking at how many elements of X are less than some number n, and then dividing that by how many elements of N are less than n, and then taking the limit as n becomes arbitrarily large. Both the numerator and the denominator are perfectly well defined for all n.

I've taken the liberty of making a graph which shows this set size ratio as a function of n, so we can get an idea of what it does for large n. Here it is:

http://forums.randi.org/imagehosting/4514e8dce1d83171.gif

Now, psionl0, looking at that graph, can you tell me how much bigger N is than X?

You can't, because the ratio doesn't converge. Ever. That's the whole point. This is your way of comparing the size of two sets, and it fails here, where the standard way succeeds.

That's not a mathematical argument. That is just a vague assertion coupled with some incorrect statements about my position based on the fact that you didn't have a clue what "number of elements" means.

One thing you need to realize is that in the standard treatment, the size of a set is not in general a "number" at all. It is a cardinality (http://en.wikipedia.org/wiki/Cardinality), a concept more general than that of a number. In the special case of finite sets, the cardinality of the set happens to be a number. For infinite sets, it is not. The cardinality of N is given by the symbol $\aleph_0$. The cardinality of the set of real numbers R is given by the symbol $\aleph_1$. We know that these two cardinalities are different because it can be proven that no 1-1 correspondence between N and R can be constructed. We know that the set of primes (for example) also has cardinality $\aleph_0$ because at least one 1-1 correspondence between the primes and N can be constructed.

Vorticity
6th October 2011, 09:48 AM
Oh, and regarding set ordering:

Here is the set of natural numbers (leave out 0 for now):

N = {1,2,3,4,5,6,7,8,9,10...}

Here is the set of even naturals:

Evens = {2,4,6,8,10,12...}

It looks like N is twice as big as the Evens since, counting from the beginning of each set, I can find two elements in N for every even. But now consider this reordering of our list of the naturals:

N = {2,4,1,6,8,3,10,12,5,14,16,7,18,20,9,...}

I take the first two evens, then the first odd, then the next two evens, and then the next odd, and so on. All the natural numbers are included in this list at some finite position. But compare this to the list of the evens! Now it looks like for every two evens, there correspond three natural numbers. This seems to be saying that the evens are now 2/3 of all natural numbers, in contradiction to the 1/2 we got before.

This method therefore gives different answers based on ordering. This conflicts our intuitive notion of the size of a set.

psionl0
6th October 2011, 09:51 AM
I appreciate all the work you put into that post Vorticity. Unfortunately, it probably wasn't warranted (unless you dispute the last couple of exchanges between W.D.Clinger and myself).

Vorticity
6th October 2011, 09:55 AM
According to the definition I gave you, the existence or non-existence of 1-3 maps is totally irrelevant.

I just want to elaborate on this. Here is a 1-3 map from the naturals to the naturals:

1 -> {1,2,3}
2 -> {4,5,6}
3 -> {7,8,9}
4 -> {10,11,12}
etc...

Does this mean that N has a different cardinality than itself?

psionl0
6th October 2011, 10:01 AM
This method therefore gives different answers based on ordering. This conflicts our intuitive notion of the size of a set.That was the problem. It appeared that the relative sizes of the two sets depended on how you did the mapping and even (as you have demonstrated) on the order that the elements were listed.

What was needed was a definition for "size" that would work for finite as well as infinite sets and which would not depend on the mapping or ordering.

The controversial bit was in saying that 1 - 1 correspondence "proves" same "number of members" without providing that definition.

Vorticity
6th October 2011, 10:02 AM
I appreciate all the work you put into that post Vorticity. Unfortunately, it probably wasn't warranted (unless you dispute the last couple of exchanges between W.D.Clinger and myself).

Ah well, you guys posted while I was writing.

At any rate, I don't think I disagree with anything in those posts. The only thing I'd add is that while it is indeed true that the "1-1 correspondence" definition of cardinality is certainly simply a definition, I'm unaware of any other definition of cardinality that is both general and consistent, i.e. it doesn't give a different answer under re-orderings of the set in question.

If anybody knows of one, I'd be interested to hear it.

W.D.Clinger
6th October 2011, 10:19 AM
Nice posts, Vorticity. I just want to clarify this point:

The cardinality of N is given by the symbol $\aleph_0$. The cardinality of the set of real numbers R is given by the symbol $\aleph_1$.
That last sentence assumes the continuum hypothesis (http://en.wikipedia.org/wiki/Continuum_hypothesis), which cannot be proved from the usual axioms of set theory and mathematics. Without that assumption, we can't say whether the cardinality of the set of real numbers is $\aleph_1$, $\aleph_{17}$, or some other infinite cardinal greater than $\aleph_0$.

We can, however, prove that the cardinality of the set of real numbers is $2^{\aleph_0}$, which means there's a one-to-one correspondence (bijection) between the set of real numbers and the set of functions that map the natural numbers to a two-element set. (One such correspondence can be constructed by starting with the function that maps real numbers to their (possibly infinite) representations in base 2, and using that function to define a tweaked version.)

We know that these two cardinalities are different because it can be proven that no 1-1 correspondence between N and R can be constructed.
And that theorem can be proved (http://en.wikipedia.org/wiki/Cantor%27s_diagonal_argument) without relying on the continuum hypothesis, axiom of choice, or any other controversial assumptions.

Vorticity
6th October 2011, 10:33 AM
That last sentence assumes the continuum hypothesis (http://en.wikipedia.org/wiki/Continuum_hypothesis), which cannot be proved from the usual axioms of set theory and mathematics. Without that assumption, we can't say whether the cardinality of the set of real numbers is $\aleph_1$, $\aleph_{17}$, or some other infinite cardinal greater than $\aleph_0$.

And here I've learned something about notation. Thank you.

I had always been under the impression that $\aleph_1$ was defined to be the cardinality of R, and that the continuum hypothesis asserted that there exists no other cardinality between $\aleph_0$ and $\aleph_1$. But now I see that you are indeed correct: $\aleph_1$ is actually defined as the next cardinality after $\aleph_0$, regardless of whether $\aleph_1$ is the cardinality of R or not.

Can it be proven (with or without reference to the AOC) that the set of all cardinalities is itself countable?

ETA: I see that this (http://en.wikipedia.org/wiki/Continuum_hypothesis#The_generalized_continuum_hyp othesis) answers this question.

sol invictus
6th October 2011, 10:58 AM
If N = {1, 2, 3, 4, ...} and X = {x1, x2, x3, x4, ...} and there exists a function f such that f(i) = xi
Then by definition the sets {1, 2, 3, 4, ...} and {f(1), f(2), f(3), f(4), ...} are said to be the same size.

I say "by definition" because it is neither a proof nor a theorem. Its truth depends upon what we mean by the "size" of an infinite set.

For finite sets, no one will disagree with this definition (i.e. it's in accord with every other sensible definition). For infinite sets, that's the only definition I'm aware of that doesn't suffer from major inconsistencies, like depending on the order of the elements of the set.

The problem is that put this way, we are basically saying that there are as many even numbers as there are natural numbers because the sets of these numbers are defined to be the same size.

No, that's wrong. We are saying they are the same size because there exists a 1-to-1 mapping. That same criterion can be applied to any other pair of sets, and so it's clearly not equivalent to a definition that applies only to a single pair. And it produces interesting results, such as the fact that there are more reals between 0 and 1 than there are even integers, but there are just as many rationals between 0 and 1 as even integers.

Well, at least we agree on that!

Since you are so determined to twist my posts, let's put it this way:
What is the number of elements in {3, -1, 10, 17}?
What is the number of elements in (10, 3, 17, -1}?
4 and 4

Go back and re-read our exchange, starting the the quote of Vorpal's you so badly misread. If you still think I "twisted" your posts, I have nothing more to say to you.

Vorpal
6th October 2011, 02:07 PM
I had always been under the impression that $\aleph_1$ was defined to be the cardinality of R, and that the continuum hypothesis asserted that there exists no other cardinality between $\aleph_0$ and $\aleph_1$. But now I see that you are indeed correct: $\aleph_1$ is actually defined as the next cardinality after $\aleph_0$, regardless of whether $\aleph_1$ is the cardinality of R or not.

Can it be proven (with or without reference to the AOC) that the set of all cardinalities is itself countable?
It's actually worse: the aleph notation in itself implicitly assumes the axiom of choice, because you need it to prove that the cardinalities are well-ordered, that (among other things) there is a well-defined 'next highest' cardinality after a particular cardinality. In ZFC, the infinite cardinals are indexed by ordinal numbers, which automatically makes them not just uncountable, but of greater 'cardinality' than any set whatsoever. (I put this in quotes because technically, it's not a cardinality, but there is a sense in which the cardinals are "as large" as the class of all sets.) Without AC, the ℵα notation has no meaning in general, though you could simply define "ℵ0" to be |N| and not worry about it whether it's meaningful for other indices.

If we understand |A|≤|B| as the existence of an injection, then the AC is what makes cardinalities have a number-like properties (and even better is the GHC, as you say), and we have nice theorems like: |A|=|B| iff |A|≤|B| and |B|≤|A|.

However, if we're talking about the natural numbers N in particular, then |N| is still the unambiguously smallest infinite cardinality: comparison to |N| works in the above sense, every infinite set has a countably infinite subset, etc. Though technically, we still need a tiny (countable) fragment of AC to prove that, it's not controversial (or at least, shouldn't be) because without it, even something as basic as calculus will be seriously crippled.

psionl0
6th October 2011, 08:10 PM
If you still think I "twisted" your posts, I have nothing more to say to you.I admit that attempts by you and Vorpal to demonstrate that I didn't know what I am talking about instead of clarifying the issue caused me to wander dangerously close to rule 12 territory myself. I apologize for any offence that I caused.

Many mathematics teachers (and even textbooks) have an annoying habit of teaching a subject (eg probability) without adequately defining it. As a result, students may be able to solve most of the problems they are given but frequently get stumped by some because they don't really understand why a formula might be applicable to one problem but not another.

Based on what I have read in this thread, the gist of my initial response to the OP might have been like this:One can not talk about the number of evens or the number of odds in any real sense of the word because both sets are infinite and you can't count up to infinity.

However, there is a 1 - 1 correspondence between each odd and each even number. The significance of this is that if two finite sets have a 1 - 1 correspondence between their elements then they both have the same number of elements or they are both the same "size". If we apply this axiom to infinite sets then we have sort of defined the "size" of an infinite set.

This does lead to some apparent paradoxes. For example, there is a 1 - 1 correspondence between the set of positive integers and many (if not all) of its infinite subsets. This means that there are as "many" integers as there are even numbers even though in any terminating sequence of natural numbers, only half of them are even. This is a consequence of the way we define the "size" of an infinite set.

If that sounds too artificial then it is worth noting that there is no 1 - 1 correspondence between whole numbers and real numbers. So, in some sense, these sets must be different "sizes".I know that you would word the response more strongly than that but if you are explaining something to an amateur then it is best not to assume that they know and agree with all the definitions used by mathematicians.

Cheers :)