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Perpetual Student · 10th June 2012, 06:35 PM

We have all seen this statement. What does it mean? What is the basis of this statement? In geometry, we can easily construct volumes and areas that are finite but unbounded, but those objects are not symmetrical, which is a supposed attribute of the universe on large scales. So, how and why is this possible?

Ray Brady · 10th June 2012, 06:41 PM

Originally Posted by Perpetual Student

In geometry, we can easily construct volumes and areas that are finite but unbounded, but those objects are not symmetrical...

I don't know what you mean here. The surface of a globe is finite but unbounded. How is that not symmetrical?

HghrSymmetry · 10th June 2012, 06:46 PM

The "simple" old hypersphere model used to be thrown around a bit as the possible geometry of the universe. In that model...the shape was completely symmetrical.

As a reduced dimension model...take a "regular" sphere, why wouldn't that be symmetrical?

You can run around the surface of the ball and never encounter any boundaries, yet the surface is finite.

ETA: Ray beat me to it.

Perpetual Student · 10th June 2012, 06:48 PM

Originally Posted by Ray Brady

I don't know what you mean here. The surface of a globe is finite but unbounded. How is that not symmetrical?

A globe certainly is bounded. It is a two dimensional surface with finite area in three dimensional space.

HghrSymmetry · 10th June 2012, 06:50 PM

Originally Posted by Perpetual Student

A globe certainly is bounded. It is a two dimensional surface with finite area in three dimensional space.

Increase the dimension by one more, and you have the same example.

shadron · 10th June 2012, 06:52 PM

Originally Posted by Perpetual Student

A globe certainly is bounded. It is a two dimensional surface with finite area in three dimensional space.

Errrr, bounds usually refer by analogy to math, where bounds exist to anchor the function at it's "boundary conditions". Bounds are places where the space stops. On a globe the space (the 2D surface) doesn't ever stop, and the fact that it repeats itself under certain circumstances is irrelevant. A 3D globe is not bounded in the 2D space of its surface. It is, of course, finite in extent; it has a calculable fixed area.

By extension, a hypersphere in 4D has an unbounded space on it's "surface", which is 3D. The question that cosmologists entertain is whether space is positively (like a hypersphere), negatively or not curved (flat). I believe that most physicists find that the space appears to be flat, though there is no definitive proof of that; it could be very gently rounded, + or -.

Perpetual Student · 10th June 2012, 06:55 PM

Originally Posted by HghrSymmetry

Increase the dimension by one more, and you have the same example.

There is no evidence of a fourth space dimension, so this is only a speculative example with no scientific basis.

HghrSymmetry · 10th June 2012, 06:58 PM

Originally Posted by Perpetual Student

There is no evidence of a fourth space dimension, so this is only a speculative example with no scientific basis.

Speculative yes, no scientific basis...no so much.

Reference: Super Symmetry, Kaluza Klein, Super String, and Membrane theory.

sol invictus · 10th June 2012, 07:05 PM

If you're asking about a universe that is finite and without boundary, there are plenty of examples. Two are both isotropic and homogeneous (the 3-sphere and real projective 3-space), and there are many more examples that are homogeneous but anisotropic (a 3-torus, for example).

None of them require a "fourth spatial dimension" since all are 3-spatial-dimensional

Perpetual Student · 10th June 2012, 07:11 PM

Originally Posted by shadron

Errrr, bounds usually refer by analogy to math, where bounds exist to anchor the function at it's "boundary conditions". Bounds are places where the space stops. On a globe the space (the 2D surface) doesn't ever stop, and the fact that it repeats itself under certain circumstances is irrelevant. A 3D globe is not bounded in the 2D space of its surface. It is, of course, finite in extent; it has a calculable fixed area.

By extension, a hypersphere in 4D has an unbounded space on it's "surface", which is 3D. The question that cosmologists entertain is whether space is positively (like a hypersphere), negatively or not curved (flat). I believe that most physicists find that the space appears to be flat, though there is no definitive proof of that; it could be very gently rounded, + or -.

The repetition is quite relevant in that it makes the sphere bounded. I can map every point on the sphere onto a finite flat two dimensional surface. I would hope the statement of "unbounded" is a bit more meaningful than that. Do we routinely refer to the earth as finite but unbounded?

Perpetual Student · 10th June 2012, 07:23 PM

Originally Posted by sol invictus

If you're asking about a universe that is finite and without boundary, there are plenty of examples. Two are both isotropic and homogeneous (the 3-sphere and real projective 3-space), and there are many more examples that are homogeneous but anisotropic (a 3-torus, for example).

None of them require a "fourth spatial dimension" since all are 3-spatial-dimensional

I'm not here to argue against the "finite but unbounded" conjecture; I would like to understand it better.
See my comment above about mapping every point on a sphere onto a bounded two dimensional surface. Can I also do that with your 3-torus and your 3-sphere and real projective three space? If not, could you explain why since I'm not familiar with those objects.

shadron · 10th June 2012, 07:25 PM

Originally Posted by Perpetual Student

The repetition is quite relevant in that it makes the sphere bounded. I can map every point on the sphere onto a finite flat two dimensional surface. I would hope the statement of "unbounded" is a bit more meaningful than that. Do we routinely refer to the earth as finite but unbounded?

It is certainly finite. What bounds do you see on its surface (as it is usually the surface which is referred to as unbounded, not its volume)?

Ray Brady · 10th June 2012, 07:27 PM

If you are moving on the surface of a globe, you can travel in any direction for any length of time without hitting a boundary. That's what is meant by unbounded.

sol invictus · 10th June 2012, 07:32 PM

Originally Posted by Perpetual Student

I'm not here to argue against the "finite but unbounded" conjecture; I would like to understand it better.
See my comment above about mapping every point on a sphere onto a bounded two dimensional surface. Can I also do that with your 3-torus and your 3-sphere and real projective three space? If not, could you explain why since I'm not familiar with those objects.

There's a big difference between "unbounded" and "without boundary", at least with their standard definitions. I can't figure out which you're asking about, but I think you may have the two confused.

Loosely speaking, "unbounded" means there are points that are infinitely far away from other points. There are no examples of finite-volume homogeneous and isotropic unbounded spaces, and no one thinks the universe is like that.

"Without boundary" means there are no boundary points (a boundary point is a point such that every region around it contains both points in the set and points outside the set). The examples I gave are finite volume and without boundary, and it is possible that the geometry of the universe is among them.

Perpetual Student · 10th June 2012, 08:21 PM

Quote:

Loosely speaking, "unbounded" means there are points that are infinitely far away from other points. There are no examples of finite-volume homogeneous and isotropic unbounded spaces, and no one thinks the universe is like that.

It is that sense of unbounded that I had in mind -- like the finite volume within a rotated curve like 1/x² which has no bound and where we can find points as far away from each other as we choose.
The other meanings of unbounded (or without boundary), like the surface of a sphere or a 3-torus, are quite boring so I'm a bit disappointed that that's all there is to it. Thanks.

SezMe · 10th June 2012, 08:28 PM

Originally Posted by Perpetual Student

I can map every point on the sphere onto a finite flat two dimensional surface.

Actually, you can't.

Perpetual Student · 10th June 2012, 08:36 PM

Originally Posted by SezMe

Actually, you can't.

I certainly can. I did not say the mapping would be continuous. Cut the sphere in half but leave it attached at one point. Now lay the two attached domes onto a surface and project the points onto the surface. Every point on the sphere has a unique point on the definitely bounded and finite surface.

SezMe · 10th June 2012, 08:53 PM

Well, then you didn't map a sphere, did you. You mapped a hemisphere.

W.D.Clinger · 11th June 2012, 12:27 AM

Originally Posted by Perpetual Student

Originally Posted by SezMe

Actually, you can't.

I certainly can. I did not say the mapping would be continuous. Cut the sphere in half but leave it attached at one point. Now lay the two attached domes onto a surface and project the points onto the surface. Every point on the sphere has a unique point on the definitely bounded and finite surface.

So long as we're speaking in trivialities...

Every nonempty topological space can be mapped onto a single point of Euclidean two-space.

(That mapping is continuous, but it isn't likely to be one-to-one.)

Perpetual Student · 11th June 2012, 06:12 AM

Originally Posted by W.D.Clinger

So long as we're speaking in trivialities...

Every nonempty topological space can be mapped onto a single point of Euclidean two-space.

(That mapping is continuous, but it isn't likely to be one-to-one.)

I don't understand the relevance of your point. The mapping I describe above is one-to-one.

dafydd · 11th June 2012, 06:15 AM

Originally Posted by Perpetual Student

I don't understand the relevance of your point. The mapping I describe above is one-to-one.

He didn't say that your mapping was not one-to-one. You seem to have missed the point of his post.

Perpetual Student · 11th June 2012, 12:24 PM

Originally Posted by dafydd

He didn't say that your mapping was not one-to-one. You seem to have missed the point of his post.

And what is that point?

Perpetual Student · 11th June 2012, 12:26 PM

It seems to me that if the universe has been expanding uniformly from a singularity, it should be more or less spherical, which would be bounded. Is there evidence to the contrary?

ynot · 11th June 2012, 01:00 PM

Originally Posted by Perpetual Student

We have all seen this statement. What does it mean? What is the basis of this statement? In geometry, we can easily construct volumes and areas that are finite but unbounded, but those objects are not symmetrical, which is a supposed attribute of the universe on large scales. So, how and why is this possible?

“Finite and unbounded” is the mother of all circular arguments.

dafydd · 11th June 2012, 01:34 PM

Originally Posted by Perpetual Student

It seems to me that if the universe has been expanding uniformly from a singularity, it should be more or less spherical, which would be bounded. Is there evidence to the contrary?

The universe could be the three dimensional surface of an expanding hypersphere.

dafydd · 11th June 2012, 01:35 PM

Originally Posted by Perpetual Student

And what is that point?

That every nonempty topological space can be mapped onto a single point of Euclidean two-space.

Perpetual Student · 11th June 2012, 01:42 PM

Originally Posted by dafydd

That every nonempty topological space can be mapped onto a single point of Euclidean two-space.

So?

Vorpal · 11th June 2012, 01:46 PM

Originally Posted by Perpetual Student

It seems to me that if the universe has been expanding uniformly from a singularity, it should be more or less spherical, which would be bounded.

You're mistaken. One can easily have an spatially infinite universe from a singularity even in the presence of perfect uniformity.

Perhaps it might be more intuitive to think of a somewhat more Newtonian example backwards: imagine a spatially infinite universe uniformly filled with dust. Around a point some distance r away, the ball of dust gravitationally attracts the dust at the edge of the ball toward its center. Because of uniformity, every point in space can be this center and sees exactly the same kind of thing: the rest of the dust collapsing towards it. With a bit of math, one can show that the density diverges in finite time. Where is this singularity? Everywhere in space, which is infinite at all times.

If you take away the condition of uniformity, and the situation can pretty obviously get much more complicated than expansion from a point. Completely disregard that mental picture; it's very, very misleading. Though ultimately the taxonomy of singularities is a much messier business than just increasing the dimensionality would suggest.

Perpetual Student · 11th June 2012, 01:47 PM

Originally Posted by dafydd

The universe could be the three dimensional surface of an expanding hypersphere.

It could also be a seven dimensional hexagon projected onto a three dimensional manifold embedded in four-space, but what is the evidence for any shape other than a spherical shape in three dimensions? I am asking about evidence -- not conjectures.

theprestige · 11th June 2012, 01:51 PM

Originally Posted by Perpetual Student

It could also be a seven dimensional hexagon projected onto a three dimensional manifold embedded in four-space, but what is the evidence for any shape other than a spherical shape in three dimensions? I am asking about evidence -- not conjectures.

What's the evidence for a spherical shape?

dafydd · 11th June 2012, 02:13 PM

Originally Posted by Perpetual Student

It could also be a seven dimensional hexagon projected onto a three dimensional manifold embedded in four-space, but what is the evidence for any shape other than a spherical shape in three dimensions? I am asking about evidence -- not conjectures.

http://ned.ipac.caltech.edu/level5/P...eacock3_1.html

GodMark2 · 11th June 2012, 03:30 PM

Originally Posted by theprestige

What's the evidence for a spherical shape?

edd · 11th June 2012, 03:32 PM

That's not evidence for a spherical shape.

dafydd · 11th June 2012, 03:36 PM

http://tinyurl.com/cxjpptx

Perpetual Student · 11th June 2012, 04:36 PM

Originally Posted by dafydd

http://ned.ipac.caltech.edu/level5/P...eacock3_1.html

Originally Posted by dafydd

http://tinyurl.com/cxjpptx

Interesting sources. Thanks.

Perpetual Student · 11th June 2012, 04:37 PM

Originally Posted by theprestige

What's the evidence for a spherical shape?

I'm not aware of any. A sphere seems like a logical default shape in the absence of any evidence.

sol invictus · 11th June 2012, 05:25 PM

Originally Posted by Perpetual Student

I'm not aware of any. A sphere seems like a logical default shape in the absence of any evidence.

The observational evidence favors at least approximate homogeneity and isotropy. There are four geometries with exact homogeneity and exact isotropy: the 3-sphere, real 3D projective space, hyperbolic 3-space, and ordinary Euclidean 3-space.

All four possibilities have a singularity in the past, and all expand uniformly. None have boundary, but the 3-sphere and projective space are bounded (i.e. finite), while the other two are infinite.

theprestige · 11th June 2012, 05:39 PM

Originally Posted by Perpetual Student

I'm not aware of any. A sphere seems like a logical default shape in the absence of any evidence.

Well, now you have at least four "logical default shapes" to choose from:

Originally Posted by sol invictus

The observational evidence favors at least approximate homogeneity and isotropy. There are four geometries with exact homogeneity and exact isotropy: the 3-sphere, real 3D projective space, hyperbolic 3-space, and ordinary Euclidean 3-space.

All four possibilities have a singularity in the past, and all expand uniformly. None have boundary, but the 3-sphere and projective space are bounded (i.e. finite), while the other two are infinite.

Sol, is "approximate homogeneity and isotropy" the same as "expand uniformly"?

Perpetual Student · 11th June 2012, 07:12 PM

Originally Posted by theprestige

Well, now you have at least four "logical default shapes" to choose from:

Indeed! Now, I'll try to make some sense of those who are more knowledge than me, who might shed some light on which might be the better model among:
the 3-sphere, real 3D projective space, hyperbolic 3-space, and ordinary Euclidean 3-space -- assuming for the moment that we can rule out other possibilities. I have no doubt there will be enough opinions on the subject to make things interesting.

Vorpal · 11th June 2012, 10:41 PM

See the Lambda-CDM model^WP. The universe is very close to spatially flat, so if forced to choose out of those four, Euclidean 3-space fits best--though of course it is also possible to have one of the others with a very large radius of curvature. But inflation can make pretty much anything look flat, so it isn't a good reason to believe that model would be a good representation of reality on the very large scale.

10th June 2012, 06:35 PM	#1
Perpetual Student Illuminator Join Date: Jul 2008 Location: USA Posts: 3,738	Is the universe finite but unbounded? We have all seen this statement. What does it mean? What is the basis of this statement? In geometry, we can easily construct volumes and areas that are finite but unbounded, but those objects are not symmetrical, which is a supposed attribute of the universe on large scales. So, how and why is this possible?
	__________________ _{It doesn't matter how beautiful your theory is, it doesn't matter how smart you are. If it doesn't agree with experiment, it's wrong. - Richard P. Feynman} ξ

10th June 2012, 06:41 PM	#2
Ray Brady Muse Join Date: Jan 2011 Posts: 589	Originally Posted by Perpetual Student In geometry, we can easily construct volumes and areas that are finite but unbounded, but those objects are not symmetrical... I don't know what you mean here. The surface of a globe is finite but unbounded. How is that not symmetrical?
Ray Brady Muse Join Date: Jan 2011 Posts: 589

10th June 2012, 06:46 PM	#3
HghrSymmetry View clearer from above. Join Date: Jul 2007 Posts: 8,510	The "simple" old hypersphere model used to be thrown around a bit as the possible geometry of the universe. In that model...the shape was completely symmetrical. As a reduced dimension model...take a "regular" sphere, why wouldn't that be symmetrical? You can run around the surface of the ball and never encounter any boundaries, yet the surface is finite. ETA: Ray beat me to it.
	__________________ HygrSym brought up some fantastic points. He's so good, he doesn't have to use pseudo words like 'chillax'... -FSM

10th June 2012, 06:48 PM	#4
Perpetual Student Illuminator Join Date: Jul 2008 Location: USA Posts: 3,738	Originally Posted by Ray Brady I don't know what you mean here. The surface of a globe is finite but unbounded. How is that not symmetrical? A globe certainly is bounded. It is a two dimensional surface with finite area in three dimensional space.
	__________________ _{It doesn't matter how beautiful your theory is, it doesn't matter how smart you are. If it doesn't agree with experiment, it's wrong. - Richard P. Feynman} ξ

10th June 2012, 06:50 PM	#5
HghrSymmetry View clearer from above. Join Date: Jul 2007 Posts: 8,510	Originally Posted by Perpetual Student A globe certainly is bounded. It is a two dimensional surface with finite area in three dimensional space. Increase the dimension by one more, and you have the same example.
	__________________ HygrSym brought up some fantastic points. He's so good, he doesn't have to use pseudo words like 'chillax'... -FSM

10th June 2012, 06:52 PM	#6
shadron Philosopher Join Date: Sep 2005 Location: Colorado Posts: 5,719	Originally Posted by Perpetual Student A globe certainly is bounded. It is a two dimensional surface with finite area in three dimensional space. Errrr, bounds usually refer by analogy to math, where bounds exist to anchor the function at it's "boundary conditions". Bounds are places where the space stops. On a globe the space (the 2D surface) doesn't ever stop, and the fact that it repeats itself under certain circumstances is irrelevant. A 3D globe is not bounded in the 2D space of its surface. It is, of course, finite in extent; it has a calculable fixed area. By extension, a hypersphere in 4D has an unbounded space on it's "surface", which is 3D. The question that cosmologists entertain is whether space is positively (like a hypersphere), negatively or not curved (flat). I believe that most physicists find that the space appears to be flat, though there is no definitive proof of that; it could be very gently rounded, + or -.
	Last edited by shadron; 10th June 2012 at 06:58 PM.

10th June 2012, 06:55 PM	#7
Perpetual Student Illuminator Join Date: Jul 2008 Location: USA Posts: 3,738	Originally Posted by HghrSymmetry Increase the dimension by one more, and you have the same example. There is no evidence of a fourth space dimension, so this is only a speculative example with no scientific basis.
	__________________ _{It doesn't matter how beautiful your theory is, it doesn't matter how smart you are. If it doesn't agree with experiment, it's wrong. - Richard P. Feynman} ξ

10th June 2012, 06:58 PM	#8
HghrSymmetry View clearer from above. Join Date: Jul 2007 Posts: 8,510	Originally Posted by Perpetual Student There is no evidence of a fourth space dimension, so this is only a speculative example with no scientific basis. Speculative yes, no scientific basis...no so much. Reference: Super Symmetry, Kaluza Klein, Super String, and Membrane theory.
	__________________ HygrSym brought up some fantastic points. He's so good, he doesn't have to use pseudo words like 'chillax'... -FSM

10th June 2012, 07:05 PM	#9
sol invictus Philosopher Join Date: Oct 2007 Location: Nova Roma Posts: 8,435	If you're asking about a universe that is finite and without boundary, there are plenty of examples. Two are both isotropic and homogeneous (the 3-sphere and real projective 3-space), and there are many more examples that are homogeneous but anisotropic (a 3-torus, for example). None of them require a "fourth spatial dimension" since all are 3-spatial-dimensional

10th June 2012, 07:11 PM	#10
Perpetual Student Illuminator Join Date: Jul 2008 Location: USA Posts: 3,738	Originally Posted by shadron Errrr, bounds usually refer by analogy to math, where bounds exist to anchor the function at it's "boundary conditions". Bounds are places where the space stops. On a globe the space (the 2D surface) doesn't ever stop, and the fact that it repeats itself under certain circumstances is irrelevant. A 3D globe is not bounded in the 2D space of its surface. It is, of course, finite in extent; it has a calculable fixed area. By extension, a hypersphere in 4D has an unbounded space on it's "surface", which is 3D. The question that cosmologists entertain is whether space is positively (like a hypersphere), negatively or not curved (flat). I believe that most physicists find that the space appears to be flat, though there is no definitive proof of that; it could be very gently rounded, + or -. The repetition is quite relevant in that it makes the sphere bounded. I can map every point on the sphere onto a finite flat two dimensional surface. I would hope the statement of "unbounded" is a bit more meaningful than that. Do we routinely refer to the earth as finite but unbounded?
	__________________ _{It doesn't matter how beautiful your theory is, it doesn't matter how smart you are. If it doesn't agree with experiment, it's wrong. - Richard P. Feynman} ξ

10th June 2012, 07:23 PM	#11
Perpetual Student Illuminator Join Date: Jul 2008 Location: USA Posts: 3,738	Originally Posted by sol invictus If you're asking about a universe that is finite and without boundary, there are plenty of examples. Two are both isotropic and homogeneous (the 3-sphere and real projective 3-space), and there are many more examples that are homogeneous but anisotropic (a 3-torus, for example). None of them require a "fourth spatial dimension" since all are 3-spatial-dimensional I'm not here to argue against the "finite but unbounded" conjecture; I would like to understand it better. See my comment above about mapping every point on a sphere onto a bounded two dimensional surface. Can I also do that with your 3-torus and your 3-sphere and real projective three space? If not, could you explain why since I'm not familiar with those objects.
	__________________ _{It doesn't matter how beautiful your theory is, it doesn't matter how smart you are. If it doesn't agree with experiment, it's wrong. - Richard P. Feynman} ξ

10th June 2012, 07:25 PM	#12
shadron Philosopher Join Date: Sep 2005 Location: Colorado Posts: 5,719	Originally Posted by Perpetual Student The repetition is quite relevant in that it makes the sphere bounded. I can map every point on the sphere onto a finite flat two dimensional surface. I would hope the statement of "unbounded" is a bit more meaningful than that. Do we routinely refer to the earth as finite but unbounded? It is certainly finite. What bounds do you see on its surface (as it is usually the surface which is referred to as unbounded, not its volume)?

10th June 2012, 07:27 PM	#13
Ray Brady Muse Join Date: Jan 2011 Posts: 589	If you are moving on the surface of a globe, you can travel in any direction for any length of time without hitting a boundary. That's what is meant by unbounded.
Ray Brady Muse Join Date: Jan 2011 Posts: 589

10th June 2012, 07:32 PM	#14
sol invictus Philosopher Join Date: Oct 2007 Location: Nova Roma Posts: 8,435	Originally Posted by Perpetual Student I'm not here to argue against the "finite but unbounded" conjecture; I would like to understand it better. See my comment above about mapping every point on a sphere onto a bounded two dimensional surface. Can I also do that with your 3-torus and your 3-sphere and real projective three space? If not, could you explain why since I'm not familiar with those objects. There's a big difference between "unbounded" and "without boundary", at least with their standard definitions. I can't figure out which you're asking about, but I think you may have the two confused. Loosely speaking, "unbounded" means there are points that are infinitely far away from other points. There are no examples of finite-volume homogeneous and isotropic unbounded spaces, and no one thinks the universe is like that. "Without boundary" means there are no boundary points (a boundary point is a point such that every region around it contains both points in the set and points outside the set). The examples I gave are finite volume and without boundary, and it is possible that the geometry of the universe is among them.

10th June 2012, 08:21 PM	#15
Perpetual Student Illuminator Join Date: Jul 2008 Location: USA Posts: 3,738	Quote: Loosely speaking, "unbounded" means there are points that are infinitely far away from other points. There are no examples of finite-volume homogeneous and isotropic unbounded spaces, and no one thinks the universe is like that. It is that sense of unbounded that I had in mind -- like the finite volume within a rotated curve like 1/x² which has no bound and where we can find points as far away from each other as we choose. The other meanings of unbounded (or without boundary), like the surface of a sphere or a 3-torus, are quite boring so I'm a bit disappointed that that's all there is to it. Thanks.
	__________________ _{It doesn't matter how beautiful your theory is, it doesn't matter how smart you are. If it doesn't agree with experiment, it's wrong. - Richard P. Feynman} ξ

10th June 2012, 08:28 PM	#16
SezMe post-pre-born Join Date: Dec 2003 Location: Santa Barbara, CA Posts: 16,651	Originally Posted by Perpetual Student I can map every point on the sphere onto a finite flat two dimensional surface. Actually, you can't.

10th June 2012, 08:36 PM	#17
Perpetual Student Illuminator Join Date: Jul 2008 Location: USA Posts: 3,738	Originally Posted by SezMe Actually, you can't. I certainly can. I did not say the mapping would be continuous. Cut the sphere in half but leave it attached at one point. Now lay the two attached domes onto a surface and project the points onto the surface. Every point on the sphere has a unique point on the definitely bounded and finite surface.
	__________________ _{It doesn't matter how beautiful your theory is, it doesn't matter how smart you are. If it doesn't agree with experiment, it's wrong. - Richard P. Feynman} ξ Last edited by Perpetual Student; 10th June 2012 at 08:49 PM.

10th June 2012, 08:53 PM	#18
SezMe post-pre-born Join Date: Dec 2003 Location: Santa Barbara, CA Posts: 16,651	Well, then you didn't map a sphere, did you. You mapped a hemisphere.

11th June 2012, 12:27 AM	#19
W.D.Clinger Master Poster Join Date: Oct 2009 Posts: 2,466	Originally Posted by Perpetual Student Originally Posted by SezMe Actually, you can't. I certainly can. I did not say the mapping would be continuous. Cut the sphere in half but leave it attached at one point. Now lay the two attached domes onto a surface and project the points onto the surface. Every point on the sphere has a unique point on the definitely bounded and finite surface. So long as we're speaking in trivialities... *Every* nonempty topological space can be mapped onto a single point of Euclidean two-space. (That mapping is continuous, but it isn't likely to be one-to-one.)

11th June 2012, 06:12 AM	#20
Perpetual Student Illuminator Join Date: Jul 2008 Location: USA Posts: 3,738	Originally Posted by W.D.Clinger So long as we're speaking in trivialities... *Every* nonempty topological space can be mapped onto a single point of Euclidean two-space. (That mapping is continuous, but it isn't likely to be one-to-one.) I don't understand the relevance of your point. The mapping I describe above is one-to-one.
	__________________ _{It doesn't matter how beautiful your theory is, it doesn't matter how smart you are. If it doesn't agree with experiment, it's wrong. - Richard P. Feynman} ξ

11th June 2012, 06:15 AM	#21
dafydd Penultimate Amazing Join Date: Feb 2008 Location: Belgium (Flatland) Posts: 31,680	Originally Posted by Perpetual Student I don't understand the relevance of your point. The mapping I describe above is one-to-one. He didn't say that your mapping was not one-to-one. You seem to have missed the point of his post.
	__________________ Yesterday upon the stairs I met a man who wasn't there He wasn't there again today I wish that he would go away.

11th June 2012, 12:24 PM	#22
Perpetual Student Illuminator Join Date: Jul 2008 Location: USA Posts: 3,738	Originally Posted by dafydd He didn't say that your mapping was not one-to-one. You seem to have missed the point of his post. And what is that point?
	__________________ _{It doesn't matter how beautiful your theory is, it doesn't matter how smart you are. If it doesn't agree with experiment, it's wrong. - Richard P. Feynman} ξ

11th June 2012, 12:26 PM	#23
Perpetual Student Illuminator Join Date: Jul 2008 Location: USA Posts: 3,738	It seems to me that if the universe has been expanding uniformly from a singularity, it should be more or less spherical, which would be bounded. Is there evidence to the contrary?
	__________________ _{It doesn't matter how beautiful your theory is, it doesn't matter how smart you are. If it doesn't agree with experiment, it's wrong. - Richard P. Feynman} ξ

11th June 2012, 01:00 PM	#24
ynot Illuminator Join Date: Jan 2006 Location: New Zealand Posts: 4,731	Originally Posted by Perpetual Student We have all seen this statement. What does it mean? What is the basis of this statement? In geometry, we can easily construct volumes and areas that are finite but unbounded, but those objects are not symmetrical, which is a supposed attribute of the universe on large scales. So, how and why is this possible? “Finite and unbounded” is the mother of all circular arguments.
	__________________ Rumours of a god’s existence have been greatly exaggerated. My post are all (IMO) unless stated otherwise.

11th June 2012, 01:34 PM	#25
dafydd Penultimate Amazing Join Date: Feb 2008 Location: Belgium (Flatland) Posts: 31,680	Originally Posted by Perpetual Student It seems to me that if the universe has been expanding uniformly from a singularity, it should be more or less spherical, which would be bounded. Is there evidence to the contrary? The universe could be the three dimensional surface of an expanding hypersphere.
	__________________ Yesterday upon the stairs I met a man who wasn't there He wasn't there again today I wish that he would go away.

11th June 2012, 01:35 PM	#26
dafydd Penultimate Amazing Join Date: Feb 2008 Location: Belgium (Flatland) Posts: 31,680	Originally Posted by Perpetual Student And what is that point? That every nonempty topological space can be mapped onto a single point of Euclidean two-space.
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11th June 2012, 01:42 PM	#27
Perpetual Student Illuminator Join Date: Jul 2008 Location: USA Posts: 3,738	Originally Posted by dafydd That every nonempty topological space can be mapped onto a single point of Euclidean two-space. So?
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11th June 2012, 01:46 PM	#28
Vorpal Extrapolate! Join Date: Jan 2005 Posts: 1,013	Originally Posted by Perpetual Student It seems to me that if the universe has been expanding uniformly from a singularity, it should be more or less spherical, which would be bounded. You're mistaken. One can easily have an spatially infinite universe from a singularity even in the presence of perfect uniformity. Perhaps it might be more intuitive to think of a somewhat more Newtonian example backwards: imagine a spatially infinite universe uniformly filled with dust. Around a point some distance r away, the ball of dust gravitationally attracts the dust at the edge of the ball toward its center. Because of uniformity, every point in space can be this center and sees exactly the same kind of thing: the rest of the dust collapsing towards it. With a bit of math, one can show that the density diverges in finite time. Where is this singularity? Everywhere in space, which is infinite at all times. If you take away the condition of uniformity, and the situation can pretty obviously get much more complicated than expansion from a point. Completely disregard that mental picture; it's very, very misleading. Though ultimately the taxonomy of singularities is a much messier business than just increasing the dimensionality would suggest.
Vorpal Extrapolate! Join Date: Jan 2005 Posts: 1,013	__________________ For every philosopher, there exists an equal and opposite philosopher. They're both wrong.

11th June 2012, 01:47 PM	#29
Perpetual Student Illuminator Join Date: Jul 2008 Location: USA Posts: 3,738	Originally Posted by dafydd The universe could be the three dimensional surface of an expanding hypersphere. It could also be a seven dimensional hexagon projected onto a three dimensional manifold embedded in four-space, but what is the evidence for any shape other than a spherical shape in three dimensions? I am asking about evidence -- not conjectures.
	__________________ _{It doesn't matter how beautiful your theory is, it doesn't matter how smart you are. If it doesn't agree with experiment, it's wrong. - Richard P. Feynman} ξ

11th June 2012, 01:51 PM	#30
theprestige Philosopher Join Date: Aug 2007 Posts: 8,905	Originally Posted by Perpetual Student It could also be a seven dimensional hexagon projected onto a three dimensional manifold embedded in four-space, but what is the evidence for any shape other than a spherical shape in three dimensions? I am asking about evidence -- not conjectures. What's the evidence for a spherical shape?
theprestige Philosopher Join Date: Aug 2007 Posts: 8,905

11th June 2012, 02:13 PM	#31
dafydd Penultimate Amazing Join Date: Feb 2008 Location: Belgium (Flatland) Posts: 31,680	Originally Posted by Perpetual Student It could also be a seven dimensional hexagon projected onto a three dimensional manifold embedded in four-space, but what is the evidence for any shape other than a spherical shape in three dimensions? I am asking about evidence -- not conjectures. http://ned.ipac.caltech.edu/level5/P...eacock3_1.html
	__________________ Yesterday upon the stairs I met a man who wasn't there He wasn't there again today I wish that he would go away.

11th June 2012, 03:30 PM	#32
GodMark2 Graduate Poster Join Date: Oct 2005 Location: Oregon, USA Posts: 1,050	Originally Posted by theprestige What's the evidence for a spherical shape?
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11th June 2012, 03:32 PM	#33
edd Graduate Poster Join Date: Nov 2007 Posts: 1,619	That's not evidence for a spherical shape.
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11th June 2012, 03:36 PM	#34
dafydd Penultimate Amazing Join Date: Feb 2008 Location: Belgium (Flatland) Posts: 31,680	http://tinyurl.com/cxjpptx
	__________________ Yesterday upon the stairs I met a man who wasn't there He wasn't there again today I wish that he would go away.

11th June 2012, 04:36 PM	#35
Perpetual Student Illuminator Join Date: Jul 2008 Location: USA Posts: 3,738	Originally Posted by dafydd http://ned.ipac.caltech.edu/level5/P...eacock3_1.html Originally Posted by dafydd http://tinyurl.com/cxjpptx Interesting sources. Thanks.
	__________________ _{It doesn't matter how beautiful your theory is, it doesn't matter how smart you are. If it doesn't agree with experiment, it's wrong. - Richard P. Feynman} ξ