Infinite Regress | Why is everyone telling me I am wrong?

[PeterDonis]

This is it, more or less. It's what we've all been saying all this time. Only with the caveat that the fourth dimension needen't actually exist as anything physical.

But the caveat means it isn't what we've been saying all this time; we have not been saying that our universe is embedded in a higher-dimensional space.

we are not on the surface of the universe, we are in the universe

There is no "surface of the universe". That's the point. The universe, spatially, is just a 3-dimensional space without a boundary. It could have a finite volume, in which case it is topologically a 3-sphere, or it could have infinite volume, in which case it is topologically what we ordinarily think of as "3-dimensional space". But either way there is no boundary and no "surface".

 

[Bandersnatch]

But the caveat means it isn't what we've been saying all this time; we have not been saying that our universe is embedded in a higher-dimensional space.

I know. I don't think I said we have. I pointed out the analogy was proper, as it gives the meaning to the shape of the 2D surface without it having boundaries, but that the embedding is not required for this to work. A 2-sphere will remain a 2-sphere whether there exists a 3rd dimension or not.

 

[Imager]

No. No boundary means no boundary, period.

For example, suppose our universe were, spatially, a 3-sphere. (And suppose, for purposes of this thought experiment, that the universe is not expanding, so the size of the 3-sphere is constant.) Then we could, in principle, send out spaceships to explore the entire 3-sphere, and measure its finite 3-volume, and none of those spaceships would ever encounter a boundary.

Hi Peter

When one of the spaceships reaches radius of sphere what happens? (Like the asteroids game does it come back in at the opposite point?)

 

[Dale]

When one of the spaceships reaches radius of sphere what happens?

I don't think that you understand the geometry being described. A normal sphere that you are used to thinking of is a 2-sphere. It is a 2D manifold. As a 2D manifold it has no boundary, and the radius doesn't even exist in the 2D manifold. I.e. there is no point in the 2D manifold where you would say that you have reached the radius. All that exists in the 2D manifold is a positive curvature which is uniform everywhere. All "spaceships" are always at the radius in a 3D embedding space where the radius would exist.

A 3-sphere is a 3D manifold with no boundary and with a positive curvature which is uniform everywhere. There is no radius in the 3-sphere.

 

[PeterDonis]

When one of the spaceships reaches radius of sphere what happens?

When you fly halfway around the Earth what happens? Do you reach a boundary?

The case of a spaceship flying around the universe is the same, just in a 3D sphere instead of a 2D sphere. The spaceship can fly in the same direction indefinitely and will never encounter a boundary: eventually, it will return to its starting point, having circumnavigated the universe.

(Like the asteroids game does it come back in at the opposite point?)

Not in a 3-sphere. If the universe had a different topology, called a 3-torus, then this would happen (or at least, this would be one way of describing what would happen). The asteroids game takes place in a 2-torus (or the computer representation of one, anyway).

 

[Imager]

I don't think that you understand the geometry being described.

You are right, I wasn't even close! One look at wiki was a mindblower! I think it is time for a beer or three...