Exploring the Homeomorphism between H^n and R^n

[StatusX]

So we're given a subset B of Rⁿ which is homeomorphic to an open subset of Rⁿ, and we want to show B is open. We've shown every point b in B has an (open) neighborhood U(b) homeomorphic to Rⁿ.

What exactly does it mean for a subset O of Rⁿ to be open? It means that every point in O is contained in an open ball which is contained in O. What we've shown above is that every point in B has an open neighborhood in B homeomorphic to an open ball. This distinction is a subtlety that I'm still not completely clear on.

We can say that the open neighborhood U(b) can be sent via the continuous inclusion map i:B->Rⁿ, and we get some set i(U(b)) in Rⁿ containing b and contained in B. But what does i(U(b)) look like? Dropping the b dependence, consider the restriction i_U:U->i(U). This is one-to-one, onto, and continuous. If it's open we're done, but i is not open, in general. It seems clear that this must be a homeomorphism, but I can't find this last piece.

 

[hypermorphism]

If we are working in Euclidean space, the theorem is just Invariance of Domain.

 

[StatusX]

Oh, I should have realized this was non-trivial by the corollary that Rⁿ is not homeomorphic to R^m unless n=m that would have followed from it. I know this problem needs algebraic topology, but the original question sounded like it could be handled by simple point set topology.

 

[hypermorphism]

Oh, I should have realized this was non-trivial by the corollary that Rⁿ is not homeomorphic to R^m unless n=m that would have followed from it.

I think that one's also provable through combining Brouwer fixed point with Borsuk-Ulam.

I know this problem needs algebraic topology, but the original question sounded like it could be handled by simple point set topology.

Yep. It seems so simple, but it seems to elude capture unless one calls on sledgehammers.

 

[AKG]

So we're given a subset B of Rⁿ which is homeomorphic to an open subset of Rⁿ, and we want to show B is open. We've shown every point b in B has an (open) neighborhood U(b) homeomorphic to Rⁿ.

Have we? In light of the "Invariance of Domain" theorem, it turns out to be true, but I don't think we showed any such thing. If you look at my previous post, we have f(U) which is contained in B, and contains b as an element. We also know that f(U) is homeomorphic to Rⁿ. So we know that f(U) is homeomorphic to the open ball. But we haven't proved that the open ball is not homeomorphic to, say, the open ball plus a single boundary point (e.g. the open ball of radius 1 centered at 0, together with the point (1, 0, 0)). If this is the case, then maybe f(U) is like this ball + point, and b is this boundary point, so although f(U) contains b, f(U) contains no open ball which contains b. This goes back to my question of whether the ball is homeomorphic to something strictly contained between the ball and its closure.

What we've shown above is that every point in B has an open neighborhood in B homeomorphic to an open ball.

I don't think we have.

Dropping the b dependence, consider the restriction i_U:U->i(U). This is one-to-one, onto, and continuous. If it's open we're done, but i is not open, in general. It seems clear that this must be a homeomorphism, but I can't find this last piece.

i(U) is, of course, open in itself. I think you mean i_U : U -> Rⁿ. In fact, we don't need to think of i as being restricted to U, we can just look at the inclusion map of U. Better yet, we can just look at U. It's already in Rⁿ, I don't think we gain anything by adding all these middlemen. By the way, the inclusion map you're looking at would just be identity, wouldn't it? It's trivially homeomorphic. This tells us nothing about the openness of U as a subset of Rⁿ.

Anyways, thanks hypermorphism, I think that theorem is all we needed.

 

[Palindrom]

If we are working in Euclidean space, the theorem is just Invariance of Domain.

Brilliant, that's exactly what I've been looking for.

Now where could I read a proof?