Is H^n Homeomorphic to R^n?

[AKG]

Well start with R² and work your way up. Also, let A and B be bounded.

 

[AKG]

Maybe an even "simpler" problem. Can $B^n$ be homeomorphic to $A$ if $B^n \subset A \subset \overline{B^n}$? We don't want to look at the case $B^n = A$ since we are interested in cases where A is not open. We don't need to look at the case $A = \overline{B^n}$ since then A would be compact, but the open ball is not compact, so it's obvious that they're not homeomorphic. So if A is strictly contained between the ball and its closure, can it be homeomorphic to the ball?

 

[Palindrom]

For R^2, I think I can prove what I need to prove, i.e. that H^n is not homeomorphic to any open subset of R^n.

Now, as for what you said, I'm not sure whether it'll help me to prove the above proposition, although it is an interesting question by itself.

Just for the record, I'm trying to show that a homeomorphism between manifolds with boundary takes the boundary onto the boundary. I'd really like to do it in the general topological case rather than in the smooth case, hence homeomorphisms and not diffeomorphisms.

That's what I have so far: If indeed H^n is homeomorphic to an open subset of R^n, then this subset must be contractible. It would then be enough to show that any contractible open subset of R^n is homeomorphic to R^n, as we've shown that R^n and H^n aren't homeomorphic.

Do you think I should still try your way, or will it not help my cause?

Any suggestions for my cause?

 

[StatusX]

I don't see why you can't just say that since, (1), for any homeomorphism f:A->B, and any subset U of A, U is homeomorphic to f(U), and (2), any point x in an open subset of Rⁿ has a neighborhood homeomorphic to Rⁿ, that any point y in B must also have a neighborhood homeomorphic to Rⁿ, and so B must be open in Rⁿ. Is there something wrong with this approach?

 

[AKG]

StatusX

Trying your approach, let A and B be subsets of Euclidean space, A is open and f : A -> B is a homeomorphism. Let b be any point in B, and let a be its preimage. Since A is open, there is a ball U such that a is in U, and U is contained in A. U is homeomorphic to Euclidean space. Thus f(U) is a subset of B containing b, homeomorphic to Euclidean space. You want to argue that f(U), being homeomorphic to Euclidean space, must be an open subset of Euclidean space. This will allow you to argue that B is open. However, how do you know it? After all, isn't that more or less what we're trying to prove in the first place? You've reduced our problem from:

Can an open subset be homeomorphic to a non-open subset?

to

Can Euclidean space be homeomorphic to a non-open subset?

or equivalently:

Can the open ball be homeomorphic to a non-open subset?

 

[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?