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Consider $\mathbb{R}^n$ with the usual Euclidean topology, or the (equivalent) product topology. Let $U \subseteq \mathbb{R}^n$ be open. Suppose that $V \subseteq \mathbb{R}^n$ is such that $V \cap \partial V \neq \emptyset$. Can there possibly exist a homeomorphism $h : U \to V$?

I have been thinking about this problem for a little while. My intuition is telling me that, no, such a homeomorphism cannot exist. After all, since $V \cap \partial V \neq \emptyset$, $V$ contains a few of its "edge" or boundary points. But, since U is open, it contains nothing of the sort. Certainly this property of being an "edge" or boundary point should be preserved (in some way) through homeomorphism?

Is there a basic topological argument that puts this question to rest, or a more advanced theorem that needs to be applied? I'm hoping we don't have bring forth the Jordan Curve Theorem here.

Hints or solutions are greatly appreciated.

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The answer is "No, there can't be a homeomorphism". It's the invariance of domain, if a subset of $\mathbb{R}^n$ is homeomorphic to an open subset of $\mathbb{R}^n$, it is itself open. I know no elementary or non-advanced proof, however. – Daniel Fischer Sep 8 '13 at 17:20
In this exact case, the answer is "no", because you forgot the case when $U$ is connected and $V$ is not. If you add the hypothesis that both $U$ and $V$ are connected, take for (counter-)example the unit ball and the punctured unit ball. – Marra Sep 8 '13 at 17:21
@Daniel: Isn't this a complete answer? Why do you post this as a comment? – Martin Brandenburg Sep 8 '13 at 17:22
@MartinBrandenburg I was hoping that somebody with a better recollection of algebraic topology could give a sketch of a proof. – Daniel Fischer Sep 8 '13 at 17:26
@DanielFischer: Thanks for the comment! I don't have a great understanding of algebraic topology, but at least I know where the proof comes from now. I should eventually have a look in Munkres about homotopy theory :) – JZShapiro Sep 8 '13 at 17:45
up vote 1 down vote accepted

The theorem on the invariance of domain says that if $U \subset \mathbb{R}^n$ is open, and $f \colon U \to \mathbb{R}^n$ is a continuous injective function, then $f$ is open, in particular $V = f(U)$ is an open subset of $\mathbb{R}^n$, and $f$ is a homeomorphism from $U$ to $V$.

So if $V \subset \mathbb{R}^n$ is homeomorphic to an open $U \subset \mathbb{R}^n$, then $V$ itself must be open - the homeomorphism provides a continuous injective function $U \to \mathbb{R}^n$.

I know of no proof of the theorem of invariance of domain that does not use advanced techniques, unfortunately.

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