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Suppose $A$ is an open subset of $\mathbb{R}^n$ considered as a topological subspace. Suppose $B$ is a another susbet of $\mathbb{R}^n$, and under the induced topology, is homeomorphic to $A$.

If $B$ is closed and bounded, then by the Heine-Borel theorem, $B$ is compact as a topological space so $A$ and $B$ cannot be homeomorhpic, because $A$ is not compact. But if $B$ is not bounded, what can be said about $B$, if any? Must $B$ be open, can it ever be closed (let's leave out the only case of being both open and closed--the entire ambient space $\mathbb{R}^n$), or can it be neither open nor closed?

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Regarding your first claim, this is not true. You could have $A=B=\emptyset$. – kahen Jan 23 '13 at 14:35
It might be instructive to consider the sets $(0,\infty)$ and $[0,\infty)$. The removal of any point in the former set disconnects the space; the same is not true for the latter set. – David Mitra Jan 23 '13 at 14:51
up vote 6 down vote accepted

A subset of $\mathbb{R}^n$ homeomorphic to an open subset of $\mathbb{R}^n$ has to be open. This is the theorem of invariance of domain.

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