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Let $X$ be a subset of $\mathbb{R^n}$. The topology on $X$ is induced by the topology of $\mathbb{R^n}$. If there is an homeomorphism from $X$ onto $\mathbb{R}^n$, is it true that $X$ is open in $\mathbb{R}^n$ ? I think yes, but how can it be shown? (of course, $X$ is open in $X$).

If $x \in X$, $H_{n-1}(\mathbb{R}^n-\{f(x)\})= \mathbb{Z}$ so $H_{n-1}(X-\{x\})= \mathbb{Z}$, but I don't know how to continue.

Thanks in advance.

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up vote 3 down vote accepted

Yes, this is a special case of Brouwer's theorem of invariance of domain.

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Is not the open half plane homeomorphic to the entire space? – leo Oct 6 '12 at 15:03
@leo: What does that have to do with anything? – Chris Eagle Oct 6 '12 at 16:06
For example in $\Bbb R^2$ let $X=\{(x,y):y\gt 0\}$. If $X$ is homeomorphic to $\Bbb R^2$ ($X$ with the relative topology), then that's an example of a $X$ homeomorphic to $\Bbb R^2$ that is not an open ball. – leo Oct 6 '12 at 17:02
@leo: At the risk of repeating myself: so what? What does that have to do with anything? – Chris Eagle Oct 6 '12 at 17:05
If I understand correctly the OP is asking for how to prove: if $X\subseteq\Bbb R^n$ with the induced topology is homeomorphic to $\Bbb R^n$ then $X$ is a ball. What I'm trying to say is: is the open half plane a counterexample? – leo Oct 6 '12 at 17:18

Suppose $X$ is not open in $\mathbb{R}^{n}$. Then you can find a point in $p\in X$ such that \begin{equation}B(p,\epsilon)\cap X^{C}\neq \emptyset,\ \forall \epsilon>0\end{equation}

where $B(p,\epsilon)$ is the open ball with center $p$ and radius $\epsilon$. Let $f:X\rightarrow\mathbb{R}^{n}$ be the homeomorphism. Choose $q\in\mathbb{R}^{n}$ such that $f^{-1}(q)=p$. Take $\delta>0$ and consider the set $f^{-1}(B(q,\delta))$.

By the invariance theorem (pointed out by Chris Eagle) $f^{-1}(B(q,\delta))$ is a open set such that $p\in f^{-1}(B(q,\delta))$. But this is an absurd because as we saw, there is some $\epsilon>0$ such that $B(p,\epsilon)\subset f^{-1}(B(q,\delta))$ and $B(p,\epsilon)\cap X^{C}\neq \emptyset$

Hence $X$ is open in $\mathbb{R}^{n}$.

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