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.