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We know that if $f:K \to X$ is continuous and injective, $K$ is compact, and $X$ is Hausdorff, then $f$ is a homeomorphism $K \cong f(K)$.

So suppose $f:U \to \mathbb{R}^n$ is continuous and injective, where $U$ is an open subset of $\mathbb{R}^n$. Choose an open ball $B$ such that $B \subset \bar{B} \subset U$. By the remark above $\bar{B} \cong f(\bar{B}) \Rightarrow B \cong f(B)$. Finally, since $B$ is open, so is $f(B)$.

My instructor said something about not knowing that $f(B)$ is open even though it is homeomorphic to an open set, but I'm not sure I understood at all.

Wilson P.
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    You only know that there exists an open set $O$ in $\mathbf{R}^n$ such that $O\cap f(\overline B) = f(B)$. That is, $f(B)$ is open in the relative topology of $f(\overline B)$ and not open in $\mathbb{R}^n$. – Rabee Tourky Jan 23 '13 at 06:11

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This tells you that $f$ is a homeomorphism between $U$ and $f(U)$, but it doesn't address whether $f(U)$ is open in $\mathbb{R}^n$.

uncookedfalcon
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