# Proving that a one-to-one continuous function on a compact subset has a continuous inverse

This is a curious problem I found in the "challenge" section of the text I'm learning real analysis from.

Suppose $A$ is a compact subset of $\mathbb{R}^{n}$ and that $f$ is a continuous function mapping $A$ one-to-one onto $B \subseteq \mathbb{R}^{m}$. Try and prove that the inverse function $f^{-1}$ is continuous on $B$.

I figure that this most likely has to do with the theorem stating that if a subset is compact in a particular vector space, then for a continuous $f$ mapping onto another vector space the function of the subset is compact in this other vector space i.e. the continuous image of a compact set is compact. I tried investigating the proof of this theorem to see if it would give me any intuition as to how to prove this question but wasn't successful, it feels required to solve the problem though.

• You probably know that a map $f$ is continuous iff the preimage of open sets is open. Taking complements, this is equivalent to the preimage of closed sets being closed. Now consdering $f^{-1}$, the preimages are the direct images of $f$. If $D\subseteq A$ is closed, it is a closed inside a compact, hence compact, so $f(A)$ is compact, hence closed. – Luiz Cordeiro Nov 19 '15 at 16:19
• @LuizCordeiro That's the topological definition of continuity, right? – snape Nov 19 '15 at 16:22
• To connect Luiz Cordeiro's comment and BigbearZzz's answer [$f(A)$ compact $\rightarrow f(A)$ closed] follows from $\mathbb{R}^m$ being Hausdorff. – rVitale Nov 19 '15 at 16:25
• Yes, that's one of the several equivalent definitions. – Luiz Cordeiro Nov 19 '15 at 16:25
• @LuizCordeiro I understand your logic and really appreciate your help, one last question would just be to clarify - when you say $D$, you're just referring to the image of $f^{-1}$ in $A$? – snape Nov 19 '15 at 16:43

There is a famous result that a continuous bijection $f$ from compact set into a Hausdorff space is a homeomorphism, i.e. $f^{-1}$ is continuous (on $\mathscr R(A)$, that is). Since the assumption stated that $A$ is compact and since $\Bbb R^n$ is a Hausdorff space for all $n\in \Bbb N$, the result follows.
• @snape Oops I'm really sorry, I should have noticed that you mentioned that you got the question from real analysis exercise. My bad. By the way, the only thing that you need to know about Hausdorff space to understand the proof is that compact sets are all closed in Hausdorff space, which is quite obvious for $\Bbb R^n$. (A set in $\Bbb R^n$ is compact iff it's closed and bounded.) – BigbearZzz Nov 19 '15 at 16:27