Let $A,B\subset\mathbb R$ and $f:A\to B$ be an invertible function (so 1-1 and onto). Prove that if $A$ is compact and $f$ is continuous, then the inverse $f^{-1}:B\to A$ is continuous. And give a counterexample when $A$ is not compact (no transcendentals).
If $f$ is continuous, then for every sequence $\{x_n\}$ in $A$ that converges to $L$, $\lim_{n\to\infty}f(x_n)=f(L)$. We need to prove that for every sequence $\{y_n\}$ in $B$ that converges to $K$, $\lim_{n\to\infty}f^{-1}(y_n)=f^{-1}(K)$ (Is this sufficient to prove that $f^{-1}$ is continuous?), I'm not sure how to proceed from there, and I'm having trouble coming up with a counterexample as well.
Best Answer
To my knowledge, the most expeditious way to prove this is to use
where $X$ and $Y$ are domain and codomain of $f$ (can be arbitrary metric space).
Based on this statement, to show $f^{-1}$ is a continuous mapping of $B$ onto $A$, it is sufficient to show that for any open set $V \subset A$, $(f^{-1})^{-1}(V)$ is an open set in $B$, that is, $f(V)$ is open in $B$.
Now since $V$ is open in $A$ and $A$ is compact, $V^c \cap A$, as a closed subset of $A$, is compact (use the theorem that any closed subset of a compact set is compact). Since $f$ is continuous, $f(V^c \cap A)$ is compact and so is closed in $B$ (use the theorem that continuous mapping of any compact set is compact). Now since $f$ is one-to-one and onto, $f(V)$ is the complement of $f(V^c \cap A)$, hence is open in $B$.
For the counterexample, consider $A = [0, 1] \cup (2, 3]$, which is not compact, and define $f$ as follows: \begin{align} f(x) = \begin{cases} x & x \in [0, 1] \\ x - 1 & x \in (2, 3] \end{cases} \end{align} $f$ is a one-to-one and onto function of $A$ to $B = [0, 2]$, in addition, $f$ is continuous at every point of $A$. It is easily to get that \begin{align} f^{-1}(x) = \begin{cases} x & x \in [0, 1] \\ x + 1 & x \in (1, 2] \end{cases} \end{align} Hence $f^{-1}$ is not continuous at $1$! (a jump occurs here).