real analysis - Show that $f, f^{-1}$ are continuous

Let $A,B \subset \mathbb{R}$ be open, and $f:A\rightarrow B$ be surjective and strictly monotonic increasing. Show that $f,f^{-1}$ are continuous.

Proof: I first show $f$ is injective. Let $x,y \in A, \mbox{and } x\neq y.$ This means either $x

To show $f$ is continuous, let $D \subset B$ be open. I need to show $f^{-1}(D)$ is open in $A$. Suppose not, i.e., $(f^{-1}(D))^{c}$ is not closed. $\quad\Rightarrow \exists$ a sequence $(x_n)_{n\in\mathbb{N}}$ in $(f^{-1}(D))^{c}$ that converges to $x$ which is in $f^{-1}(D)$. As $f$ is bijective, $\exists$ a unique $y_n,y$ for each $x_n$ such that $f(x_n)=y_n,f(x)=y,\forall n\in\mathbb{N}$, where $(y_n)_{n\in\mathbb{N}}$ is in $D^{c}, y\in D$. Here, $y_n\rightarrow y$. If it does not, this means $f^{-1}(y_n)=x_n$ does not converge to $f^{-1}(y)=x$, contradiction. $\quad \Rightarrow D^c$ is not closed. $\Rightarrow D$ is not open. Contradiction.

$f^{-1}$ can also be proved to be continuous in the same way as above.

I somehow get a feeling that I am not allowed to argue $y_n \rightarrow y$ because a priori, my argument, which I think, assumes $f$ is continuous, which I have not proven yet!

Is my proof correct or my doubt? Please help me get out of this situation!

Answer

A conceptual proof follows from the material in $\S$ 6.3 of my honors calculus notes:

Step 1: We are given that $f$ is bijective and increasing. So $f^{-1}$ exists and is moreover increasing: suppose not; then there are $y_1 < y_2 \in B$ with $f^{-1}(y_1) \geq f^{-1}(y_2)$. Then applying $f$ we get $y_1 \geq y_2$, contradiction. Thus the situation is perfectly symmetrical with respect to $f$ and $f^{-1}$, so it suffices to show that $f$ is continuous.

Step 2: We use the fact that every monotone function defined on $A$ has a left hand limit at every $c \in A$ -- namely $\sup \{ f(x) \mid x < c\}$ and a right hand limit -- namely $\inf \{ f(x) \mid x > c\}$, and the value $f(x)$ lies in between. (This is part of the Monotone Jump Theorem in $\S$ 6.3 of my notes.) Thus the only way we can have a discontinuity is if $\lim_{x \rightarrow c^-} f(x) < f(c)$ or $f(c) < \lim_{x \rightarrow c^+} f(c)$. But if either of these occurs, then $f(c)$ is not an interior point of $f(A)$, contradicting the hypothesis that $f(A)$ is open.

This answers the OP's question. I claim that it also proves that the inverse of a continuous function is continuous, at least in the case that the domain of $f$ is an interval. This is because every injective continuous function $f: I \rightarrow \mathbb{R}$ must be monotone: see $\S$ 5.6.3 of the notes. (There is a bit of combinatorial trickiness here.) Using the fact that every open subset of $\mathbb{R}$ is a disjoint union of open intervals -- which is not in the notes (I don't do any explicit topology whatsoever there) but is well known and not hard to show -- and that for every open interval $I$ and continuous $f$, $f(I)$ is an interval ($\S$ 6.2 of the notes) one sees that this extends to continuous functions on any open subset of $\mathbb{R}$, but this seems to be the longer way around this particular question.

Added: It is certainly not the case that any continuous bijection $f: X \rightarrow Y$ of topological spaces must have a continuous inverse. To get a counterexample, let $Y$ be your favorite non-discrete topological space, let $X$ be the same set endowed with the discrete topology, and let $f$ be the identity map. From this perspective the "automatic continuity" property of the inverse for continuous bijections on open subsets of $\mathbb{R}$ is surprising. It can be generalized to open subsets of $\mathbb{R}^n$ and then becomes a quite famous (and rather deep) theorem, Brouwer's Invariance of Domain. This can be generalized to topological manifolds. There are other "Open Mapping Theorems" in mathematics -- famously in complex analysis and Banach space theory -- but such results are highly prized, as they are the exception rather than the rule.