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.

  • $\begingroup$ Do you know the following two statements? 1. The image of a compact set under a continuous function is compact. 2. A function is continuous if and only if for every open set in the codomain, its pre-image is open. // Your assumptions plus statement 1 implies that $f$ is an open map. So using 2 you have that its inverse function is coninuous. $\endgroup$ – Willie Wong Nov 11 '15 at 5:04
  • $\begingroup$ @Willie Wong No I have neither of those statements $\endgroup$ – BCho Nov 11 '15 at 5:06
  • $\begingroup$ What statements do you have at your disposal? Do you have Heine-Borel theorem? What is your operating definition of continuity and compactness? $\endgroup$ – Willie Wong Nov 11 '15 at 5:09
  • $\begingroup$ @Willie Wong Yes I have Heine-Borel theorem for compactness, continuity is the normal $\varepsilon, \delta$ $\endgroup$ – BCho Nov 11 '15 at 5:13

To my knowledge, the most expedient way to prove this is to use

$f$ is continuous on $X$ if and only if for any open set $V \subset Y$, $f^{-1}(V)$ is an open set in $X$.

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).

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  • $\begingroup$ How do I use the statement to prove my statement? $\endgroup$ – BCho Nov 11 '15 at 5:44
  • $\begingroup$ I'm not sure how you got that last step, why if $f(V)$ the complement of $f(V^c\cap A)$? $\endgroup$ – BCho Nov 11 '15 at 6:05
  • $\begingroup$ Because $V \cap (V^c \cap A) = \varnothing$, so $f(V) \cap f(V^c \cap A) = \varnothing$, by the injectivity of $f$. The fact that $f(V) \cup f(V^c \cap A) = B$ is obvious by surjectivity. $\endgroup$ – Zhanxiong Nov 11 '15 at 6:08
  • $\begingroup$ @Solitary: mind that the OP claimed he is not aware of the theorem stating continuous maps compact to compact math.stackexchange.com/questions/1523622/… $\endgroup$ – Willie Wong Nov 11 '15 at 6:10
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    $\begingroup$ @AnthonyPeter: I suspect the real reason is that there is a higher chance "random math student" remember the open set version than the closed set version, especially if he/she has taken a standard course in point set topology. But a facetious answer could be sometimes it just makes more sense to use open sets $\endgroup$ – Willie Wong Nov 11 '15 at 13:55

What you have:

  • $f$ is bijective $A\to B$
  • $\forall x\in A ~\forall \epsilon > 0 ~\exists \delta > 0 ~\forall x' \in A (|x - x'| < \delta \implies |f(x) - f(x') | < \epsilon)$.

What you want:

  • $\forall x \in A~ \forall \epsilon > 0 ~\exists \delta > 0 ~\forall x' \in A (|f(x) - f(x')| < \delta \implies |x - x'| < \epsilon)$.

Proof by contradiction (sketch): assume for contradiction

$$ \exists x\in A ~ \exists \epsilon > 0 ~\forall \delta > 0 ~\exists x' \in A ~(|f(x) - f(x')| < \delta ~\wedge ~ |x - x'| \geq \epsilon )$$

Then letting $\delta_k = \frac{1}{k}$ we can find a sequence $\{x_k\}$ such that $|f(x) - f(x_k)| < 1/k$ while $|x - x_k| \geq \epsilon$.

By the Heine-Borel we have that $A$ is bounded. By Bolzano-Weierstrass we have that $\{x_k\}$ has a converging subsequence which we call $\{y_k\}$. Since $A$ is closed the limit, which we call $y$, is in $A$.

Since $f$ is continuous and $y_k \to y$ you have $f(y_k) \to f(y)$. On the other hand since by assumption $f(x_k) \to f(x)$, the subsequence $f(y_k)$ must also converge to $f(x)$. This implies $f(x) = f(y)$. But we know that $|y - x| \geq \epsilon$ by triangle inequality and so $y \neq x$. This contradicts injectivity.

I am assuming since you already have Heine-Borel you also have Bolzano-Weierstrass.

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  • $\begingroup$ Why does $|y-x|\geq \varepsilon$? $\endgroup$ – BCho Nov 11 '15 at 5:45
  • $\begingroup$ $|y-x| - |y - x'| \geq |x - x'|$ by triangle inequality. Sub in $y_k$ for $x'$ and note $y - y_k \to 0$. $\endgroup$ – Willie Wong Nov 11 '15 at 5:47
  • $\begingroup$ Can you sub in a sequence for an element? $\endgroup$ – BCho Nov 11 '15 at 6:03
  • $\begingroup$ doesn't $|x-x'|$ also go to 0? $\endgroup$ – BCho Nov 11 '15 at 6:26
  • $\begingroup$ Of course not. $\forall k$ we have $|x - x_k| \geq \epsilon$ by assumption. $\endgroup$ – Willie Wong Nov 11 '15 at 13:52

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