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Let $f$ be one-to-one and continuous on $[a,b] \subset \mathbb{R}$. I want to show that the inverse function $f^{-1}$ is continuous.

I know the possibly relevant fact that $f([a,b]) = [c,d]$. The approach I tried was to show that if you have any sequence $(f(x_n)) \rightarrow f(s)$, then $(x_n) \rightarrow s$. But I can't even see why $(x_n)$ converges, much less to $s$.

rorty
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  • This should help http://math.stackexchange.com/questions/541082/proving-the-inverse-of-a-continuous-function-is-also-continuous?rq=1 – SquirtleSquad May 25 '16 at 06:10
  • Would you prefer to use sequence argument? – 3x89g2 May 25 '16 at 06:26
  • @Misakov Yes, thanks. – rorty May 25 '16 at 06:41
  • @Merlinsbeard I saw that post, but I am not familiar with the "if and only if condition that continuous function bring back closed sets to closed sets" that the answer depends on. It's unclear to me what this means... I thought closedness was not preserved by continuous functions. I also prefer a sequential solution. – rorty May 25 '16 at 06:47

1 Answers1

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We can try to argue by contradiction.

Suppose $f^{-1}$ is not continuous. Then there exists a sequence $\{y_n\} \in [c,d]$ such that $y_n \to y$, yet $f^{-1}(y_n)$ does not converge to $f^{-1}(y)$.

Consider the sequence $x_n = f^{-1}(y_n)$. Also define $x = f^{-1}(y)$. We do not know whether $x_n$ converge or not (actually we do), but we know that it has a convergent subsequence. Let the convergent subsequence be $x_{n_j}$ and let $x_{n_j}\to x'$.

Since $f$ is continuous, then $\lim f(x_{n_j})=f(x')$. Notice that $f(x_{n_j})$ is also a subsequence of $y_n$. So we have found a subsequence of $y_n$ that converges to $f(x')$. We know that $y_n$ converges to $y$, so it has to be the case that $y=f(x')$. But this is impossible.

3x89g2
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  • I think you're onto something, but I see why this isn't there yet. It could be that $(x_n)$ doesn't converge, but $x'=x$. If we show that $(x_n)$ must converge, then we're done. – rorty May 25 '16 at 07:05
  • @rorty Yes, you are right. That is actually easy to show though. See https://math.la.asu.edu/~dajones/class/371/ch4.pdf and http://math.stackexchange.com/questions/776899/if-every-convergent-subsequence-converges-to-a-then-so-does-the-original-boun – 3x89g2 May 25 '16 at 07:15
  • @rorty By the way, you might want to look into the "mapping open set" definition of continuous function. That one is much easier to use. – 3x89g2 May 25 '16 at 07:16