Let $D\subset \mathbb{R}$ and D is bounded and closed. $$ f:D \rightarrow f(D)\in \mathbb{R}$$ is continuous and injective. Then, the inverse $ f^{-1}:f(D) \rightarrow D$ is continuous in the interval $f(D)$.

What I've got so far: Because the function is injective and is mapped onto $f(D)$ it is also bijective. Furthermore, I've proven that when it is injective, it must be either strictly increasing or strictly decreasing, therefore $f^{-1}$ is,too.

From the definition of continuity, I know:

$ \forall \varepsilon > 0 \; \; \exists \delta > 0$ such that $|f(x)-f(x_0)|< \varepsilon \; \; \forall x \in D : |x-x_0| < \delta $

My thoughts are, I need to pick some $y \in f(D)$ and use the above mentioned definition along with the strictly increasing monotonicity and the fact, that the interval is bounded and closed. I was told, this was a pretty important part, yet I am very unsure, as to what that even tells me.

Could somebody provide me with some hints, as to what useful information I can extract from the fact, that it is bounded and closed? I'd assume, I need the monotonicity mainly for the purpose of having a well defined inverse, so I'm not sure, if I still need it at this point.

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    $\begingroup$ You don't know (unless there's more information) that $D$ is an interval. You do know, however, that it is compact. $\endgroup$ – David Bowman Dec 21 '16 at 21:56
  • $\begingroup$ - deleting comment until I reassure myself that it is relevant. $\endgroup$ – Unwisdom Dec 21 '16 at 22:00

Hint: First prove that for any closed subset $K$ of $D$, $f(K)$ is closed in $\mathbb{R}$.

To prove this you should use the fact that a subset of $\mathbb{R}$ is compact if and only if it is closed and bounded. (See: show that a subset of $\mathbb{R}$ is compact iff it is closed and bounded)

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  • $\begingroup$ I appreciate it, but I have already proven in another exercise, that if $f: [a,b] \rightarrow \mathbb{R}$ and $[a,b]$ is continuous, that $f([a,b])$ is, too. This proof is just one of many other exercises and I honestly can't see it needing so many other proofs in order to be proven itself $\endgroup$ – Jonathan Dec 21 '16 at 22:08
  • $\begingroup$ This is a stronger result than what you just wrote - I did not assume that $K$ is an interval. $\endgroup$ – Natha Dec 21 '16 at 22:30
  • $\begingroup$ I do appreciate how that you are being rigorous, but I do not see how this is getting me any closer to the proof. Like I said in my post, I do not know what the result $f(K) \in R$ being closed would tell me. This is coming from a standpoint of:" I know what the definition is, but I do not see the value of knowing that it applies, how is this useful" What information does it provide that makes it relevant to the proof? $\endgroup$ – Jonathan Dec 21 '16 at 22:33

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