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I need to prove that if $X$ is a subset of $\mathbb{R}^n$ and $Y$ is a subset of $\mathbb{R}^m$, and $X$ and $Y$ are closed and bounded, then if $f:X \rightarrow Y$ is continuous and has a inverse function, than the inverse function is also continuous. In the previous exercise I was asked to show that if $f$ has a inverse function, the inverse function is continuous $\iff f$ sends open groups to open groups $\iff f$ sends closed groups to closed groups, This I've shown, and I should probably use it for this exercise, but I didn't find a way to use the fact that $X$ and $Y$ are closed and bounded(which is necessary)

Thanks!

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  • $\begingroup$ groups = subsets, I suppose? $\endgroup$
    – Henno Brandsma
    Commented Apr 23, 2016 at 7:52
  • $\begingroup$ What have you proved about closed and bounded subsets of $\mathbb{R}^n$ specifically? $\endgroup$
    – Henno Brandsma
    Commented Apr 23, 2016 at 7:55
  • $\begingroup$ Yes, groups means subsets. $\endgroup$
    – Amit
    Commented Apr 23, 2016 at 8:02
  • $\begingroup$ We learned that a sub-space of a complete normed vectoric space is a complete metric space if and only if it is a closed group, and we learned that all the norms on R^n are equivalent $\endgroup$
    – Amit
    Commented Apr 23, 2016 at 8:03
  • $\begingroup$ Nothing about sequences and the existence of convergent subsequences? $\endgroup$
    – Henno Brandsma
    Commented Apr 23, 2016 at 8:06

1 Answer 1

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$X$ and $Y$ are - as bounded and closed subsets of $\mathbb R^n$ and $\mathbb R^m$ respectively - compact Hausdorff spaces.

In such spaces a set is compact if and only if it is closed. If $f:X\to Y$ is continuous and $F\subseteq X$ is closed then $F$ is compact so that $f(F)\subseteq Y$ is compact as well. The next conclusion is that $f(F)$ is closed. So apparantly $f$ is a closed function, i.e. sends closed sets to closed sets. A continous and closed bijection (i.e. a map that has an inverse) is a homeomorphism. Its inverse will also be a homeomorphism.

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  • $\begingroup$ Hi, thank you but we did not learn in the course about Hausdorff spaces and compact groups, but from what I understood you are using a theorome which is more general, can you try to prove this specific case without the more general theorome? Thanks! $\endgroup$
    – Amit
    Commented Apr 22, 2016 at 19:39
  • $\begingroup$ Hmm..., it is beyond doubt that there are persons here who can do that. I posed this question on this. Have a look. $\endgroup$
    – drhab
    Commented Apr 23, 2016 at 7:40
  • $\begingroup$ Thanks! you might want to add the fact that this question has a previous exercise, which is to prove that if f has an inverse function, than f is an open function if and only if f is a closed function if and only if f^-1 is continuous $\endgroup$
    – Amit
    Commented Apr 23, 2016 at 7:44
  • $\begingroup$ I think persons who react will find that out automatically. This because they will have a look at your comments following the link in my question. $\endgroup$
    – drhab
    Commented Apr 23, 2016 at 7:45

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