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 $f:X\to Y$ is continuous and has a inverse function, than the inverse function is also continuous.

Kyle Sutton
2022-07-02
Answered

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 $f:X\to Y$ is continuous and has a inverse function, than the inverse function is also continuous.

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amanhantmk

Answered 2022-07-03
Author has **17** answers

$X$ 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.

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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