Let $U$ and $V$ be open sets in $\mathbb{R}^n$ and let $f$ be a one-to-one mapping from $U$ onto $V$ (so that there is an inverse mapping $f^{-1}:V\rightarrow U$). Suppose that $f$ and $f^{-1}$ are both continuous. Show that for any set $S$ whose closure is contained in $U$ we have $f(\partial S)=\partial(f(S))$.

According to http://www.math.washington.edu/~folland/Homepage/advcalc.pdf,

Assume also that the closure of $f(S)$ is contained in $V$.

I think this assumption is redundant. Here is my reasoning.

Since $f(\overline{S})=(f^{-1})^{-1}(\overline{S})$, by the continuity of $f^{-1}$ and the closedness of $\overline{S}$, $f(\overline{S})$ is closed in $V$. Since $f(S)\subset f(\overline{S})$, $\overline{f(S)}\subset f(\overline{S})\subset V$, as desired.

Am I right?

EDIT: There is an error in my reasoning. In the last statement, I used the fact that the closure of any set is the smallest closed set containing that set. But the closure of a set depends on the underlying metric space. Hence what I actually proved is really a triviality that the close of $f(S)$ in $V$ lies in $V$.

Consider $U=\mathbb{R}$, $V=(-\pi/2,\pi/2)$, $f=\arctan$. Then $\overline{f(R)}=[-\pi/2,\pi/2]\not\subset V$.

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    $\begingroup$ Your proof looks fine to me. $\endgroup$ May 16, 2022 at 4:56
  • $\begingroup$ $f $ is a homeomorphism from $U$ to $V$ so for any $S\subset U$ we have $f(Cl_U(S))=Cl_V(f(S))\subset V.$ But "any set $S$ whose closure is contained in $U$" means that $S\subset U$ and that $Cl_{\Bbb R^n}(S)=U\cap Cl_{\Bbb R^n}(S)=Cl_U(S). $ So you are CORRECT that $Cl_{\Bbb R^n}f(S))\subset V$ is redundant as it is implied by the other conditions. +1 for finding an improvement to a textbook. $\endgroup$ May 16, 2022 at 10:08


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