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