# Error in exercise 1.3.9 of Folland's advanced calculus

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 $$f(\partial S)=\partial(f(S))$$.

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

• Your proof looks fine to me. May 16, 2022 at 4:56
• $f$ 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. May 16, 2022 at 10:08