# Proving $f\in\mathbb{C}(D)$ if and only if for every closed set $F\subseteq R^m$, $f^{-1}(F)$ is closed (relatively) in $D$.

I am confident in my first inclusion but I am quite lost in my second. Can I please have help?

$$\def\x{{\mathbf x}} \def\R{{\mathbb R}} \def\C{{\mathbb C}} \def\f{{\mathbf f}}$$

Let $$\f\colon D\to \R^m$$, $$D\subseteq\R^n$$. Prove $$\f\in\C(D)$$ if and only if for every closed set $$F\subseteq \R^m$$, $$\f^{-1}(F)$$ is closed (relatively) in $$D$$.

$$\textbf{Solution:}$$

$$(\rightarrow)$$ Suppose $$\f\in\C(D)$$ and $$F$$ be any arbitrary closed set in $$\R^m$$. Our claim is that $$\f^{-1}(E)$$ is relatively closed in $$D$$. $$F$$ is a closed set in $$\R^m$$ if and only if $$\R^m \setminus F$$ is open set in $$\R^n$$. As $$\f$$ is continuous on $$D$$, $$\f^{-1}(\R^m\setminus E) \text{ is open (relatively) in } D$$ $$\f^{-1}(\R^m \setminus E) = \f^{-1}(\R^m)\setminus \f^{-1}(E) = D\setminus \f^{-1}(E)$$ because $$\f^{-1}(A\setminus B) = \f^{-1}(A)\setminus \f^{-1}(A) \setminus \f^{-1} (B)$$ and $$\f^{-1}(\R^n) = D$$.

Then, $$D\setminus \f^{-1}(E)$$ is open (relatively) in $$D$$. Thus, $$D\setminus (D\setminus \f^{-1}(E)) = \f^{-1}(E)$$ is relatively closed in $$D$$. This is for all $$E$$ closed in $$\R^n$$.

$$(\leftarrow)$$ Conversely, suppose for every $$F$$ closed in $$\R^m, \f^{-1}(F)$$ is relatively closed in $$D$$. Our claim is that $$\f$$ is continuous, $$\f \in \C(D)$$, i.e. for every open set $$E \subseteq \R^m, \f^{-1}(E)$$ is relatively open in $$D$$. $$E$$ is any open set in $$\R^m$$ which is equivalent to saying $$\R^m \setminus E$$ is closed in $$\R^m$$. Then $$\f^{-1}(\R^m\setminus E)$$ is closed (relatively) set in $$D$$. So, $$\f^{-1}(\R^m\setminus E) = \f^{-1}(\R^m)\setminus \f^{-1}(E) = D\setminus \f^{-1}(E)$$ is closed (relatively) set in $$D$$. So, $$\f^{-1}(E)$$ is relatively open in $$D$$ and this is for all $$E$$ open in $$\R^m$$. Thus, $$\f\in \C(D)$$.

• Your proof seems fine. This is a general fact about arbitrary topological spaces: $f:X\to Y$ is continuous iff $f^{-1}(E)$ is closed in $X$ for every closed $E\subset Y$. Apr 21, 2020 at 15:17
• @Reveillark Thank you for the feedback. I was really confused about the converse proof to the problem so this reassures me of that direction. Also, I meant to switch out the $E$'s with $F$. Apologies if that was any confusion there. Apr 21, 2020 at 15:20

As Reveillark noted: "Your proof seems fine. This is a general fact about arbitrary topological space $$f : X \to Y$$ is continuous iff $$f^{-1}(E)$$ is closed in $$X$$ for every closed $$E \subseteq Y$$."