# continuous function: preimage of closure of a set is closure of preimage

i am struggeling with the following problem: let $f:X\to Y$ be a continuous mapping bet ween two metric spaces.

Does it hold that $f^{-1}(\overline{A})=\overline{f^{-1}(A)}$ and $f^{-1}(int(A))=int(f^{-1}(A))$, where the overline denotes the closure of a set and $int()$ the interior of a set.

One direction is obvious: $f^{-1}(A)\subset f^{-1}(\overline{A})$ and by continuity of $f$ the set$f^{-1}(\overline{A})$ is closed. So we proved $\overline{f^{-1}(A)}\subset f^{-1}(\overline{A}$. Any idea or reference where I can find the proof or an counter example?

## 2 Answers

Let it be that $X$ and $Y$ both have $\mathbb R$ as underlying set, that the topology on $X$ is discrete and the topology on $Y$ is the usual one.

Both topologies are metrizable and $f:X\to Y$ prescribed by $x\mapsto x$ is continuous.

If $A\subset Y$ is not closed then: $$\overline{f^{-1}(A)}^X=f^{-1}(A)=A\varsubsetneq\overline{A}^Y=f^{-1}(\overline{A}^Y)$$

If $B\subset Y$ is not open then: $$f^{-1}\left(\text{int}_Y(B)\right)=\text{int}_YB\varsubsetneq B=f^{-1}(B)=\text{int}_X(f^{-1}\left(B\right))$$

You can, on the other hand, prove that a map $f: X\rightarrow Y$ is closed if and only if $\forall A\subseteq Y$ you have $f^{-1}(\overline{A})\subseteq\overline{f^{-1}(A)}$.

So $f^{-1}(\overline{A})=\overline{f^{-1}(A)}$ if and only if f is continuous and closed.

(you get analogous results for the interior and open maps)