Let $f: X \to Y$ be a continuous function, where $B\subset Y$.
Then, $\overline {f^{-1}(B)} \subset f^{-1}(\bar B) $ holds, here's the proof.
I am looking for an example to illustrate that the above containment is proper in general metric spaces.
In $\mathbb R$, I could not think of an example.
Kindly help. Thanks in advance.
Ok, what would an example have to look like? Say $x\in f^{-1}(\overline{B})\setminus\overline{f^{-1}(B)}$. We know every neighbourhood $U$ of $f(x)$ intersects $B$, pulling back to a neighbourhood $f^{-1}(U)$ of $x$ intersecting $f^{-1}(B)$. As $x$ fails to be in the closure, there must be a neighbourhood $V$ of $x$ which does not contain $f^{-1}(U)$ for any neighbourhood $U$ of $x$; that is to say, $f(V)$ does not contain any neighbourhood of $f(x)$.
So we need $x$ to have neighbourhoods whose $f$-images are hollow (or at least hollow at $x$). The easiest way to arrange this is to make $f$ collapse some neighbourhood of $x$ to a point, to have $f$ constant (to a point of $\overline{B}$) on some neighbourhood $V$ of $x$ which avoids $f^{-1}(B)$. However we can't really do that and have $f$ continuous if our space has good connectivity properties, so my natural instinct was to look at the rationals - which are totally separated.
Indeed, let $f:\Bbb Q\to\Bbb R$ map $x\mapsto\begin{cases}\pi&x>\pi\\x&x<\pi\end{cases}$. $f$ is obviously continuous, and with $B:=(-\infty,\pi)\subset\Bbb R$ we find a counterexample. The entirety of $(\pi,\infty)\cap\Bbb Q$ lies in $f^{-1}(\overline{B})\setminus\overline{f^{-1}(B)}$; $f^{-1}(\overline{B})=\Bbb Q$ but $f^{-1}(B)=(-\infty,\pi)\cap\Bbb Q$ is closed (and properly contained) in $\Bbb Q$.