I'm struggling to prove or disprove that the continuity of $f$ implies $f^{-1}(\bar A)\subset\overline{f^{-1}(A)}$.
$f:X\to Y$ is a map between metric spaces $(X,d),(Y,d')$ while $\bar M$ denotes the closure of $M$.
The definition of continuity I'm supposed to use for this exercise is:
$f$ is continuous$\ \Leftrightarrow\ f^{-1}(M)$ is open if $M\subset Y$ is open$\ \Leftrightarrow\ f^{-1}(M)$ is closed if $M\subset Y$ is closed.
Best Answer
Let $Y = \mathbb{R}$ with usual metric and $X = (0,1) \cup \{2\}$ with metric inherited from the standard metric on $\mathbb{R}$. let $f(x) = x$ on $(0,1)$ and $f(2) = 1$ and take $A = (0,1)$. it is then easy to see that for this particular case your statement doesn't hold