Closure operation is not necessarily preserved

Question

If $(X,d)$ and $(Y,d')$ are metric spaces and if $f:X\rightarrow Y$, show that $f$ is continuous if and only if for every $A\subset X$, $f(\overline{A})\subset\overline{f(A)}$, and construct an example to show that the closure operation is not necessarily preserved by a continuous mapping. Moreover, if $f$ is a $1:1$ onto, show that $X$ and $Y$ are homeomorphic is and only if $f(\overline{A})=\overline{f(A)}$ for all $A\subset X$.

Answer 1

Suppose first that $f$ is continuous and let $x\in\overline{A}$, which means that for any open set $f(x)\in O\subset Y$ we have $f^{-1}(O)\cap A\ne\varnothing$ because $x\in f^{-1}(O)$ and $f^{-1}(O)$ is open, but $f^{-1}(O)\cap A\ne\varnothing\Rightarrow O\cap f(A)\ne\varnothing$ indeed if $O\cap f(A)=\varnothing$ then $$\varnothing=f^{-1}(O\cap f(A))=f^{-1}(O)\cap f^{-1}(f(A))\supset f^{-1}(O)\cap A$$ which is a contradiction with our initial assumption, so
$O\cap f(A)\ne\varnothing$ for any open $O$ containing $f(x)$ so $f(x)\in\overline{f(A)}\Rightarrow f(\overline{A})\subset \overline{f(A)}$.
Now suppose that for any $A\in X$, $f(\overline{A})\subset \overline{f(A)}$, Let $U=f^{-1}(O)\in X$ where $O$ is arbitrary open set of $Y$ we claim that $U$ is open. But for this note that $x\in \overline{U^{c}} \Rightarrow x \notin U$, indeed if $x\in U$ when we note that \begin{eqnarray*} &&f(U^{c})=f(X)\cap O^{c}\\ &&f(\overline{U^{c}})\subset\overline{f(U^{c})}=\overline{ f(X)\cap O^{c}}\subset\overline{f(X)}\cap\overline{O^{c}}= \overline{f(X)}\cap O^{c} \end{eqnarray*} which is certainly a contradiction, because $f(x)\in f(\overline{U^{c}})$ but as we assume that $x\in U\Rightarrow x\notin U^{c}\Rightarrow f(x)\notin O^{c}$. So for every open $O$ $f^{-1}(O)$ is open and so $f$ is continuous. Now define, \begin{eqnarray*} f : \mathbb{R} & \rightarrow & \mathbb{R}\\ x &\mapsto & e^{x} \end{eqnarray*} with the usual metric $f(\mathbb{R})=(0,\infty)$ which is not closed. Nos suppose that $f$ is one-to-one and that $f(\overline{A})=\overline{ f(A)}$ for all $A\subset X$ then by what we have just proven $f$ is continuous, furthermore as $f$ is one-to-one we can rewrite $\overline{A}=f^{-1}(\overline{f(A)})$ now call $B=f(A)$ and recall that $f$ is a one-to-one map onto $Y$ so $$\overline{f^{-1}(B)}=f^{-1}(\overline{B})$$ and so $f^{-1}$ is continuous too, and so $f$ is a homeomorphism.
Conversely let $f$ be a homeomorphism. The as $A\subset\overline{A} \Rightarrow f(A)\subset f(\overline{A})$ recall that $f$ is a homeomorphism so $f(\overline{A})$ is closed and so $f(A)\subset f(\overline{A}) \Rightarrow \overline{f(A)}\subset\overline{f(\overline{A})}= f(\overline{A})$, now let $y=f(x)\notin \overline{f(A)}$ then for some $\varepsilon>0$, $S_{\varepsilon}(y)\cap f(A)=\varnothing \Rightarrow f^{-1}(S_{\varepsilon}(y))\cap A=\varnothing \Rightarrow f^{-1}(y)\notin\overline{A}\Rightarrow y\notin f(\overline{A})$ so we have just proven that \begin{eqnarray*} \overline{f(A)}^{c}\subset f(\overline{A})^{c}\Rightarrow \overline{f(A)}\supset f(\overline{A}) \end{eqnarray*}
then $f(\overline{A})=\overline{f(A)}$, which finishes the solution.