Closure of continuous image of closure

Solution 1:

Yes. Recall that $f$ is continuous if $f[\overline{S}]\subseteq\overline{f[S]}$ for all $S\subseteq X$. Applying this, we get that $f[\overline{A}]\subseteq\overline{f[A]}$. Since the set on the right is closed we also get $\overline{f[\overline{A}]}\subseteq\overline{f[A]}$. For the other inclusion, simply note that $f[A]\subseteq f[\overline{A}]$ and take closures.

Solution 2:

HINT: $\newcommand{\cl}{\operatorname{cl}}$$A\subseteq\cl A$, so $f[A]\subseteq f[\cl A]$, and therefore $\cl f[A]\subseteq\cl f[\cl A]$ whether $f$ is continuous or not. The real question, therefore, is whether continuity of $f$ implies that $\cl f[\cl A]\subseteq\cl f[A]$. It suffices to show that $f[\cl A]\subseteq\cl f[A]$. (Why?)

To show that $f[\cl A]\subseteq\cl f[A]$, suppose not; then there is an $x\in\cl A$ such that $f(x)\notin\cl f[A]$. Thus, there is an open $V\subseteq Y$ such that $f(x)\in V$ and $V\cap f[A]=\varnothing$. And here’s where you use the continuity of $f$: $V$ is open in $Y$, so $f^{-1}[V]$ is open in $X$. For convenience let $U=f^{-1}[V]$; then $U$ is an open neighborhood of $x$. Can you use $U$ to derive a contradiction with the assumption that $x\in\cl A$?

Solution 3:

Clearly $\overline{f(A)} \subseteq \overline{f(\overline{A})}$. Now suppose $x \in \overline{f(\overline{A})} \backslash \overline{f(A)}$. Then there is an open set $U$ containing $x$ that is disjoint from $f(A)$, but every such $U$ intersects $f(\overline{A})$. Thus there is $y \in \overline{A}$ with $f(y) \in U$. Now $f^{-1}(U) = V$ is an open set containing y but disjoint from $A$, contradicting the fact that $y \in \overline{A}$.