TITLE: If $\overline{A}=\overline{B}$, is it true that $\overline{f(A)}=\overline{f(B)}$?
QUESTION [8 upvotes]: For a continuous function $f:X\longrightarrow Y$ of topological spaces with subsets $A$ and $B$ of $X$ satisfying $\overline{A}=\overline{B}$, is it true that $\overline{f(A)}=\overline{f(B)}$? Intuitively it seems so, but if someone could walk me through how this could be proved, I would be very grateful, regards.

REPLY [6 votes]: You might also observe that

PROPOSITION $f:(X,\mathcal I)\to (Y,\mathcal L)$ is continuous if and only if for each $A\subseteq X$, $f(\bar A)\subseteq \overline{f(A)}$

PROOF Suppose that $f$ is continuous. Given a subset $A$, we have $f(A)\subseteq \overline{f(A)}$, which means $A\subseteq f^{-1}f(A)\subseteq f^{-1}\overline{f(A)}$. This last set is closed, so $\bar A\subseteq  f^{-1}\overline{f(A)}$, which means $f(\bar A)\subseteq \overline{f(A)}$.
Conversely, suppose that $f(\bar A)\subseteq \overline{f(A)}$ for each $A\subseteq X$. Let $F$ be closed in $Y$. Then $f(\overline{f^{-1}(F)})\subseteq \overline{f({f^{-1}(F)})}\subseteq \bar F=F $. Thus $\overline{f^{-1}(F)}\subseteq f^{-1}(F)$. Since the converse always holds, we have $f^{-1}(F)=\overline {f^{-1}(F)}$, so this set is closed and $f$ is continuous.
NOTE From the above proposition, we can see the condition $f(\bar A)\subseteq \overline{f(A)}$ for each $A\subseteq X$ can be taken as the definition of continuity in closure spaces.