for any $A\subseteq X$, $f(\overline{A})\subseteq\overline{f(A)}$ , if and only if $f: X \to Y$ is continuous.

Let $X$ and $Y$ be two topological spaces.

Prove that for any $A\subseteq X$, $f(\overline{A})\subseteq\overline{f(A)}$ , if and only if $f: X \to Y$ is continuous.

I am stuck on the converse. Suppose $f: X \to Y$ is continuous. Then for every closed set C in Y, $f^{-1}(C)$ is closed in X.

WTS for any $A\subseteq X$, $f(\overline{A})\subseteq\overline{f(A)}$.

Since $\overline{f(A)}$ is a closed set in Y, $f^{-1} \circ \overline{f(A)}$ = $\overline{ f^{-1} \circ \overline{f(A)} }$

Also, $f(\overline{A}) \subseteq \overline{ f(\overline{A})}$

• how would you show a map is continuous? Inverse image of open (closed) is open(closed)... what should be the good choice here? closed set or open set? – user87543 Dec 6 '13 at 2:49
• closed set? For the converse, I was able to show that for any element in X, and any open set V in Y containing f(p), there is an open set U in X containing p, such that f(U) is contained in V. Should I try to show this, or that the image of a closed set is closed? – sarah Dec 6 '13 at 2:53
• Does image of closed set is closed implies continuity?? I guess you mean Inverse image of closed set is closed implies continuity... As you have guessed, it would be easy to take a closed set.. and show its inverse is closed.... how do you prove some set is closed?? Does $\bar{A}\subset A$ imply $A$ is closed? – user87543 Dec 6 '13 at 2:55
• yes since a set that equals is closure is always closed. – sarah Dec 6 '13 at 2:56
• yes.. so, now the question would be to prove $\bar{f^{-1}(A)}\subset f^{-1}(A)$ for closed subset $A$.... Can you see why this is true using given condition $f(\bar{A})\subset \bar{f(A)}$? – user87543 Dec 6 '13 at 2:58

Suppose than for any $A\subseteq X$, we have $f(\overline A)\subseteq \overline{f(A)}$. Pick a closed set $F\subseteq Y$. We want to show $f^{-1}(F)$ is closed. Using the above, can you show that $\overline {f^{-1}(F)}\subseteq f^{-1}(F)$ must hold true? You need to use $F$ is closed.
ADD (Spoiler) We use $F=\overline F$ and $ff^{-1}(F)\subseteq F$. Then using $f(\overline A)\subseteq \overline{f(A)}$ with $A=f^{-1}(F)$, we get $f\left(\overline{f^{-1}(F)}\right)\subseteq \overline{ff^{-1}(F)}\subseteq \overline F=F$. Thus $\overline{f^{-1}(F)}\subseteq f^{-1}f\left(\overline{f^{-1}(F)}\right)\subseteq f^{-1}(F)$, so $f^{-1}(F)$ is closed and $f$ is continuous.