TITLE: $f\colon M\to N$ continuous iff $f(\overline{X})\subset\overline{f(X)}$
QUESTION [8 upvotes]: Possible Duplicate:
Continuity and Closure 

$f\colon M\to N$ is continuous iff for all $X\subset M$ we have that $f\left(\overline{X}\right)\subset\overline{f(X)}$.
I only proved $\implies$.
If $f$ is continuous then for any $X\subset M$,
$$X\subset f^{-1}[f(X)]\subset f^{-1}\left[\overline{f(X)}\right]=\overline{f^{-1}\left[\overline{f(X)}\right]}$$
therefore
$$\overline{X}\subset f^{-1}\left[\overline{f(X)}\right]\implies f\left(\overline{X}\right)\subset \overline{f(X)}.$$
The other side must be the same idea but I don't know why I can't prove it.
Added: With exactly same idea when I proved $\implies$ I did proved $\Longleftarrow$,
Let $F\subset N$ any closed set then:
$$f\left[f^{-1}(F)\right]\subset f\left[ \overline{f^{-1}(F)}\right]\subset \overline{f\left[ f^{-1}(F)\right]}\subset \overline{F}=F$$
in particular
$$f\left[ \overline{f^{-1}(F)}\right]\subset F\implies f^{-1}(F)\supset\overline{f^{-1}(F)}$$
then $f^{-1}(F)=\overline{f^{-1}(F)}$ and $f$ is continuous.

REPLY [4 votes]: First, a general fact about maps of sets:
$$X \subseteq f^{-1} Y \iff f X \subseteq Y$$
Now, suppose for all $X$ in $M$ we have $f \overline{X} \subseteq \overline{f X}$. Let $Y$ be a closed subset of $N$ and let $X = f^{-1} Y$. A map $f : M \to N$ is continuous if and only if the preimage of every closed set is closed, so we need to show $X$ is closed. Clearly, $f X \subseteq Y$ and $X = f^{-1} f X$. Consider $\overline{X}$: we have $X \subseteq \overline{X}$, so $f X \subseteq f \overline{X} \subseteq \overline{f X} \subseteq Y$, hence $\overline{X} \subseteq f^{-1} Y = X$, i.e. $\overline{X} = X$.

REPLY [2 votes]: Let $V\subset N$ be an open set around $f(x)$. Then its complement $V^c$ is closed. Let $U=cl({f^{-1}(V^c)})^c$. Then it is an open set. Because of the property of the function:
$$
f(cl({f^{-1}(V^c)}))\subset cl(f({f^{-1}(V^c)}) \subset V^c
$$
we see that $x\in U$, and that $f(U)\subset V$. Then f is continuous.