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.