$f\colon M\to N$ continuous iff $f(\overline{X})\subset\overline{f(X)}$ 
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.
 A: Suppose that $f$ is not continuous. Then there is a closed set $C\subseteq N$ such that $f^{-1}[C]$ is not closed in $M$. Let $X=f^{-1}[C]$. Since $X$ is not closed, there is some $p\in\operatorname{cl}X\setminus X$. Then $f(p)\in f[\operatorname{cl}X]$, but $f(p)\notin C\supseteq\operatorname{cl}f[X]$, so $f[\operatorname{cl}X]\nsubseteq\operatorname{cl}f[X]$.
A: 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. $\overline{X} = X$.
A: 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.
A: I think that I have the answer. 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.
A: Suppose $f$ is not continuous. Then for some $U$ open in $N,$ $f^{-1} (U)$ contains no neighborhood $N_x$ about some point $x\in f^{-1} (U).$ In other words, $N_x \cap M / f^{-1} (U) $ is nonempty for each $N_x,$ so that $x \in \overline{M / f^{-1} (U)}$ - thus $f(x) \in f(\overline{M / f^{-1} (U)} ).$ However, $\overline{ f( M / f^{-1} (U) )} \subset \overline{N / U} = N/U,$ and $x\notin N/U.$ Thus $f(\overline{M / f^{-1} (U)} )$ is not contained in $\overline{ f( M / f^{-1} (U) )}.$ 
