Continuity via Closure Operator I'm having a rather simple question:
Lets say a function preserves neighborhoods iff: $N\in\mathcal{N}_x \Rightarrow f^{-1}(N)\in\mathcal{M}_x$
and a function preserves closeness iff: $x\parallel A \Rightarrow f(x)\parallel f(A)$
I want to show that, in fact, these are equivalent.
So far the second property is the same as: $f(\overline{A})\subseteq \overline{f(A)}$
I already passed a proof that a function preserves neighborhoods iff it is continuous in the usual sense: $V\in\mathcal{T} \Rightarrow f^{-1}(V)\in\mathcal{S}$
...while $x\parallel A$ is meant to mean $x\in\overline{A}$.
 A: For the $\implies$ direction, I suggest a proof by contradiction. Assume that $x\in\bar A$ but $f(x)\notin\overline{f(A)}$. Then there is a neighborhood $V$ of $f(x)$ disjoint from $f(A)$. This is equivalent to $f^{-1}(V)\cap A=\emptyset$, contradicting $x\in\overline A$, as $f^{-1}(V)$ is a neighborhood by hypothesis.
For the other direction, assume that $V$ is a neighborhood of $f(x)$, but no neighborhood of $x$ has an image in $V$. Then $x$ is in the closure of $f^{-1}(Y\setminus V)$, thus $f(x)$ is in the closure of $f(f^{-1}(Y\setminus V))$ by hypothesis, hence it is also in the closure of the larger set __ ... (I'll leave it to you to finish the argument.)
This actually proves a stronger statement, as it works for a fixed point $x$ while $A$ ranges over the subsets of $X$. So it shows that continuity at a point $x$ is equivalent to the implication
"$x\in\overline A\implies f(x)\in\overline{f(A)}$ for all $A\subset X$"
A: ...ok, sry, I got it by myself ...thanks, Stefan, anyway! #thumbs-up# =)
The assertion is equivalent to:
$\overline{A}\subseteq f^{-1}(\overline{f(A)})$
So, the assertion follows from:
$\overline{A}\subseteq\overline{f^{-1}(f(A))}\subseteq\overline{f^{-1}(\overline{f(A)})}=f^{-1}(\overline{f(A)})$


*

*Inclusion: $A\subseteq f^{-1}(f(A)) \Rightarrow \overline{A}\subseteq\overline{f^{-1}(f(A))}$

*Inclusion: $f(A)\subseteq\overline{f(A)} \Rightarrow f^{-1}(f(A))\subseteq f^{-1}(\overline{f(A)}) \Rightarrow \overline{f^{-1}(f(A))}\subseteq \overline{f^{-1}(\overline{f(A)})}$

*Equality: $\overline{f(A)} \text{ closed} \Rightarrow f^{-1}(\overline{f(A)}) \text{ closed} \Rightarrow \overline{f^{-1}(\overline{f(A)})}=f^{-1}(\overline{f(A)})$


The converse assertion is equivalent to:
$\overline{B}=B \Rightarrow \overline{f^{-1}(B)}=f^{-1}(B)$
So, the converse assertion follows from:
$f^{-1}(B)\subseteq\overline{f^{-1}(B)}\subseteq f^{-1}(f(\overline{f^{-1}(B)}))\subseteq f^{-1}(\overline{f(f^{-1}(B))}) \subseteq f^{-1}(\overline{B}) =f^{-1}(B)$
That gives:
$f^{-1}(B)=\overline{f^{-1}(B)}$


*

*Inclusion: $A\subseteq \overline{A} \text{ in general}$

*Inclusion: $A\subseteq f^{-1}(f(A)) \text{ in general}$

*Inclusion: $f(\overline{A})\subseteq \overline{f(A)} \text{ by assumption}$

*Inclusion: $f(f^{-1}(B))\subseteq B \text{ in general} \Rightarrow \overline{f(f^{-1}(B))}\subseteq \overline{B} \Rightarrow f^{-1}(\overline{f(f^{-1}(B))})\subseteq f^{-1}(\overline{B})$

*Equality: $\overline{B}=B \Rightarrow f^{-1}(\overline{B})=f^{-1}(B)$

A: So I misunderstood the question, here is an update:
if $f$ is continuous then $f^{-1}(X\setminus \overline{f(A)})$ is open. So its complement is closed and contains $A$. Thus it contains $\overline{A}$. So
$f(\overline{A}) \subset \overline{f(A)}$. 
Other direction: Let $U$ be an open set and $V=f^{-1}(U)$. We want to show that $V$ is open.  Take $B=X \setminus U, A=f^{-1}(B)$. Then $\overline{f(A)} \subset B$ so $f(\overline{A}) \cap U= \emptyset$ . Thus $\overline{A}=A$ and $V=X \setminus A$ is open.
Previous version:
The first property is, indeed, continuity, but the second is not.
Functions, that satisfy this property are called closed. For example, map the open unit interval to the circle by ``folding'' the ends together. Then the image of the interval itself (it is closed in induced topology) is not closed in the topology of circle, however the map is continuous.
See: http://en.wikipedia.org/wiki/Open_and_closed_maps 
