Two questions on topology and continous functions I have two questions:
1.) I have been thinking a while about the fact, that in general the union of closed sets will not be closed, but I could not find a counterexample, does anybody of you have one available?
2.) The other one is, that I thought that one could possibly say that a function f is continuous iff we have $f(\overline{M})=\overline{f(M)}$(In the second part this should mean the closure of $f(M)$). Is this true?
 A: Or to be even lazier, consider that generally singletons are closed (for varying definitions of "generally"). A set is always the union of its points, so you can write $(0,1)$ as $\bigcup_{x \in (0,1)} \{ x \}$.
A: (1) Take the closed sets
$$\left\{\;C_n:=\left[0\,,\,1-\frac1n\right]\;\right\}_{n\in\Bbb N}\implies \bigcup_{n\in\Bbb N}C_n=[0,1)$$
A: (2) is not true. Because it implies that $f$ is a closed map, i.e, maps closed sets to closed sets. However, there are continuous maps which are not closed. For example consider the projection $\pi: \mathbb{R}^2 \rightarrow \mathbb{R}$ given by $(x,y) \mapsto x$. Observe that the set $$C = \{(x,1/x) : x\in \mathbb{R} - \{0\}\}$$ is closed in $\mathbb{R}^2$. However, $\pi(C) = \mathbb{R} - \{0\}$ is not closed in $\mathbb{R}$. Yet $\pi$ is definitely continuous.
The correct equivalence for continuity given by the closure operator is $f(\overline{M}) \subseteq \overline{f(M)}$ for every subset $M$ of $X$.
Edit: Here's my argument for why $C$ is closed. Suppose $(a,b) \in \mathbb{R}^2$ lies inside $\overline{C}$. We will show that $(a,b)$ must be in $C$. Since $\mathbb{R}^2$ is first countable, there exists a sequence $y_n$ in $C$ converging to $(a,b)$. By the definition of $C$, we can write $y_n = (x_n,1/x_n)$. Therefore $x_n$ converges to $a$ and $1/x_n$ converges to $b$. Now note that if $x_n$ were to converge to $0$, $1/x_n$ would diverge. Hence we conclude that $a \neq 0$. Thus $1/x_n$ converges to $1/a$. Since we are in a Hausdorff space, the sequence $1/x_n$ cannot converge to two different points. Thus $b = 1/a$ and hence $(a,b) = (a,1/a) \in C$.
A: A function $f:X \to Y$ between topological spaces sends closed sets to closed sets (what I call $f$ is closed) iff 
$$ \forall A \subset X:  \overline{f[A]} \subset f[\overline{A}]$$
If $f$ is closed and $A \subset X$, then $\overline{A}$ is closed, so $f[\overline{A}]$ is also closed. As $A \subset \overline{A}$, $f[A] \subset f[\overline{A}]$ and so also $\overline{f[A]} \subset \overline{f[\overline{A}]} = f[\overline{A}]$ as the latter set is closed.
On the other hand, if $f$ satisfies the closure property, and $C \subset X$ is closed, then $$ f[C] \subset \overline{f[C]} \subset f[\overline{C}] = f[C]$$ 
as $C$ is closed. It follows that $f[C]$ equals its closure, hence is closed. So $f$ is a closed map.
Sort of dually: $f$ is continuous iff
$$\forall A \subset X : f[\overline{A}] \subset \overline{f[A]}$$
So the other inclusion then holds.
If $f$ is continuous, and $A \subset X$, $$A \subset f^{-1}[f[A]] \subset f^{-1}[\overline{f[A]}]$$
and the latter set is closed by continuity of $f$ (inverse images of closed sets are closed).
So as it contains $A$, it also contains $\overline{A}$, which is the smallest closed set containing $A$. So 
$$ \overline{A} \subset f^{-1}[\overline{f[A]}]$$
which implies $f[\overline{A}] \subset \overline{f[A]}$. On the other hand, if $f$ satisfies the second closure property, let $C \subset Y$ be closed. Then taking $A = f^{-1}[C]$ then 
$$f[\overline{f^{-1}[C]}] \subset \overline{f[f^{-1}[C]]} \subset \overline{C} = C$$
This implies (by definition of $f^{-1}$) that $\overline{f^{-1}[C]} \subset f^{-1}[C]$ which means that $f^{-1}[C]$ is closed. So $f$ is continuous, as inverse images of closed sets are closed.
So the equality $\overline{f[A]} = f[\overline{A}]$ for all subsets $A$ of $X$ is exactly saying that $f$ is both closed and continuous. 
