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?

(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 $(a,b) = (a,1/a) \in C$.
