example of a continuous function that is closed but not open Give an example of a continuous function $f:\mathbb{R} \to \mathbb{R}$ that is closed but not open.
$ f(x)=x^2$ is continuous and not open but It's not closed. What is an example?
Thanks in advance.
 A: Consider the constant function $f(x)\equiv 0$, which is clearly continuous. Then for any set $E\subset \mathbb R$, either $f(E)=\varnothing$ or $f(E)=\{0\}$, each of which is a closed set. Hence $f$ is closed.
To show that $f$ is not open, just observe that $\mathbb R$ is open but $f(\mathbb R)=\{0\}$ is not open.
A: As several people have noted, there are much simpler examples, but in fact the squaring map $f:\Bbb R\to\Bbb R:x\mapsto x^2$ is closed. 
Let $F$ be a closed set in $\Bbb R$. Let $F_0=F\cap[0,\to)$; $F_0$ is closed in $[0,\to)$, and $f\upharpoonright[0,\to)$ is a homeomorphism of $[0,\to)$ onto itself, so $f[F_0]$ is a closed set in $[0,\to)$ and hence in $\Bbb R$. Similarly, if $F_1=F\cap(\leftarrow,0]$, then $f[F_1]$ is closed in $\Bbb R$, because $f\upharpoonright(\leftarrow,0]$ is a homeomorphism of $(\leftarrow,0]$ onto $[0,\to)$. But then $f[F]=f[F_0]\cup f[F_1]$ is closed in $\Bbb R$, and hence $f$ is a closed map.
(In case some of the notation is unfamiliar, $[0,\to)$ is another notation for $[0,\infty)$, and $f\upharpoonright A$ is the restriction of the function $f$ to the set $A$, also sometimes written $f|A$ or $f|_A$.)
