Closure and continuous map

Question

Let $S_i$ be a countable family of sets and $f$ a continuous map (say everything is in the complex plane). Is it true that $$ f( \overline{ \cup_i S_i } ) = \overline{ \cup_i f( S_i ) } $$ or perhaps the inclusion $\subseteq$?

Answer 1

The inclusion holds. You have$$f\left(\overline{\bigcup_iU_i}\right)\subset\overline{f\left(\bigcup_iU_i\right)}$$and this has nothing to do with unions. It's just that, if $f$ is continuous and $A\subset D_f$, $f\left(\overline A\right)\subset\overline{f(A)}$.

However, the equality doesn't hold in general. Just take, say, $f(z)=\frac1{1+|z|^2}$. Then $f(\mathbb{C})=(0,1]$, which is not closed. But$$f(\mathbb{C})=f\left(\bigcup_{z\in\mathbb C}\overline{\{z\}}\right)=(0,1]$$whereas $\overline{\bigcup_{z\in\mathbb{C}}f\bigl(\{z\}\bigr)}$ is obviously a closed set (it turns out that it is $[0,1]$).

Answer 2

The identity is not even true for one set, because some continuous functions simply do not send closed sets to closed sets. Take, for instance, $f(z)=i\Im z+\arctan\Re z$, which sends $\Bbb C$ to $\left\{z\in\Bbb C\,:\, \lvert\Re z\rvert<\frac\pi2\right\}$. The $\subseteq $ is true, because $\overline{\bigcup_i f[S_i]}$ is a closed set containing $f\left[\bigcup_i S_i\right]=\bigcup_i f[S_i]$, and so by continuity $f^{-1}\left[\overline{\bigcup_i f[S_i]}\right]$ must be a closed set. Since $f^{-1}[A]\subseteq f^{-1}[B]$ whenever $A\subseteq B$, it must also contain $f^{-1}\left[f\left[\bigcup_i S_i\right]\right]=\bigcup_i S_i$. Finally, since $f^{-1}\left[\overline{\bigcup_i f[S_i]}\right]\supseteq \bigcup_i S_i$, $$\bigcup_i f[S_i]=f\left[\bigcup_i S_i\right]\subseteq f\left[f^{-1}\left[\overline{\bigcup_i f[S_i]}\right]\right]=\operatorname{im}f\cap\overline{\bigcup_i f[S_i]}\subseteq \overline{\bigcup_i f[S_i]}$$