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Let $(M,d)$ and $(N,d')$ be metric spaces and let $f:M \rightarrow N$ be a function.

I want to show that $f$ is continuous and closed $\Longleftrightarrow \overline{f(E)} = f(\overline{E})$ for all $E \subseteq M$.

I have already proved that $f$ is continuous if and only if $f(\overline{E}) \subseteq \overline{f(E)}$ for all $E \subseteq M$, and also that $\overline{f(E)} = f(\overline{E})$ for all $E \subseteq M \Longrightarrow f$ is continuous and closed.

I am only missing $f$ is continuous and closed $\Longrightarrow \overline{f(E)} \subseteq f(\overline{E})$ for all $E \subseteq M$.

I would really appreciate some help. Thanks!

Sebastián
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  • When you say closed, you mean $f$ is a closed map right? – IAmNoOne Oct 06 '16 at 01:37
  • Yes :) I don't mean this in a rude way: is there another meaning? I am just asking to know if there are other definitions with which I can get confused in the future. – Sebastián Oct 06 '16 at 01:44
  • Or did you mean that "$f$ is a closed map" is the correct expression for what I meant by "$f$ is closed"? – Sebastián Oct 06 '16 at 02:21

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As I remember, a closed map is a function which maps closed sets to closed sets. So, you have $f(\bar{E})$ is a closed set.

You already proved that $f(\bar{E}) \subseteq \overline{f(E)} \subseteq \overline{f(\bar{E})}$. But, we know that $\overline{f(\bar{E})}$ is the smallest closed set contain $f(\bar{E})$, then $\overline{f(\bar{E})} \subseteq f(\bar{E})$.

GAVD
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