# Prove that $\overline{f(A)}\subseteq f(\overline{A})$ where $f: X \rightarrow Y$ is continuous, $X$ is compact and $A \subseteq X$

Suppose that $X$ and $Y$ are topological spaces, $f: X \rightarrow Y$ is a continuous map and $A \subseteq X$. It's not very hard to prove that $f(\overline{A})\subseteq \overline{f(A)}$, where $\overline{A}$ denotes the closure of $A$.

Now assume that $X$ is compact. I'm trying to prove the inclusion $\overline{f(A)}\subseteq f(\overline{A})$.

We know that $\overline{A}$ is compact so $f(\overline{A})$ is compact too. If $Y$ were a Hausdorff space, $f(\overline{A})$ would be close and $\overline{f(\overline{A})}=f(\overline{A})$, so $\overline{f(A)}\subseteq \overline{f(\overline{A})}=f(\overline{A})$, as required.

Nevertheless, I don't have the hypothesis that $Y$ is Hausdorff so I'm struggling either trying to prove the inclusion or to find out a counterexample.

Thank you

• $X$ is compact, then $\bar{A}$ is compact. Then $f(\bar{A})$ is compact. You are correct. Probably is missing that $f$ is surjective, or $Y$ Hausdorff....... – L.F. Cavenaghi Jun 22 '16 at 18:01

The claim can be made false without the assumption that $Y$ is Hausdorff. To see this, let $Y\equiv\{1,2\}$ be endowed with the indiscrete topology (that is, the only open sets are the empty set and the whole set). Let $X\equiv\{1,2\}$ be endowed with the discrete topology (all subsets are open) and $f(1)\equiv 1$ and $f(2)\equiv 2$. Clearly, $f$ is continuous. But if $A\equiv\{1\}$ (a finite set is compact in any topological space), then $$\overline{f(A)}=\overline{\{1\}}=\{1,2\},$$ yet $$f(\overline{A})=f(\{1\})=\{1\}.$$

• You beat me by one minute. +1 – PhoemueX Jun 22 '16 at 18:07
• Thank you so much, nice counterexample. – Math35 Jun 22 '16 at 19:42
• @Math35 My pleasure! – triple_sec Jun 22 '16 at 19:51

This is false in general if $Y$ is not Hausdorff.

Take any map $f : X \to X$, where on the left, $X$ has the discrete topology and on the right the indiscreet topology (only $\emptyset$ and $X$ are open), where $X$ is any finite set.

Finally, let $A=\{x\}$. Is is not hard to see that this is a counterexample (if $X$ has more than one element).

• Thanks! Good generalization of triple_sec 's answer. – Math35 Jun 22 '16 at 19:45