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

Question

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 closed, 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.

Answer 1

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 $$f(\overline{A})=f(\{1\})=\{1\}.$$

Answer 2

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 indiscrete 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).

Answer 3

Let $f : X \to Y$ be a map. Your claim follows from two well-known ingredients:

1. $f$ is a closed map if and only $\overline{f(A)}\subset f(\overline{A})$ for all $A \subset X$.

2. If $X$ is compact and $Y$ is Hausdorff, then $f$ is a closed map.

Obviously we cannot drop the assumption "$Y$ Hausdorff" in 2. As a counterexample take any compact $X \ne \emptyset$ and any map $f : X \to Y$ to a space $Y$ having the trivial topology such that $f(X) \subsetneqq Y$.