7
$\begingroup$

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

$\endgroup$
  • $\begingroup$ $X$ 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....... $\endgroup$ – L.F. Cavenaghi Jun 22 '16 at 18:01
7
$\begingroup$

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

$\endgroup$
  • 1
    $\begingroup$ You beat me by one minute. +1 $\endgroup$ – PhoemueX Jun 22 '16 at 18:07
  • 1
    $\begingroup$ Thank you so much, nice counterexample. $\endgroup$ – Math35 Jun 22 '16 at 19:42
  • $\begingroup$ @Math35 My pleasure! $\endgroup$ – triple_sec Jun 22 '16 at 19:51
5
$\begingroup$

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).

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.