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Prove that $f$ is closed if and only if $f(\text{cl}(A)) \supseteq \text{cl}(f(A)).$ I can show the forward directions by saying:
Suppose $f$ is closed. Then, $f(\text{cl}(A)) = f(A) \cup f(\partial A)$ which is closed implying the inclusion of the smallest closed set containing $f(A)$ in $f(\text{cl}(A)).$

But I am struggling going in the other direction. I want to say take a closed set $B \subset X.$ $B = \text{cl}(A)$ for some open $A \subset X.$ But then I am stuck.

Thanks in advance

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For the forward direction there’s no need to deal with the boundary of $A$: just observe that if $f$ is closed, then $f[\operatorname{cl}A]$ is a closed set containing $f[A]$ and must therefore contain $\operatorname{cl}f[A]$, since that’s the intersection of all closed sets containing $f[A]$.

For the other direction it’s probably easiest to suppose that $f$ is not closed and try to find a set $A\subseteq X$ such that $f[\operatorname{cl}A]\nsupseteq\operatorname{cl}f[A]$. If $f$ is not closed, there is a closed set $F\subseteq X$ such that $f[F]$ is not closed. Clearly $f[\operatorname{cl}F]=f[F]$, but $f[F]$ is not closed, so ... ?

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  • $\begingroup$ on an intuitive level, it seems this theorem is saying closed maps expand the boundaries of a set. Am i correct in this intuition? $\endgroup$ – masszz Sep 29 '13 at 22:34
  • $\begingroup$ @sergey: In general I don’t find it useful to think in terms of boundaries at all. If anything it says that a closed map shrinks the closure (and hence the boundary). $\endgroup$ – Brian M. Scott Sep 29 '13 at 22:41
  • $\begingroup$ Another way to do the other direction: Let $A$ be closed. $\operatorname{cl} A = A$. Thus $f(A) \supset \operatorname{cl}(f(A))$. Hence $f(A)$ is closed. $\endgroup$ – Ayman Hourieh Sep 29 '13 at 22:56
  • $\begingroup$ @BrianM.Scott Brian doesn't it say that f takes a closed set and maps it something larger than the closure of the interior, hence EXPANDING the original closed set? $\endgroup$ – masszz Sep 30 '13 at 0:56
  • $\begingroup$ @sergey: It says that the closure of the image of $A$ may be smaller than the image of the closure of $A$, which I see as a kind of shrinkage. $\endgroup$ – Brian M. Scott Sep 30 '13 at 1:01

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