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Let $X$ be a compact space and $f: X \to X$ be a continuous map. Show that there is a non-empty closed set $E \subseteq X$ such that $E = f(E)$.

First, something is a bit fishy about this problem. I think I am missing the condition that this should be Hausdorff.

My original idea is as follows. Since $f$ is continuous, any open $U \in Y$ has $f^{-1}(U)$ is open. So $(f^{-1}(U))^c$ is closed. It seems like it's easy to find a closed set, but I'm not sure how to fulfill the second property.

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  • $\begingroup$ It is a duplicate but it will take me hours to find it if I ever find it. Meanwhile, note that you want to add "nonempty" closed set. Otherwise... $\endgroup$ – Julien Apr 18 '13 at 4:19
  • $\begingroup$ Good call. It's always those small words that make a difference. $\endgroup$ – emka Apr 18 '13 at 4:26
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HINT: Let $K_0=X$, and for $n\in\Bbb N$ let $K_{n+1}=f[K_n]$. Now consider $\bigcap_{n\in\Bbb N}K_n$. (You don’t need $X$ to be Hausdorff to get $E$ to be non-empty, but you do need it to ensure that $E$ is closed. Some people define compact to include Hausdorff; that may be why Hausdorffness isn’t mentioned in the statement of the problem.)

Added: You will need to assume that $X$ is Hausdorff. To see why, let $\tau$ be the cofinite topology on $\Bbb N$; then $\langle\Bbb N,\tau\rangle$ is a compact $T_1$ space, but it’s not Hausdorff. Let $f:\Bbb N\to\Bbb N:n\mapsto n+1$; this map is continuous, but there is no non-empty $E\subseteq\Bbb N$ such that $f[E]=E$, even if we don’t require $E$ to be closed. To see this, let $E$ be a non-empty subset of $\Bbb N$, and let $m=\min E$; then $m\notin f[E]$, so $f[E]\ne E$.

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  • $\begingroup$ It seems like you are leading me down the direction of the finite intersection property. However, how do I guarantee the fact that $K_n$ is closed? $\endgroup$ – emka Apr 18 '13 at 4:28
  • $\begingroup$ @abet: That’s where you need $X$ to be Hausdorff. $\endgroup$ – Brian M. Scott Apr 18 '13 at 4:34
  • $\begingroup$ I left off non-empty when I first wrote out the question. $\endgroup$ – emka Apr 18 '13 at 4:36
  • $\begingroup$ @abet: I assumed that you intended it; and you get it automatically from the approach that I suggested, even if $X$ is not Hausdorff. $\endgroup$ – Brian M. Scott Apr 18 '13 at 4:38
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    $\begingroup$ @abet: No, the theorem in question is that a compact subset of a Hausdorff space is closed. A companion theorem is that a closed subset of a compact space is compact (even if the space is not Hausdorff), but we don’t need that one. And you will need to assume that $X$ is Hausdorff, because the theorem is false otherwise. The cofinite topology on $\Bbb N$ is compact, the map $f:\Bbb N\to\Bbb N:n\mapsto n+1$ is continuous, and the intersection of the sets $K_n$ in this space is empty. $\endgroup$ – Brian M. Scott Apr 18 '13 at 6:17

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