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Let $Y$ be compact and $f: Y\to Y$ continuous. Prove that there exists a non empty closed set $A\subseteq Y$ such that $A=f(A)$.

I saw this problem in James Dugundji topology book. Can anyone give some(tiny) hint to solve this problem?

Edit: I don’t know how to solve this “kind” of problem, showing two sets are equal, $A=f(A)$. First(& only) thing coming to mind is $f(Y)$ is compact, $A$ is compact since $A$ is closed in $Y$, $f(A)$ is also compact.

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Hint : set $A_0=Y$, $A_1=f(A_0)$, ... , $A_n=f(A_{n-1})$ for all $n\in\mathbb{N}$ and then set $$A=\bigcap_n A_n.$$

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  • $\begingroup$ Thank you for the answer. I will try to solve this problem using your hint. $\endgroup$
    – user264745
    Apr 4 at 13:29
  • $\begingroup$ Your hint is absolutely amazing. Thanks @SacAndSac. Your hint gave direction to work on. Without this info, it’s not possible to solve this problem by beginner like me, IMO. First I showed $A_n$ is compact, $\forall n\in \Bbb{N} \cup \{0\}$ by induction. I also assume(inspired by David Mitra comment) $Y$ is Hausdorff to make $A_n$ closed in $Y$ $\forall n$, by theorem 26.3. So $\bigcap A_n$ is closed in $Y$. I also showed $A_n \neq \emptyset$ $\forall n$, by induction. I knew, to show $A=\bigcap A_n \neq \emptyset$ I have to use definition of compactness, theorem 26.9. $\endgroup$
    – user264745
    Apr 4 at 14:59
  • $\begingroup$ But before that I have to show $\{A_n | n\in \{0\}\cup \Bbb{N}\}$ have finite intersection property. I took arbitrary finite subset $\bigcap_{i=1}^{k} A_{n_i}$ and tryed to show it’s non empty. Defined $p=\max \{n_i\}$ and $q=\min \{n_i \}$. After that it gave me headache. Then I saw $A_{n+1} \subseteq A_n$ claim from Dr. Sundar answer. I proved that claim using induction. So $A\neq \emptyset$. By exercise 8 section 2, $A=f(A)$. $\endgroup$
    – user264745
    Apr 4 at 14:59

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