# Chapter 11, Section 2 Exercise 10 of James Dugundji Topology [duplicate]

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.

• Do you require $Y$ to be Hausdorff? (Re, this.) Apr 4 at 13:08
• @DavidMitra It is not given in the question. Apr 4 at 13:27
• Link of Chapter 11 Exercise 8 Section 2: math.stackexchange.com/q/4014558/861687. Apr 6 at 10:55

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 $$A=\bigcap_n A_n.$$
• 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. Apr 4 at 14:59
• 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, $A=f(A)$. Apr 4 at 14:59