Assume $f: X \rightarrow X$ is a continuous map where X is a compact metric space. Prove that there exists a non-empty set $A \subset X$ such that $f(A) = A$.
(Hint: Set $F_1 = f(X), F_{n+1} = f(F_n), ... $)
So, using Cantor's intersection theorem: If ($F_n$) is a decreasing sequence of closed sets then the intersection is non-empty. I also know that f is a continuous map so I could use that (topology definition or metric space def).
I'm not actually sure if each $F_n$ is closed, or how to show this. Also, if they are closed what would this non-empty intersection even show? The closest I can come to is that using the sequential continuity property - one could possibly show that $X, f(X), f(f(X)), ... $ converges to $A$ implies that $F_n$ converges to $f(A)$. Again, just seeing what I have really. Any help?
Set $F_1=X$. Then $F_1$ is compact.
Set $F_2=f(F_1)=f(f(X))$. Then $F_2$ is compact because continuous image of a compact set is compact and also $F_2\subset F_1=X$.
$F_3=f(F_2)$ and $F_3= f(F_2)\subset f(F_1)=F_2$
By induction prove that there exists a decreasing sequence ($F_n$) of compact sets. Then $\bigcap_{n} F_n=A\neq \emptyset$.
Then $A=\bigcap_{n+1} F_{n+1}=\bigcap_{n} f(F_n)=f(\bigcap_{n}F_n)=f(A)$.