Let $$f:X\rightarrow X$$ be a continuous map on the compact metric space $X.$ Show that there is a subset $A\subset X$ such that $f(A)=A$.
Now the given hint is that to consider $$A_1=f(A)\\and\ \ \ A_{n+1}=f(A_n)$$
Now I guess , the case would be the sequence $\{A_n\}$ is decreasing satisfying $A_{n+1}\subset A_n$ $\forall n\in \mathbb N.$ Then obviously they would satisfy Finite Intersection Property and compactness would assure existence of a non-empty intersection , say $A$ . and that would be our answer . But my problem is that I cannot show that $$A_{n+1}\subset A_n$$ holds .
Please help me with that .
Thanks.
$A_1=f(X)$.| $A_2=f(A_1)$ .
Now $$f:X\rightarrow X$$ and we have $$A_1\subset X\\or,\ f(A_1)\subset f(X)=A_1\\i.e.\ A_2\subset A_1\\or,\ \ f(A_2)\subset f(A_1)\\i.e.A_3\subset A_2\\.\\.\\.\\.\\.\\so\ \ \ \ on.$$
Assume that this holds upto the integer $n$. i.e. $A_n\subset A_{n+1}.$
Then we see $$f(A_n)\subset f(A_{n+1}\\i.e. A_{n+1}\subset A_{n+2}$$ And our result is proved for all $n\in \mathbb N$.
Now this decreasing sequence and each is compact being continuous image of compact spaces. And since for any $"n"$ , the set $A_n$ is non-empty we have that $\{A_n\}_n$ satisfies Finite Intersection Property . Now compact means closed in the metric space. In a compact metric space Every family of closed sets satisfying F.I.P. has a non-empty intersection. So $$\cap_{n=1}^{\infty}A_n\neq \Phi$$. Let $$\cap_{n=1}^{\infty}A_n = A$$
Now $$f(\cap_{n=1}^{\infty}A_n)=f(A)\\ \\or,\ \ \cap_{n=1}^{\infty}f(A_n)=f(A)\\ \\or,\ \ \cap_{n=1}^{\infty}A_n=f(A)$$
So $$f(A)=A.$$
Proved.