compact metric space and the existence of f(A)=A

Question

Let $X$ be a compact metric space , $f: X\to X$ be a continuous map. Prove that there exists a non-empty subset $A\subset X$ s.t $f(A)=A$.

My attempt:
Set $A_1=f(X), A_n=f(A_{n-1}), n\geq 2$. Then $\{A_n\}$ is a decreasing sequence of compact set. My point is to prove $A=\bigcap\limits_{n=1}^{\infty}A_n$ is the satisfying set.
For $y\in f(\bigcap\limits_{n=1}^{\infty}A_n),\exists x\in\bigcap\limits_{n=1}^{\infty}A_n$ s.t $y=f(x)$.
Since $x\in\bigcap\limits_{n=1}^{\infty}A_n\subset X$, hence $y\in f(X)=A_1$.
Since $x\in\bigcap\limits_{n=1}^{\infty}A_n\subset A_1$, hence $y\in f(A_1)=A_2$.
So, $y\in\bigcap\limits_{n=1}^{\infty}A_n\Rightarrow f(\bigcap\limits_{n=1}^{\infty}A_n)\subset\bigcap\limits_{n=1}^{\infty}A_n $.
Now, let $y\in\bigcap\limits_{n=1}^{\infty}A_n$. I don't know how to show that $y\in f(\bigcap\limits_{n=1}^{\infty}A_n)$. Could anyone give me any hints? Thanks in advance

Answer 1

1. You can simplify your proof of $f(\bigcap\limits_{n=1}^{\infty}A_n) \subset \bigcap\limits_{n=1}^{\infty}A_n$:

$$f(\bigcap\limits_{n=1}^{\infty}A_n) \subset \bigcap\limits_{n=1}^{\infty}f(A_n) =\bigcap\limits_{n=1}^{\infty} A_{n+1} = \bigcap\limits_{n=2}^{\infty} A_n = \bigcap\limits_{n=1}^{\infty} A_n$$ It is a well-known fact of elementary set theory that the first inclusion is true for any function $f : X \to Y$ and any family of subsets $A_\alpha$ of $X$. The last eqaution is true because $A_2 \subset A_1$.

Note that 1. is true for any set $X$ and any function $f :X \to X$; we do not need topology here.

2. $\bigcap\limits_{n=1}^{\infty}A_n \subset f(\bigcap\limits_{n=1}^{\infty}A_n)$:

Let $y \in \bigcap\limits_{n=1}^{\infty}A_n$. Then there exist $x_n \in A_n$ such that $f(x_n) = y$. The sequence $(x_n)$ has a convergent subsequence $(x_{n_k})$ with limit $\xi \in X$. Note that for subsequences we have $n_k \ge k$. Let $k_0 \in \mathbb N$. For $k \ge k_0$ we have $x_{n_k} \in A_{n_k} \subset A_k \subset A_{k_0}$. Since $A_{k_0}$ is compact, thus closed in $X$, we get $$\xi = \lim_{k \to \infty} x_{n_k} \in A_{k_0} \text{ for all } k_0 .$$ Hence $\xi \in \bigcap\limits_{n=1}^{\infty}A_n$. Since $f$ is continuous, we get $$f(\xi) = f(\lim_{k \to \infty} x_{n_k}) = \lim_{k \to \infty} f(x_{n_k}) = \lim_{k \to \infty} y = y .$$ Thus $y \in f(\bigcap\limits_{n=1}^{\infty}A_n)$.

3. $\bigcap\limits_{n=1}^{\infty}A_n \ne \emptyset$:

Pick $x_0 \in X$ and define recursively $x_{n+1} = f(x_n) \in A_n$. As in 2. the sequence $(x_n)$ has a convergent subsequence $(x_{n_k})$ whose limit $\xi$ is contained in $\bigcap\limits_{n=1}^{\infty}A_n$.