A question about a continuous function on a compact metric space suppose that (X,d) is a compact metric space and $f:X\to X$ is a continuous function.
Define $X_1 = f(X), X_2 = f(X_1),...,X_{i+1} = f(X_i),...$ and let $A = \bigcap_{i=1}^\infty X_i$. Is $A \subseteq F(A)$?
 A: Yes, it is true that $A\subseteq f(A)$.
Let $a\in A$, so that $a\in X_i$ for every $i$.  Since $X_i = f^i(X)$, for each $i$ choose an element $x_i$ of $X$ such that $f^i(x_i)=a$.  Now consider the sequence $f(x_2), f^2(x_3), f^3(x_4), \ldots$, and let $x$ be the limit of one of its convergent subsequences (which exist, because $X$ is compact).  Since $f^i(x_{i+1})$ is in $X_n$ for every $i\geq n$, we must have $x\in X_n$ for each $n$ because $X_n$ is closed (being the image of a compact set in a Hausdorff space).  Therefore $x\in A$.  On the other hand, $f(x)$ is the limit of a convergent subsequence of $\{f^{i+1}(x_{i+1})\}$, which is constantly $a$.  Therefore $f(x)=a$ and $x\in A$, so $a\in f(A)$.
On the other hand, it's true without any assumptions on $X$ and $f$ that $f(A)\subseteq A$, so in fact we have $f(A)=A$.
A: Yes. In fact, slightly more general statement is true. 
Let $X, Y$ be metric spaces and $f:X\rightarrow Y$ be continuous. Let $(K_n)$ be a decreasing sequence of non-empty compact subsets of $X$. Then,
$$
f(\cap_{n\in\mathbb{N}}K_n)=\cap_{n\in\mathbb{N}}f(K_n).
$$ 
Proof follows from the nested set theorem as follows.
Let us set 
$$
C=\cap_{n\in\mathbb{N}}K_n.
$$
Evidently.
$$
f(C)\subseteq \cap_{n\in\mathbb{N}}f(K_n).
$$
Conversely, let $y\in \cap_{n\in\mathbb{N}}f(K_n)$. Then, $(f^{-1}({y})\cap K_n)_{n\in\mathbb{N}}$ is  decreasing sequence of non-empty compact sets. Therefore,
$$
\emptyset\neq\cap_{n\in\mathbb{N}}f^{-1}({y})\cap K_n=f^{-1}({y})\cap C,
$$
which proves the result.
