TITLE: Must a map on a compact space be surjective on $\cap_{n=1}^\infty f^n(X)$?
QUESTION [8 upvotes]: Maybe I'm just being a bit dense here, but this has me stumped right now.
A fairly well-know thm is the following:  Let $X_0$ be a compact metric space and $f:X_0\to X_0$ be continuous.  For each $n\ge1$, let $X_n=f(X_{n-1})$, i.e., the range of the nth "iterate" of $f$.  Since $X_0,X_1,\dots$ is a decreasing sequence of compact sets, $X_\infty=\cap_{n=1}^\infty X_n$ is a non-empty compact set.  It is easily seen that $f$ maps $X_\infty$ into itself. But, in fact, $f(X_\infty)=X_\infty$.  The only argument that I know for this last fact uses sequential compactness.  Briefly: fix $x\in X_\infty$, choose a sequence such that each $x_n\in X_n$ and each $f(x_n)=x$, and pick a convergent sub-sequence.  So, this theorem ($f(X_\infty)=X_\infty$) actually holds for any compact and sequentially compact topological space.
My question is whether or not sequential compactness is needed.  There may be
a simple argument that I'm missing (as I said at the start).  Or, maybe there
is a counterexample I'm not seeing - perhaps a mapping on $\beta\omega$?
In summary:

QUESTION: If $f$ is a continuous map of a compact (but not sequentially compact) space $X_0$ into itself, and $X_\infty$ is defined as above, must $f$ map $X_\infty$ onto itself?
Thanks in advance for any help or pointers.

-Jeff Norden
PS: by "compact" I really mean "compact and Hausdorff".

REPLY [2 votes]: $\bf{Method\  1:}$
For $x\in X_{\infty}$ consider the projective system 
$$(f^{-n}(x) \overset{f}{\leftarrow} f^{-n-1}(x))_{n\in \mathbb{N}}$$
The projective limit $\lim_{\leftarrow} f^{-n}(x)$
is nonvoid ( $(f^{-n}(x))_{n \in \mathbb{N}}$ nonvoid compacts),  so take an element in it : $(x, x_1, x_2, \ldots )$. We have $x_1 \in X_{\infty}$ and $f(x_1) = x$
$\bf{Method\  2:}$
Let $x \in X_{\infty}$. The sets $f^{-(n+1)}(x)$ are nonvoid, therefore $f^{-1}(x) \cap f^n(X) \ne \emptyset$ and so 
$$\cap_{n\ge 0} (f^{-1}(x) \cap f^n(X)) \ne \emptyset$$ therefore 
$$f^{-1}(x) \cap (\cap_{n\ge 0} f^n(X) ) = f^{-1}(x) \cap X_{\infty}\ne \emptyset$$