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".



Must a map on a compact space be surjective on $\cap_{n=1}^\infty f^n(X)$?
Don't pick a convergent subsequence: pick a point in $ \cap\_{n \in \mathbb{N}} \overline{\{ x\_n,x\_{n+1}, \dots \}}$ (the overline meaning closure) which is non empty as an intersection of decreasing family non empty compact set, and your argument will work.


