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 n^{th} "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".