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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".

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    $\begingroup$ I really think this question is reasonable for MO. $\endgroup$ – Todd Trimble Oct 8 '14 at 2:17
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    $\begingroup$ @ToddTrimble: I just cast the fourth vote. So you can go ahead and cast the fifth if you want. $\endgroup$ – Willie Wong Oct 8 '14 at 11:25
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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.

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  • $\begingroup$ Thanks for the quick response! My initial feeling (that I was just being a bit dense) was obviously correct. $\endgroup$ – Jeff Norden Oct 7 '14 at 21:08
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$\bf{Method\ 1:}$

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 $$f^{-1}(x) \cap (\cap_{n\ge 0} f^n(X) ) = f^{-1}(x) \cap X_{\infty}\ne \emptyset$$

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