MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I saw a statement about the Birkhoff center. Namely let $X$ be a compact metric space and $f:X\to X$ be a homeomorphism on $X$. Then for each ordinal $\alpha$ we define

  1. for $\alpha=0$, let $\Omega_0(f)=X$,

  2. for a successor ordinal $\alpha=\beta'$, let $\Omega_{\alpha}(f)=\Omega(f,\Omega_\beta(f))$,

  3. for a limiting ordinal $\alpha$, let $\Omega_\alpha(f)=\bigcap_{\beta<\alpha}\Omega_\alpha(f)$.

The proposition is

For each homeomorphism $f:X\to X$ on a compact space $X$, there exists a countable ordinal $\alpha$ such that $\Omega_{\alpha'}(f)=\Omega(f,\Omega_\alpha(f))$.

This implies $\Omega_{\beta}(f)=\Omega(f,\Omega_\alpha(f))$ for all $\beta>\alpha$. The least of such $\alpha$ is called the center depth of $(X,f)$ and the corresponding $\Omega_\alpha(f)$ is called the Birkhoff center of $(X,f)$.

I tried several times to find a proof. Have you seen this before? Thanks!

share|cite|improve this question
I don't know what $\Omega(f,{\rm whatever})$ means. – Gerry Myerson Dec 23 '10 at 15:26
I am sorry that I forgot the definition. Thanks Todd for clarifying that. – Pengfei Dec 24 '10 at 5:18
up vote 2 down vote accepted

First, to attempt to answer Gerry's question: I think that if $C$ is a closed set for which $f(C) = C$, then $\Omega(f, C)$ is supposed to be the set of non-wandering points of the restriction of $f$ to $C$, that is

$$\Omega(f, C) = \{x \in C: \forall U \in Nbhd(x) \forall n_0 \exists n \geq n_0 s.t. f^{(n)}(U) \cap U \neq \emptyset\}$$

where $Nbhd(x)$ consists of open neighborhoods of $x$ relative to $C$. I had to look this up myself and it's conceivable that I misunderstood something somewhere.

But I think we only need the fact that in a compact metric space, a strictly decreasing transfinite sequence of closed sets must terminate after countably many steps. That is to say, if $\gamma$ is an ordinal and $F: \gamma \to P(X)$ has the property: for $\alpha < \beta < \gamma$, there is a strict inclusion $F(\beta) \subsetneq F(\alpha)$ of closed subsets, then $\gamma$ is countable.

For this, we just need (1) compact metric spaces are separable, and (2) the conclusion holds for separable spaces. (1): a compact metric space has a countable dense subset (for each $n$, there is a finite cover by open balls of radius $1/n$, and the centers of these balls for $n$ ranging over natural numbers gives the countable dense subset). (2): Each closed subset in a separable space (with given countable dense set $D$) is characterized by the set of elements of $D$ is contains. Since each difference $F(\alpha) - F(\alpha+1)$ has nonempty interior and therefore contains at least one element of $D$, there can be at most countably many $\alpha < \gamma$.

Edit: Andreas Blass pointed out that this suggested proof is flawed, and also the easy fix in his comment below. The argument given for (1) shows that compact metric spaces have a countable base for the topology (second countability). Then replace (2) by "each closed subset in a second-countable space (with given countable base) is characterized by the open sets in the base it is disjoint from", and proceed similarly as above.

share|cite|improve this answer
@Todd: I'm not convinced about item (2) in your answer. The real line is separable, and the rationals are dense in it, but there are nonempty closed subsets of the reals that contain no rationals. Also, $F(\alpha)-F(\alpha+1)$ is open in $F(\alpha)$ but, as far as I can see, not in the whole space. The problems are easily fixable, though. Separable metric spaces are 2nd countable, and, once you fix a countable base for the topology, a closed set is determined by the basic open sets disjoint from it. As $\alpha$ increases, you get more basic open sets disjoint from $F(\alpha)$. – Andreas Blass Dec 23 '10 at 20:28
Whoops, that was dumb. Thanks for catching that, Andreas. – Todd Trimble Dec 23 '10 at 20:46
What's the MO etiquette for fixing this? Edit out the errors with an attribution of credit? Delete the answer? – Todd Trimble Dec 23 '10 at 20:52
Todd, different people use different approaches to editing; my preference is for what you've done. Thanks for clearing up $\Omega$. – Gerry Myerson Dec 23 '10 at 21:20

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.