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 $\Omega_{\alpha}=\Omega(f,\Omega_\beta)$,

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

The proposition is

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

This implies $\Omega_{\beta}=\Omega(f,\Omega_\alpha)$ 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!

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!