# Poincare recurrence theorem and convergence on compact metric spaces

I am looking for a proof of the following theorem:

Let $X$ be a compact metric space with metric $d$, endow $X$ with the Borel $\sigma$-algebra and a probability measure $\mu$. Let $T\colon X\to X$ be a continuous map.

EDIT (after martini's answer): $\mu$ has to be $T$-invariant.

Then for $\mu$-almost every $x\in X$ there is a sequence $n_k\to\infty$ in $\mathbb{N}$ such that $T^{n_k}x\to x$ as $k\to\infty$.

I tried already for some time and looked for references, unfortunately unsuccessful. It seems that one needs to find the right formulation to be able to use the Poincare recurrence theorem. Any ideas?

• How is this not simply Poincare's recurrence theorem? For $\mu$-a.a. $x$ you can find $n_k$ with $d(x,T^{n_k}x) < 1/k$. Now because there are just countably many conditions, for $\mu$-a.a. for all $k$ you can find suitable $n_k$. – Jakub Konieczny Nov 23 '13 at 22:07
• @Feanor I believe there is a small difference with this and the theorem. In the theorem we find that the orbit of almost any point of a measurable set returns to the set (infinitely many times), but now we are trying to show a 'return' to the original seed. – Alp Uzman Mar 9 '15 at 17:17
• On second thought, applying the theorem to the ball centered at $x$ might be sufficient.. But why would we need the compactness of $X$? – Alp Uzman Mar 9 '15 at 17:28

This is wrong. Let $X = \{x,y\}$ with $x\ne y$ and the discrete metric, let $\mu = \delta_x$ and $T$ the constant map to $y$. Then the only elment $z$ for which $(T^n z) = (z, y,y,y,\ldots)$ has a subsequence converging to $z$ is $z = y$, but $\delta_x(\{y\}) = 0 \ne 1$.
• Uh, I am sorry, I forgot the condition that $\mu$ has to be $T$-invariant. – Maik Köster Oct 15 '13 at 14:28