Recurrent point of continuous transformation in a compact metric space

Question

Given a metric space $(X,d)$ and a transformation $T:X\rightarrow X$, a point $x\in X$ is said to be recurrent iff it belongs to the closure of its orbit $\{T(x), T^2(x),...\}$: more precisely, there exists an increasing sequence $(n_k)$ of natural numbers with $n_k \rightarrow \infty$ such that $T^{n_k}(x)\rightarrow x$ when $k \rightarrow \infty$.

Show that when $X$ is compact and $T$ is continuous, recurrent points always exist.

Note: It's possible to prove this using the Recurrence Poincaré's theorem, wich involves considering some measure structure in $X$. But I'm asking for a direct proof, without considering such additional structure.

Thanks.

Answer 1

$\newcommand{\cl}{\operatorname{cl}}\newcommand{\orb}{\operatorname{orb}}\newcommand{\corb}{\overline{\operatorname{orb}}}$For $x\in X$ let $\orb(x)=\{T^n(x):n\in\Bbb N\}$, and let $\corb(x)=\cl_X\orb(x)$. A set $K\subseteq X$ is invariant if $T^n[K]\subseteq K$ for all $n\in\Bbb N$. A closed invariant $K\subseteq X$ is minimal if $K$ and $\varnothing$ are the only closed invariant subsets of $K$. It’s not hard to check that $X$ is minimal iff $\corb(x)=X$ for all $x\in X$.

Let $\mathscr{I}=\{K\subseteq X:K\ne\varnothing\text{ is closed and invariant}\}$; $X\in\mathscr{I}$, so $\mathscr{I}\ne\varnothing$. If $\mathscr{C}$ is a chain in the partial order $\langle\mathscr{I},\supseteq\rangle$, $\bigcap\mathscr{C}$ is easily seen to be an upper bound for $\mathscr{C}$ in $\langle\mathscr{I},\supseteq\rangle$. By Zorn’s lemma there is a $\supseteq$-maximal $K\in\mathscr{I}$. If $x\in K$, we must have $\corb(x)=K$, as otherwise $\corb(x)\in\mathscr{I}$ with $K\supsetneqq\corb(x)$, contradicting the $\supseteq$-maximality of $K$. Thus, $K$ is minimal.

Now fix any $x\in K$; $\corb(x)=K$ by the minimality of $K$. If $T^n(x)=x$ for some $n\ge 1$, then certainly $x$ is recurrent. If not, $x\in\corb\left(T(x)\right)$, so there is a strictly increasing sequence $\langle n_k:k\in\Bbb N\rangle$ in $\Bbb Z^+$ such that $\langle T^{n_k}(x):k\in\Bbb N\rangle\to x$, and again $x$ is recurrent.

For those who care about such things, one can prove without Zorn’s lemma that $X$ has a minimal subset, though one does still need countable choice.

Answer 2

Are you sure that this is true?

Let $X$ be the circle, and let $d(x,y)$ be the measure of the smallest angle between $x$ and $y$. Let $f$ be a rotation of the circle by $\alpha$, and take $\alpha \notin \{2 \pi n : n \in \mathbb{Z}\}$, so that the rotation is non-trivial. $f$ is continuous. But none of the orbits will be convergent, so the won't have limit points.

Did you mean to say that a point is recurrent if it is a limit point of some subsequence of its orbit?