If the non wandering set is the whole space, then the recurrent set is dense?

Question

I tried to prove following statement and use some techniques but I couldn't get result :

Question: If Non wandering Set is whole space then Recurent set is dense??

when $T:X \to X$ is hemeomorphism on compact metric space

Thanks for any hint

$ \text{My attempt}:$ we want to show that for all $x$ and any neighborhood $U_x$ about $x$ intersect $R(T)$ $$R(T)\cap U_x\neq\emptyset$$

or equivalently aim is to find element $y$ in nbhd which and there exist subseqquence$\{n_i\}$ such that $T^{n_i}(y)\to y$ as $n_i\to \infty$

Since $x$ is in Non wandering set every nbhd comes back in some natural number

$$T^{m}(U_x)\cap U_x\neq\emptyset\to \exists y,z\in U_x \quad,\quad T^mz=y $$

to make sub sequence I assume this nhbhd $B(x,\frac{1}{n})$ that is goes to $x$ as $n \to\infty$ also $$T^{m_n}(B(x,\frac{1}{n}))\cap B(x,\frac{1}{n})\neq\emptyset\to \exists y_n,z_n\in B(x,\frac{1}{n}) \quad,\quad T^{m_n}z_n=y_n $$

Answer 1

In Khanickus's answer, it is not clear if $T^{n_i}v_i=v$ with $v_i\to v$ can guarantee $T^{n_i}v\to v$, since $n_i\to\infty$, and the family $\{T^n:n\ge 1\}$ may not be uniformly continuous. We need a small modification:

1. Given an open set $V_1\subset X$, pick $n_1\ge 1$ such that $T^{-n_1}V_1\cap V_1\neq\emptyset$.

2. Pick a smaller open set $V_2\subset T^{-n_1}V_1\cap V_1$, and then $n_2\ge n_1+1$ such that $T^{-n_2}V_2\cap V_2\neq\emptyset$.

3. Induction: pick a smaller open set $V_{k+1}\subset T^{-n_k}V_k\cap V_k$, and then $n_{k+1}\ge n_k+1$ such that $T^{-n_{k+1}}V_{k+1}\cap V_{k+1}\neq\emptyset$.

Note that $n_k\ge k$. Without loss of generality we assume $\overline{V_{k+1}}\subset V_k$ and $\text{diam} (V_k)\to 0$.

Let $x\in\bigcap_{k\ge 1}\overline{V_{k+1}}=\bigcap_{k\ge 1}V_{k}$. We claim $x$ is recurrent.

This is clear since

• $x\in V_{k+1}\subset V_k$;

• $T^{n_k}x\in T^{n_k}V_{k+1}\subset V_k$;

• $d(T^{n_k}x,x)\le \text{diam}(V_k)\to 0$.

Answer 2

I tried using closed balls and think it helped.

Let $V_0=\overline B_{1/n}(x)\subset U_x$.

Since $NW(T)=X$ we get nested closed sets:

$V_1=T^{n_1}(V_0)\cap V_0$

$V_2=T^{n_2}(V_1)\cap V_1$

...

for $n_1<n_2<...$

Now since $T$ is a homeomorphism we can assume $V_i$ are closed balls $V_i=\overline B_{1/m_i}(x_i)$ so that $\cap_{i=0}^\infty V_i=\{ v\}\subset V_0$ with

$v=T^{n_1}v_0=T^{n_2}v_1=T^{n_3}v_3=...$

for $v_i\in V_i$.

Hence $v_i\to v$ and so $T^{n_k}v\to v$.

Answer 3

When the underlying space is compact, a proof can be found in the book of Furstenberg, Recurrence in Ergodic Theory and Combinatorial Number Theory.

This is a direct consequence of the Baire Category Theorem, hence holds in metric spaces with the Baire property. Define $$ U_{k,n} = \bigcup_{m\geq n}\{x \in X \mid d(x, T^m(x)) < 1/k\} $$ The set of recurrent points is exactly the intersection of the $U_{k,n}$ for all $k,n \geq 1$. If there are no wandering points, the open sets $U_{k,n}$ are dense and the result follows by the Baire Category Theorem.