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


Question: If [Non wandering Set](https://en.wikipedia.org/wiki/Wandering_set) is whole space then [Recurent set](https://en.wikipedia.org/wiki/Recurrent_point)  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 
$$T^{m\_n}(B(x,\frac{1}{n}))\cap B(x,\frac{1}{n})\neq\emptyset\to \exists y\_n,z\_n\in B(x,\frac{1}{n}) \quad,\quad T^{m\_n}z\_n=y\_n $$ 



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


* $x\in V\_{k+1}\subset V\_k$;
* $T^{n\_k}x\in T^{n\_k}V\_{k+1}\subset V\_k$;
* $d(T^{n\_k}x,x)\le \text{diam}(V\_k)\to 0$.


