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Let $f:X\to X$ be a continuous map and $x\in X$. A point $y\in X$ is called an $\alpha$-limit point of $x$ under $f$ if and only if there is a strictly increasing sequence of positive integer $\{k_n\}_{n=0}^\infty$ and a sequence of points $\{y_n\}_{n=0}^\infty$ such that

1) $f^{k_n}(y_n)= x$

2) $\lim_{n\to \infty} y_n= y$.

Also, a point $y\in X$ is called a non-wandering point, if for every open set $U$ of $y$, there is $n\in\mathbb{N}$ such that $f^n(U)\cap U\neq \emptyset$.

I know that if $f$ is a homeomorphism, then every $\alpha$- limit point of $x$ is a non-wandering point. But I do not know if $f$ is a continuous map, then every $\alpha$- limit point of $x$ is a non-wandering point?

Please help me to know it.

user479859
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  • I've never encountered this way of defining $\alpha$-limit point (maybe it's specific for case when $f$ is not invertible), but shouldn't $k_n$ be decreasing? What part of definition corresponds to "time going backwards"? – Evgeny Sep 22 '18 at 21:45
  • @Evgeny, $f^{k_n}(y_n)=x$ means that $y_n$ is in backwards orbit of $x$, but $f$ is not invertible . – user479859 Sep 23 '18 at 05:22
  • Thank you very much, now I think I got this definition! – Evgeny Sep 23 '18 at 06:42
  • The answer to your question is yes. Hint: Notice that after taking $m>n$ with $f^{k_n}(y_n)=f^{k_m}(y_m)$ you can assume without loss of generality that $y_n$ is such that $f^{m-n}(y_m)=y_n$. – John B Sep 23 '18 at 16:26

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