Let $X$ be a set and $f: X \to X$ a function. A point $x \in X$ is, of course, said to be **periodic** for $f$ if $x \in \{f(x), f^2(x), \ldots\}$.


Now suppose that $X$ is a topological space and $f$ is continuous. In private, I've been saying that a point $x \in X$ is **topologically periodic** for $f$ if $x \in \overline{\{f(x), f^2(x), \ldots\}}$.


But "topologically periodic" is just a name I made up.



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> What is the standard name for "topologically periodic" point?
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Note that I'm looking for an adjective that applies to *points*, not *functions*. I'm aware of terms such as quasiperiodic function and almost periodic function, but they don't answer my question.


In case it helps, I'm working in less generality than arbitrary topological spaces and continuous maps. In my situation, $X$ is a compact metric space and $f$ is distance-decreasing ($d(f(x), f(y)) \leq d(x, y)$). Under those hypotheses, my definition of "topologically periodic" may have other equivalent formulations that are more familiar.



What is the name for a point that is periodic to within $\varepsilon$?
Such a point is called **recurrent**.




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It was stated by Alp Uzman in a comment that the usual definition of recurrent is that $x$ is in its own $\omega$-limit set. This is the definition I had in mind, and under a suitable separation axiom it's obvious that this is equivalent to what was asked, which is why I just wrote the statement. Actually you don't need any separation axiom. I'll just write a full proof.


The **$\omega$-limit set** $\omega(x, f)$ of $x$ is the set of accumulation points of the sequence $I\_x = (x, f(x), f^2(x), \ldots)$. (In the reference given by Uzman, read literally the first definition of the $\omega$-limit set given is that it's the accumulation points of the set $\{f^i(x) \;|\; i \in \mathbb{N}\}$. This would not be equivalent, and is "wrong", but it's clear from the discussion that the author means accumulation points of the sequence.)


We need an auxiliary notion. For two points $x, y$, write $y \leq x$ if every open set containing $x$ also contains $y$. We need two lemmas, which are left as exercise.



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> Lemma A. Continuous functions preserve $\leq$ in the sense that $y \leq x \implies f(y) \leq f(x)$.
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> Lemma B. The $\leq$-relation is transitive.
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> Theorem. Let $X$ be a topological space and $f : X \to X$ a continuous function. Then $x$ is in $\omega(x, f)$ if and only if $x \in \overline{\{f^i(x) \;|\; i \geq 1\}}$.
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Proof. Clearly $\omega(x, f) \subset \overline{\{f^i(x) \;|\; i \geq 1\}}$, so it suffices to show $x \in \overline{\{f^i(x) \;|\; i \geq 1\}} \implies x \in \omega(x, f)$. Suppose thus $x \in \overline{\{f^i(x) \;|\; i \geq 1\}}$.


First, suppose that $f^n(x) \leq x$ for some $n \geq 1$. Then by Lemma A, $f^{(k+1)n}(x) \leq f^{kn}(x)$ for all $k$. By Lemma B, $f^{kn}(x) \leq x$ for all $k$. Then clearly $x$ is even an accumulation point of the subsequence $(x, f^k(x), f^{2k}(x), \ldots)$ -- all points of this set are in every neighborhood of $x$.


Next, suppose $f^n(x) \not\leq x$ for all $n \geq 1$. For all $n \geq 1$ let $V\_n$ be a neighborhood of $x$ not containing $f^n(x)$, and $W\_n = \bigcap\_{i = 1}^n V\_i$. Let $U \ni x$ be any neighborhood. By the assumption on $x$, for any $n$ we have $f^m(x) \in U \cap W\_n$ for some $m \geq 1$. Necessarily $m > n$, so in particular we find infinitely many $m$ such that $f^m(x) \in U$. Square.


