The orbit closure of a periodic point in a dynamical system is the same as the orbit itself

dynamical systemsgeneral-topologytopological-groups

Suppose that $X$ is a topological space and consider $\mathbb R$ as an addictive topological group. Let $\mathbb R$ act continuously on $X$ on the right. We call the pair $(X,\mathbb R)$ a dynamical system.

By definition, the orbit of a point $x\in X$ is the set of all $x.r$ as $r$ ranges over $\mathbb R$. $x$ is called a periodic point if there is nonzero $r$ with $x.r=x$, provided that $x$ is not a “fixed” point. The smallest such $r$ is called the period of $x$.

It is easy to see that if $x$ is a periodic point, then its orbit is the set $O:=\{x.r: r\in[0,P)\}$, where $P$ is the period of $x$. I want to show that $O$ is a closed set. How can we show that? Intuitively it is a circle.

Best Answer

With your notations, its orbit is also $O=\lbrace x.r : r \in [0,P] \rbrace$. So it is the continuous image of the compact $[0,P]$ by the application $r \mapsto x.r$, so it is compact, so it is closed.

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