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Suppose $(X,f)$ is a dynamical system which consists of a discrete space $X$ with $|X| \geq 2$ and consisting of a single periodic orbit. I need to show that $(X,f)$ is topologically transitive.
Is this obvious? If $(X,f)$ consists of a single periodic orbit, then does it have a transitive point?
Also, how do I show that the system $(X\times X,F)$, where $F:X\times X \rightarrow X\times X$, $$F(x_1,x_2)=(f(x_1),f(x_2))$$ is not topologically transitive?

wwinters57
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  • Since the system is transitive (is this correct?), there is a transitive point $x$. Let $U,V \subseteq X$ be open and non-empty. Since $x$ is transitive, there are infinitely many $m \in \mathbb{N}$ such that $f^m(x) \in U$. Now, $f^m(x)$ is also transitive, so there are infinitely many $n \in \mathbb{N}$ such that $f^n(f^m(x)) \in V$. Hence, $f^n(U) \cap V \neq \emptyset$, so $(X,f)$ is topologically transitive. Is this correct? – wwinters57 May 22 '20 at 19:25
  • Also, I don't see how $(X \times X,F)$ is not topologically transitive. – wwinters57 May 22 '20 at 19:26
  • Sorry, I meant since the system is only one periodic orbit, there is a transitive point. – wwinters57 May 22 '20 at 19:37
  • See my answer please. – John B May 22 '20 at 20:15
  • Well, I know it is not true in general, as the example here https://math.stackexchange.com/questions/139011/product-of-transitive-systems suggests. – wwinters57 May 22 '20 at 20:24

1 Answers1

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It is correct what you wrote, but in your case it is somewhat simpler because you have a single periodic orbit.

Notice that all elements of $X$ are transitive and in fact all orbits of all elements of $X$ intersect all nonempty open sets (because $X$ is a single periodic orbit).

Now iterate any element $(x,y)$ under $F$. Since $X$ is s single period orbit, say $$\{f^k(p):k=0,\ldots,n-1\},$$ then $$(x,y)=(f^n(p),f^m(x))\quad\text{for some} \ n,m\in\{0,\ldots,n-1\}.$$ Notice that the $F$-orbit never intersects all open sets (for all integers $n$ and $m$). Since $X$ has the discrete topology, this means that no point is transitive.

John B
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