showing system is topologically transitive

Question

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?

Answer 1

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.