I have been studying the proof of the following theorem:
Theorem: Let's suppose that $X$ is some metric space and $X$ is a infinite set. If $f:X\to X$ is transitive and has dense periodic points then $f$ has sensitive dependence on initial conditions.
It was proved by J. Banks, Here's a link
Almost all the proof is clear, but I can't understand how it's possible to choose two arbitrary periodic points $q_{1}$ and $q_{2}$ with disjoint orbits $\mathcal{O}(q_{1},f)$ and $\mathcal{O}(q_{2},f)$. I know that this is related to the fact that $X$ is a infinite set, but i don't see it!
Also, I have the following question:
Question 2: Suppose that $q_{1},q_{2}\in Per(f)$, i.e, exists $n,m\in\mathbb{N}$ such that $f^{n}(q_{1})=q_{1}$ and $f^{m}(q_{2})=q_{2}$. If $X$ is infinite or finite, which is the relationship between $\mathcal{O}(q_{1},f)$ and $\mathcal{O}(q_{2},f)$?.
- $\mathcal{O}(q_{1},f)\cap \mathcal{O}(q_{2},f)\neq \emptyset$.
- If $n<m$ or $n>m$, $\mathcal{O}(q_{1},f)\subset \mathcal{O}(q_{2},f)$
- etc,..
I've been thinking on and off about this questions for a couple of days,...
Take any periodic point $q_1$. Its orbit is a finite set, and therefore is not dense. So there must be another periodic point $q_2$ whose orbit is not $\mathcal O(q_1, f)$. But the orbits of periodic points are either the same or disjoint.