# Doubts in proof of topologically transitive + dense periodic points = Devaney Chaotic

Let $(X,d)$ be an infinite metric space with no isolated points and $f : X \to X$ be topologically transitive with dense periodic points. Then $f$ exhibits sensitive dependence on initial data.

The first in the proof given here say that there is $\delta>0$ such that for each $x \in X$ there exists a periodic point $q$ whose orbit $O(q)$ is of distance at least $\frac {\delta}{2}$ from $x.$

We choose two periodic points $q_1,q_2 \in X$ such that $O(q_1) \cap O(q_2) =\emptyset$ and let $\delta = d(O(q_1),O(q_2))>0.$ Then for $x \in X$ and all $i,j$ we have $$\delta\leq d(f^i(q_1),f^j(q_2))<d(f^i(q_1),x)+d(x,f^j(q_2))$$ So, either $$d(f^i(q_1),x) \geq \frac {\delta}{2} \text{ or } d(f^j(q_2),x) \geq \frac {\delta}{2}\tag{1}$$

My question:

How does it follow from here that either $$d(O(q_1),x)\geq \frac{\delta}{2} \text{ or } d(O(q_2),x)\geq \frac{\delta}{2}\tag{2}$$ How do we get that for each $i$ and $j$ the same inequality in $(1)$ holds?

Edit: I think my question lacked clarity. I understand how $(1)$ follows. But I don't understand how $(2)$ follows from $(1)?$ It may happen that when I take $i,j=1,$ then inequality in $(1)$ holds for $q_1$ but when I take different $i,j$ inequality holds for $q_2.$

## 2 Answers

Since the untagged equation is true for all $i,$ therefore

$$\delta\leq d(O(q_1),x)+d(x,f^j(q_2)).$$

And this is true for all $j.$ Thus,

$$\delta \leq d(O(q_1),x)+d(x,O(q_2)).$$ It follows that $$d(O(q_1),x)\geq \frac{\delta}{2} \text{ or } d(O(q_2),x)\geq \frac{\delta}{2}$$

If they were both less than $\delta/2$, their sum couldn’t exceed $\delta$

• Please check my edit. You misunderstood my question. – Sahiba Arora Jan 16 '18 at 12:35
• @SahibaArora You misunderstood the answer. – orole Jan 16 '18 at 12:36
• @SahibaArora I will not do the thinking for you. Your job. – orole Jan 16 '18 at 12:38
• @orole I think I got it, thanks. – Sahiba Arora Jan 16 '18 at 12:43