1
$\begingroup$

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 $q_2.$

$\endgroup$
1
$\begingroup$

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 $$d(O(q_1),x)\geq \frac{\delta}{2} \text{ or } d(O(q_2),x)\geq \frac{\delta}{2}$$

$\endgroup$
0
$\begingroup$

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

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.