TITLE: Devaney's definition of sensitive dependence on initial conditions
QUESTION [2 upvotes]: Devaney defines that 

a funtion $f:J \to J$ has sensitive dependence on initial conditions
  if there exists $\delta > 0$ such that, for every $x \in J$ and any
  neighborhood $N$ of $x$, there exists $y \in N$ and $n \geq 0$ such
  that $|f^n(x)-f^n(y)| > \delta$.

Here I already see a misspell... I would rather say "exist $y \in N$ and $n \geq 0$", without the "s". But probably here I am wrong.
I was told that this definition is equivalent to 

there exists $\delta > 0$ such that for every open $U \subset J$ there
  exist two points $x, y\in U$ and a number $n \in \mathbb{Z}^+$ such
  that $|f^n(x)- f^n(y)| \geq \delta$

But I cannot see how. Any hint on a formal proof of the equivalence?

REPLY [3 votes]: It is clear that the first definition implies the second because there is the freedom to choose $x$ while in the second definition we have only the existence of $x$.
To prove that the second definition suppose we have a neighborhood $N$ of $x$. There exists $x'$ and $y'$ in $N$ and $n\in \mathbb{Z}^+$ such that $|f^n(x')-f^n(y')|>\delta '$ therefore we have that $|f^n(x')-f^n(x)|+|f^n(x)-f^n(y')| \geq|f^n(x')-f^n(y')| \geq \delta$  by the triangle inequality and from this we have that either $|f^n(x')-f^n(x)| \geq \frac{\delta'}{2}$ or $|f^n(x)-f^n(y')|\geq \frac{\delta'}{2}$ therefore we can set $\delta = \frac{\delta'}{2}$
{"set_name": "stack_exchange", "score": 2, "question_id": 1492784}