Let $Dom$ be a uniform space, and $\hspace{.04 in}f$ be a continuous function from $Dom$ to itself satisfying:


1. For all non-empty open subsets $U$ and $V$ of $Dom$, there exists a natural
  
number $n$ and a member $x$ of $U$ such that $f^n(x)$ is a member of $V$.
2. The periodic points of $f$ are dense in $Dom$.


  

Does it follow that $f$ satisfies (3)?


$\;\;$3.$\:\:$There exists an entourage $E$ of $Dom$ such that for all members $x$ of
  

$\quad \;\;$ $Dom$ and all neighborhoods $U$ of $x$, there exists a member $y$ of $U$ and
  

$\quad \;\;$ a natural number $n$ such that $\:\langle \hspace{.05 in}f^n(x)\hspace{.02 in},\hspace{.03 in}f^n(\hspace{.03 in}y)\rangle\:$ is not a member of $E$.


  

According to [this paper](http://pb.math.univ.gda.pl/chaos/pdf/on_intervals.pdf), the implication holds in metric spaces.



Chaos in uniform spaces
I believe the following works. Follow the proof in the reference supplied by Matthew Daws: choose two distinct periodic orbits and choose a compatible pseudometric $\rho$ such that all points in those orbits are at least $1$ apart under this pseudometric.
The proof establishes that the entourage $\lbrace(x,y):\rho(x,y)<\frac18\rbrace$ is as required.


