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

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 $V$.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, the implication holds in metric spaces.