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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, the implication holds in metric spaces.

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I believe in condition $(3)$, the set $U$ should be a neighborhood of $U \times U$. – rpotrie Sep 14 '10 at 8:04
A better reference that (1)+(2) => (3) in a metric space is or – Matthew Daws Sep 14 '10 at 8:43

1 Answer 1

up vote 2 down vote accepted

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.

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I haven't checked the details, but I was wondering if a pseudo-metric approach to uniform spaces might allow one to lift the proof... so I think this is a great idea! – Matthew Daws Sep 14 '10 at 9:59

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