Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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.

share|improve this question
    
I believe in condition $(3)$, the set $U$ should be a neighborhood of $U \times U$. –  rpotrie Sep 14 '10 at 8:04
1  
A better reference that (1)+(2) => (3) in a metric space is ams.org/mathscinet-getitem?mr=1157223 or jstor.org/stable/2324899 –  Matthew Daws Sep 14 '10 at 8:43
add comment

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.

share|improve this answer
    
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
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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