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Let Dom be a uniform space, and 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 U of Dom such that for all members x of Dom and all neighborhoods N of x, there exists a member y of N and a natural number n such that ((f^n)(x),(f^n)(y)) is not a member of U.

According to pb.math.univ.gda.pl/chaos/pdf/on_intervals.pdf, the implication holds in metric spaces.

edited Sep 14 2010 at 23:08

asked Sep 13 2010 at 21:04

Ricky Demer
6,401●8●31

I believe in condition $(3)$, the set $U$ should be a neighborhood of $U \times U$. – rpotrie Sep 14 2010 at 8:04

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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 2010 at 8:43

KP Hart · Accepted Answer · 2010-09-14 09:31:40Z UTC

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.

Chaos in uniform spaces

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