Let S 1 be a circle of perimeter equal to 1 (not radius 1) .Let...

Donna Flynn

Donna Flynn

Answered

2022-07-17

Let S 1 be a circle of perimeter equal to 1 (not radius 1) .Let f : S 1 S 1 be an arbitrary homeomorphism. By uniform continuity, it always possible to find an ϵ > 0 such that d ( x , y ) < ϵ implies
d ( f ( x ) , f ( y ) ) < ϵ ,
where d denote the shortest distance between two points on the circle.
My question is: Is it always possible to find an ϵ > 0 such that for any arc I (or call it interval if you like) on S 1 such that l e n g t h ( I ) < ϵ implies that
l e n g t h ( f ( I ) ) < 1 4 ?
In other words, do homeomorphisms always send very short arcs to short arcs, rather than long arcs?
Since a homeomorphism can be expanding or contracting, I am really puzzled by this. There must be some tricks I don't know.

Answer & Explanation

penangrl

penangrl

Expert

2022-07-18Added 17 answers

In general, the property you claim holds for compact spaces and is a vital part of the no small subgroup argument.

For each point p S 1 , let q = f ( p ), and let J q be the arc of length 0.25 around q. Then f 1 ( J q ) is open, so contains an open interval I p around p. Let ϕ ( p ) > 0 be the largest radius of such interval I p . Then we note that
| ϕ ( p ) ϕ ( p ) | d ( p , p )
so ϕ is continuous. As S 1 is compact, we conclude that ϕ 1 is upper bounded. Let N be the upper bound, and take ϵ = N 1 .

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