Donna Flynn

2022-07-17

Let ${S}^{1}$ be a circle of perimeter equal to 1 (not radius 1) .Let $f:{S}^{1}\to {S}^{1}$ be an arbitrary homeomorphism. By uniform continuity, it always possible to find an $ϵ>0$ such that $d\left(x,y\right)<ϵ$ implies
$d\left(f\left(x\right),f\left(y\right)\right)<ϵ,$
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 $length\left(I\right)<ϵ$ implies that
$length\left(f\left(I\right)\right)<\frac{1}{4}?$
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.

penangrl

Expert

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

For each point $p\in {S}^{1}$, let $q=f\left(p\right)$, and let ${J}_{q}$ be the arc of length 0.25 around $q$. Then ${f}^{-1}\left({J}_{q}\right)$ is open, so contains an open interval ${I}_{p}$ around $p$. Let $\varphi \left(p\right)>0$ be the largest radius of such interval ${I}_{p}$. Then we note that
$|\varphi \left(p\right)-\varphi \left({p}^{\prime }\right)|\le d\left(p,{p}^{\prime }\right)$
so $\varphi$ is continuous. As ${S}^{1}$ is compact, we conclude that ${\varphi }^{-1}$ is upper bounded. Let $N$ be the upper bound, and take $ϵ={N}^{-1}$.

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