Donna Flynn

Answered

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 $\u03f5>0$ such that $d(x,y)<\u03f5$ implies

$d(f(x),f(y))<\u03f5,$

where d denote the shortest distance between two points on the circle.

My question is: Is it always possible to find an $\u03f5>0$ such that for any arc $I$ (or call it interval if you like) on ${S}^{1}$ such that $length(I)<\u03f5$ implies that

$length(f(I))<\frac{1}{4}?$

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.

$d(f(x),f(y))<\u03f5,$

where d denote the shortest distance between two points on the circle.

My question is: Is it always possible to find an $\u03f5>0$ such that for any arc $I$ (or call it interval if you like) on ${S}^{1}$ such that $length(I)<\u03f5$ implies that

$length(f(I))<\frac{1}{4}?$

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.

Answer & Explanation

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 $p\in {S}^{1}$, 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 $\varphi (p)>0$ be the largest radius of such interval ${I}_{p}$. Then we note that

$|\varphi (p)-\varphi ({p}^{\prime})|\le d(p,{p}^{\prime})$

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 $\u03f5={N}^{-1}$.

For each point $p\in {S}^{1}$, 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 $\varphi (p)>0$ be the largest radius of such interval ${I}_{p}$. Then we note that

$|\varphi (p)-\varphi ({p}^{\prime})|\le d(p,{p}^{\prime})$

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 $\u03f5={N}^{-1}$.

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