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 &gt; 0 such that d ( x , y ) &lt; ϵ implies d ( f ( x ) , f ( y ) ) &lt; ϵ ,where d denote the shortest distance between two points on the circle.My question is: Is it always possible to find an ϵ &gt; 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 ) &lt; ϵ implies that l e n g t h ( f ( I ) ) &lt; 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.

Question

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  &amp;gt;  0 such that   d  (  x  ,  y  )  &amp;lt;  ϵ implies  d  (  f  (  x  )  ,  f  (  y  )  )  &amp;lt;  ϵ  ,where d denote the shortest distance between two points on the circle.My question is: Is it always possible to find an   ϵ  &amp;gt;  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  )  &amp;lt;  ϵ implies that  l  e  n  g  t  h  (  f  (  I  )  )  &amp;lt;      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&#039;t know.

penangrl · Accepted Answer

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  )  &amp;gt;  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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