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Circle is not homeomorphic

by joao_pimentel
Tags: circle, homeomorphic
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Feb17-13, 12:46 AM
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P: 1,172
Assume there is a homeomorphism h between S^1 and E , where E is a subset

of the real line. Since connectedness is a topological property--i.e., connectedness

is preserved by homeomorphisms ( continuous maps will do)-- h(S^1) is a connected

subset of the real line. Then h(S^1) is an interval ; by compactness of S^1 (which must be preserved by h), h(S^1) is also compact in the real line, it then follows by

Heine-Borel, h(S^1) is closed and bounded in the real line , so h(S^1)=[a,b].

Now, if h is a homeomorphism, the restriction of h to any subset of

S^1 is a homeomorphism (into its image). Consider x in S^1 with h(x) not an endpoint , i.e.,

h(x)≠ a,b; say h(x)=c . Now consider the restriction of h to S^1 -{x}. This is

a homeomorphism from the connected space S^1-{x} to the disconnected space

[a,b]-{c} . This is not possible , so no such h can exist.

Moral of the story/ general point: disconnection number is a homeomorphism invariant.
Feb19-13, 06:42 AM
P: 68
Thank you so very much :)
Feb19-13, 08:58 PM
Sci Advisor
P: 1,723
It might be fun to try to prove this by approximating a homeomorphism of the circle in R^n by a sequence of smooth maps to see how far off from a diffeomorphism the homeomorphism can get. the image of a smooth approximation must have measure zero so it can not produce a homeomorphism between the circle and euclidean space.

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