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Circle is not homeomorphic 
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#19
Feb1713, 12:46 AM

Sci Advisor
P: 1,169

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 propertyi.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 HeineBorel, 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. 


#20
Feb1913, 06:42 AM

P: 68

Thank you so very much :)



#21
Feb1913, 08:58 PM

Sci Advisor
P: 1,716

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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