## Circle is not homeomorphic

ok $S^1$ is compact, i.e., closed and limited and the line would have to be if there were a homeomorphism between both.

On the other hand, if I remove a point, I wouldn't have a compact set, on both, but one is still conncted and the other is not.

And I suppose, that if one is path connected, and there is a homeomorphis to another set, the other set must be path connected...

and then we have a contradiction....
 Recognitions: Science Advisor 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.
 Thank you so very much :)
 Recognitions: Science Advisor 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.