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You are not logged in. #1 20101029 23:39:29
Homeomorphism of Euclidean plane to sphere with point removed
It is clear that is not homeomorphic to a sphere (by which I mean the surface only, not the interior). The latter is compact whereas the former isn’t. However if you remove a single point from the sphere, then the two spaces are homeomorphic.
Let us take the unit sphere centred at , given by , and let us remove the north pole . Call this “pointless” sphere . Then is the disjoint union of circles formed by the intersection of with the plane as t varies from 0 to 2 (including 0 but not 2). Now consider itself. This is the disjoint union of origincentred circles of all possible nonnegative radii (counting as a circle of radius 0). Let be any continuous bijection (e.g. or ). Then if we define by and for where , we have a homeomorphism! I was thinking about this last night. Thinking about math problems is a great way to pass the time when you’re having insomnia. Last edited by JaneFairfax (20101030 00:50:05) #2 20110115 22:58:13
Re: Homeomorphism of Euclidean plane to sphere with point removed
The surface of the sphere is homeomorphic to the extended complex plane .
#3 20110116 06:58:17
Re: Homeomorphism of Euclidean plane to sphere with point removedsince homeomorphism is defined by continuity but does not require smoothness we can map the punctured sphere to sharpened pencil shape: a truncated halfcone with apex at the south pole intersecting the sphere at the equator, glued to a half infinite cylinder. then the cylinder is easily mapped to the remaining (infinite) portion of the cone. finally flatten out the cone. Last edited by coprime (20110116 07:14:11) 