Both are connected. Both have no boundary. Both are closed, since both contain their accumulation points. Both have uncountable elements. So how can they be distinguished within the context of point-set topology? TIA, AG
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Both are connected. Both have no boundary. Both are closed, since both contain their accumulation points. Both have uncountable elements. So how can they be distinguished within the context of point-set topology? TIA, AG
A closed curve on a sphere with a point not on the curve can be contracted to a point without crossing the point not the curve no matter where that point is.
Brent
On 1/24/2020 5:38 PM, Alan Grayson wrote:
Both are connected. Both have no boundary. Both are closed, since both contain their accumulation points. Both have uncountable elements. So how can they be distinguished within the context of point-set topology? TIA, AG--
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On Friday, January 24, 2020 at 6:58:29 PM UTC-7, Brent wrote:A closed curve on a sphere with a point not on the curve can be contracted to a point without crossing the point not the curve no matter where that point is.
Brent
Doesn't seem right. If we have a circle on the sphere, and a point at its center, your claim will fail. AG
You have to go beyond point-set topology and consider homology or homotopy theory, With a sphere a curve is a loop that is contractible to a point. A line in flat spacetime is not contractible. This might make one thinkthere is a homology, or cohomology, of H^1(R^2) = ker(M)/im(M) for M a map. The homotopy π_1 for the sphere is zero, contractible, but is Z for the Euclidean space. One might think the homology H_1(R^2) is the same, but Euclidean plane and 2-sphere have a trick up their sleeve with the stereographic projection so the pole of S^2 gets mapped to "infinity" So the middle homology group or ring is zero. The homotopy fundamental form π_1 has commutators that make it not zero. However, that stereographic projections means the point at the pole is mapped away so while the sphere has H_0(S) = H_2(S) = Z the Euclidean plane has H_0(R^2) = 0.LC