I Full-cone as topological space

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Summary
Full-cone topological space but not topological manifold
Hello,

consider a full-cone (let me say a cone including bottom half, upper half and the vertex) embedded in ##E^3##. We can endow it with the topology induced by ##E^3## defining its open sets as the intersections between ##E^3## open sets (euclidean topology) and the full-cone thought itself as subset of ##E^3##. This way it has topological space structure.

Nevertheless I believe it is not a topological manifold because any neighborhood of the vertex cannot be homeomorphic to ##E^3## or one of its open subset.

Is that correct ? Thanks
 

Orodruin

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The space that the cone embedded in ##E^3## is locally homeomorphic to everywhere but at the apex is ##E^2##, not ##E^3##.

It is true that the cone is not locally homeomorphic to either ##E^2## or ##E^3## at the apex.
 
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The space that the cone embedded in ##E^3## is locally homeomorphic to everywhere but at the apex is ##E^2##, not ##E^3##.
yes, sure.
 

fresh_42

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Summary: Full-cone topological space but not topological manifold

Hello,

consider a full-cone (let me say a cone including bottom half, upper half and the vertex) embedded in ##E^3##. We can endow it with the topology induced by ##E^3## defining its open sets as the intersections between ##E^3## open sets (euclidean topology) and the full-cone thought itself as subset of ##E^3##. This way it has topological space structure.

Nevertheless I believe it is not a topological manifold because any neighborhood of the vertex cannot be homeomorphic to ##E^3## or one of its open subset.

Is that correct ? Thanks
The cone (one half) is homeomorphic to a ball or a cube. It is topologically a compact subset of ##\mathbf{E}^3##. Inner points are locally homeomorphic to ##\mathbf{E}^3##, boundary points to ##\mathbf{E}^2##.

The vertex is no specific point from a topological point of view, except that it is on the boundary.

The cone is no analytic manifold, but it is a topological manifold, or a compact, connected subset of one to be precise.
 

Orodruin

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The cone (one half) is homeomorphic to a ball or a cube. It is topologically a compact subset of ##\mathbf{E}^3##. Inner points are locally homeomorphic to ##\mathbf{E}^3##, boundary points to ##\mathbf{E}^2##.

The vertex is no specific point from a topological point of view, except that it is on the boundary.

The cone is no analytic manifold, but it is a topological manifold, or a compact, connected subset of one to be precise.
He is considering a double cone, joined at the apex. I.e., style light-cone:
243947
 

lavinia

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Summary: Full-cone topological space but not topological manifold

Hello,

consider a full-cone (let me say a cone including bottom half, upper half and the vertex) embedded in ##E^3##. We can endow it with the topology induced by ##E^3## defining its open sets as the intersections between ##E^3## open sets (euclidean topology) and the full-cone thought itself as subset of ##E^3##. This way it has topological space structure.

Nevertheless I believe it is not a topological manifold because any neighborhood of the vertex cannot be homeomorphic to ##E^3## or one of its open subset.

Is that correct ? Thanks
Have you come up with a proof?
 
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Have you come up with a proof?
Basically removing a point from an ##\mathbb R^2## disk we get a connected set. Yet the set attained removing the apex from a double cone is not.
 

WWGD

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Yes, I think in general an n-manifold must have an n- or (n-1)-dimensional cutset. A 1-ball ( interval) can be separated by a single point., a 2-ball must be separated by a line(segment), etc.
 

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