# I Full-cone as topological space

#### cianfa72

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

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#### 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.

#### cianfa72

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

Mentor
2018 Award
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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2018 Award
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:

#### lavinia

Gold Member
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?

#### cianfa72

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

Gold Member
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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