How to define the vertex in a cone consistent with the polyhedron vertex I'm sorry if this question is too simple but I'm unable to find a good answer. I searched through the  site and could  not find a similar question.
A vertex in a 3D polyhedron is where 3 edges meet. As an extension a vertex in N dimensions polyhedron is the intersection of N hyperplanes.
How to define the vertex on a cone?  More importantly how to reconcile it with the definition of a polygon vertex.
In practice we can make a cone from paper which is a 2D polyhedron with vertices. Is there is a way to formalize this into proving the vertex of cone is indeed similar to the vertex of a rectangle? If so can it be extended to N dimensions?
Or am I thinking about it all wrong and is there a much simpler answer.
 A: There is a general concept which incorporates both polyhedra and cones, namely the concept of a convex body. And the concept of a vertex can be defined in that general setting.
By definition, a subset $C \subset \mathbb R^3$ is a convex body if there exists a set of half-spaces $\{H_i\}_{i \in I}$ such that
$$C = \bigcap_{i \in I} H_i
$$
And, to be complete, a (closed) half space $H$ is the solution set of an equation of the form $ax + by + cz + d \ge 0$, for some constants $a,b,c,d \in \mathbb R$ such that at least one of $a,b,c$ is nonzero:
$$H = \{(x,y,z) \in \mathbb R^3 \mid ax + by + cz + d \ge 0\}
$$
Some extra terminology: the boundary plane of $H$ is
$$\partial H =  \{(x,y,z) \in \mathbb R^3 \mid ax + by + cz + d =0\}
$$
If $H$ is a half-space, if $C \subset H$, and if $C \cap \partial H \ne \emptyset$, then we say that $H$ is a supporting half-space of $C$. In this terminology, $C$ is equal to the intersection of the set of all supporting half-spaces of $C$.
So, for example, a finite sided (convex) 3-D polyhedron can be defined as a convex body $C \subset \mathbb R^3$ which is the intersection of a finite set of closed half spaces (this allows for finite sided polyhedra to be unbounded; one could also restrict ones attention just to finite sided polyhedra that are bounded).
Let me also define a cone in this setting, more specifically a right circular cone (this is just to keep things simple; one could study more general cones). First, one is given a 2-dimensional plane $P \subset \mathbb R^3$, a line $L \subset \mathbb R^3$ perpendicular to $P$, a point $Q = P \cap L$, another point $R \in L - \{Q\}$, and a radius $s > 0$. One takes $C$ to be the circle in $P$ of radius $s$ centered on $Q$.
For each $x \in C$ one takes $H_x$ to be the half space such that its boundary plane $\partial H_x$ passes through $R$ and intersects the plane $P$ in a line that is tangent to $C$ at the point $x$, and such that the half space $H_x$ entirely contains the circle $C$.
With that general definition and these two special examples, here's the definition of a vertex. First, given a cone $C \subset \mathbb R^3$ and a point $p \in c$, to say that $p$ is an extreme point of $C$ means that there exists a closed half-space $H \subset \mathbb R^3$ such that $H \cap C = \{p\}$ (this implies that $p$ is contained in the boundary plane $\partial H$; this also implies that the "opposing" half space of $H$ is a supporting half-space of $C$). Let's also say that any half-space $H$ satisfying the above property is a witness to the extremity of $p$.
A half space that witnesses the extremity of $p \in C$ might not be unique, and we need to quantify the non-uniqueness. For each half-space $H$ that witnesses the extremity of $p$, let $v^\perp_H$ be the unique unit vector based at $p$ which is orthogonal to $\partial H$ and points into $H$. The half-space $H$ determines and is determined by the vector $v^\perp_H$, so I'll extend the terminology to say that the vector $v^\perp_H$ witnesses the extremity of $p$.
To say that an extreme point $p$ is a vertex means that there exist three non-coplanar vectors $v^\perp_1,v^\perp_2,v^\perp_3$ each based at $p$, and each witnessing the extremity of $p$.
The unique vertex of the right circular cone defined above is the point $R$. Also, in a finite sided polyhedron, a point is a vertex if and only if it is a common endpoint of a set of three edges. (This begs the definition of edges. First, let's say that an extreme point $x \in C$ is an edge point if the set of vectors $v_H$ which witness the extremity of $x$ traces out a circular arc in the unit sphere centered on $x$; and then let's say that an edge is a set of edge points all of which share the same set of witnessing vectors.)
One other interesting example is the 3-dimensional ball $B^3 = \{(x,y,z) \in \mathbb R^3 \mid x^2 + y^2 + z^3 \le 1\}$, which is a convex body. It's set of extreme points is the unit sphere $S^2 = \{(x,y,z) \in \mathbb R^3 \mid x^2 + y^2 + z^2 = 1\}$, but $B^3$ has no vertices and no edges, because the witnessing half-space of each extreme point is unique.
A: You start by talking about the vertices of a polyhedron and then switch to a cone.  But by any definition I know, a cone is not a polyhedron.  Exactly what is your definition of "cone"?
