Set-theoretic definition of polytopes

Discussion of tapertopes, uniform polytopes, and other shapes with flat hypercells.

Set-theoretic definition of polytopes

Postby ubersketch » Thu Jul 11, 2019 9:12 pm

Polytope definition:
The 0-polytopes 0{},1{},2{}... (vertices)
The n+1-polytope 0{x,y,z...}, 1{x,y,z...}, 2{x,y,z...} where x,y,z are n-polytopes.

Define the set of elements E_P where P is a polytope as:
e∈P
e∈E_P

Two polytopes n{x,y,z...} (denoted as X) and m{w,v,u...} (denoted as Y) are distinct iff they satisfy one of the following:
n=/=m
There exists x such that x∈X and y∉Y
There exists x such that x∈Y and y∉X

Define the set of elements E_P where P is a polytope as:
e∈P
e∈E_P

Two elements a and b, of a polytope P are incident iff:
a∈b
or
there exists c such that c∈a and c∈b

A polytope P is vertex-transitive iff:
there is an isomorphism from a given vertex and pseudoelements (i have not figured out a way to define pseudoelements yet) and elements incident to it to every other vertex.

Note this definition also includes strange degenerate cases such as the 3-pointed edge, compounds, etc.
Last edited by ubersketch on Fri Jul 12, 2019 1:36 pm, edited 1 time in total.
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Re: Set-theoretic definition of polytopes

Postby wendy » Fri Jul 12, 2019 9:31 am

The set-theory definition fails on a number of grounds.

Firstly, a polygon is not a 'set of points', because the only valid surtopes of a polygon are its edges. Suppose a hexagon is vertices 1-6. The subsets of vertices with meaning is the consecutive pairs 12, 23, 34, 45, 56, 61. A set like 123 or 135 is pretty meaningless, except set theory demands it.

Second. Set theory encloses a dual operation of 'intersection', ie ~(A+B) = (~A)&(~B) ie, the complemnent of the union is the intersection of the union. The complement set is meaningless in polytopes. Instead, the upwards incidence (which mostly functions as the dual of an element) contains the element itself.

Third. The null set ø is not a universal, but each polytope has its own nulloid. Of course, it is possible to imagine if S is the named space, and Ø is the universal null set, then ø = S&Ø might apply.

Polytopes are in the first instance, solids bounded by flat planes. A cone is a solid, and one can construct an incidence table for it, but it's not bounded by flat planes. Since this definition is regressive, we get 'polytopes are bound by polytopes'. and then 'there are polytopes in every dimension'. A point in 3D is a third-order member, ie the crossing of three planes.

pseudopolytopes

These correspond to what Coxeter calls 'j-circuits that do not bound'. In a regular icosahedron, a circuit of edges divides the surface into two parts. So any 'j-circuit' bounds. The great dodecahedron {5,5/2} has non-bounding circuits, such as the triangles of the icosahedron. The surface is presented as 20 nonbounding circuits, and 12 bouning curcuits, giving an excess of 8, which leads to its genus of 4.

In reality, non-bounding circuits are not part of the polytope, but fall into the iffy part that a polytope is a flat-surface thing, does not preclude that the sum of the flat-surfaces is not topologically a convex object, and therefore non-bounding j-circuits are not precluded.

Even the dyadic rule or the single-density does not preclude it. A tower of prisms, bent into a ring, still remains single-density, but it contains j-circuits that do not bound. But to close these curcuits, would involve a spanning polygon inside and outside the tube, which would produce a non-dyadic figure. In effect, such spanning surtopes are themselves non-bounding to the interior, and thus non-surface things (since they do not separate areas of different densities).
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Re: Set-theoretic definition of polytopes

Postby ubersketch » Fri Jul 12, 2019 1:28 pm

Do you not know set theory?

I'm treating polytopes like how graphs in graph theory are formalized in set theory as unordered pairs. Let's take an example

{0{0,1},1{1,2},2{2,0}}

In graph theory, edges are essentially sets of points. Besides, this is for abstract polytopology anyways.
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