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# "Faces" of elements in partially ordered sets

edited June 2018

What follows is a generalization of extreme subsets in vector spaces to partially ordered sets. If someone has seen a version of this material that uses category theory I would appreciate a reference. If the constructions herein seem similar to some constructions in category theory I would appreciate knowing about that as well.

Recall that in a real vector space if $$\varnothing \neq E \subseteq A$$ is an extreme subset of $$A$$ provided whenever $$a_{0}, a_{1} \in A$$ and there exists a $$t \in (0, 1)$$ with $$ta_{0} + (1 - t)a_{1} \in E$$ then $$a_{0}, a_{1} \in E$$. The two points $$a_{0}$$ and $$a_{1}$$ are used because the convex hull operation on sets is defined in terms of line segments. Suppose that $$C$$ is a cube in the $$\mathbb{R}^{3}$$ . The convex extreme subsets of $$C$$ are each vertex, each edge, each face, and $$C$$ . The non-convex extreme subsets of $$C$$ are unions of collections of convex extreme subsets. If $$E$$ is an extreme subset of $$C$$, convex or not, then $$C \smallsetminus E$$ is convex.

Suppose that $$\langle P; \leq, \bot \rangle$$ is a partially ordered set with least element $$\bot$$. Let $$\mathsf{C} \subseteq P$$ be a complete lower semilattice containg $$\bot$$. Suppose also that if $$A \subseteq \mathsf{C}$$ is not empty then the greatest lower bound of $$A$$ in $$\mathsf{C}$$ ( $$\wedge A$$ ) is also the greatest lower bound of \A \) in \P \). For any nonempty $$a \subseteq P$$ define $${\downarrow}A = \lbrace p \in P \colon p \leq a \text{ for some } a \in A \rbrace$$.

Let $$\mathsf{C} \subseteq P$$ be a complete lower semilattice whose meet operation agrees with the greatest lower bound partial operation in $$P$$. Let $$\mathsf{Bnd} = \lbrace B \subseteq P \colon B \neq \varnothing \text{ and } B \leq c \text{ for some } c \in \mathsf{C} \rbrace \text{.}$$ Define $$\mathsf{con} \colon \mathsf{Bnd} \rightarrow \lbrace c \rbrace \colon c \in \mathsf{C} \rbrace$$ for each $$B \in \mathsf{Bnd}$$ by the assignment $$\mathsf{con} \colon B \mapsto \wedge \lbrace c \in \mathsf{C} \colon B \leq \lbrace c \rbrace \rbrace \text{.}$$ \end{equation} The function $$\mathsf{con}$$ is a closure operator that we will consider an abstraction of the convex hull operator.

Suppose that $$e, a \in P$$ satisfy, for all nonempty $$D \subseteq {\downarrow}a$$, we have $${\downarrow}e \cap {\downarrow}\mathsf{con}(D) = {\downarrow}e \cap {\downarrow}\mathsf{con}({\downarrow}D \cap {\downarrow}e)$$ then we will say that $$e$$ is an \emph{extreme} subregion of $$a$$. If we choose to restrict these notions to elements of $$\mathsf{C}$$ and remove the braces on the elements in the codomain this equation can be simplified to $$e \wedge \mathsf{con}(D) = \mathsf{con}({\downarrow}D \cap {\downarrow}e) \text{.}$$ If $$e$$ is an extreme subregion of $$a$$ we will write $$e \sqsubseteq a$$ . Abusing notation we will write $${\sqsubseteq} (a) = \lbrace e \leq a \colon e \sqsubseteq a \lbrace \text{.}$$ Informally this formula says we can change an extreme subregion into a down set and then move it inside the convex hull operation.

We will say that $${\sqsubseteq}(a)$$ satisfies the \emph{peripheral} property for $$a$$ if, for all nonempty $$E \subseteq {\sqsubseteq}(a)$$ and all $$\bot \neq e^{\ast} \in E$$ if $$e^{\ast} \leq \mathsf{con}(E)$$ then there exists an $$e \in E$$ and $$e^{\prime} \in {\sqsubseteq}(a)$$ with $$\bot < e^{\prime} \leq \lbrace e, e^{\ast} \rbrace$$. Informally the peripheral property ties every extreme subregion to some edge of $$a$$ .

We will say that $$\mathsf{con}$$ is \emph{algebraic} if, for all $$B \in \mathsf{Bnd}$$, we have $$\mathsf{con}(B) = {\vee}{ \left[ \cup \lbrace \mathsf{con}(F) \colon F \subseteq B \text{ where } F \text{ is nonempty and finite} \rbrace \right]}\text{.}$$ In this equation we are asserting the existence of a least upper bound.

The relation $$\sqsubseteq$$ is transitive. Suppose that $$e_{0} \sqsubseteq a_{0}$$, $$e_{1} \sqsubseteq a_{1}$$. If both $$e_{0} \wedge e_{1}$$ and $$a_{0} \wedge a_{1}$$ exist then $$e_{0} \wedge e_{1} \sqsubseteq a_{0} \wedge a_{1}$$.

For all $$a \in P$$ the set $${\sqsubseteq}(a)$$ satisfies the peripheral property for $$a$$.

Suppose that $$\mathsf{con}$$ is algebraic. Then, for $$e \leq a$$, we have $$e \sqsubseteq a$$ iff for all nonempty finite $$F \subseteq {\downarrow}a$$ we have $${\downarrow}e \cap {\downarrow}\mathsf{con}(F) = {\downarrow}e \cap {\downarrow}\mathsf{con}({\downarrow}F \cap {\downarrow}e)$$ Furthermore, if $$\varnothing \neq E \subseteq {\sqsubseteq}(a)$$ and $$\wedge E$$ exists then $$\wedge E \sqsubseteq a$$.

This post has some related posts concering the Krein-Milman theorem. The Krein-Milman material is longer because proofs are included.

https://forum.azimuthproject.org/discussion/2222/generalized-krein-milman-theorem-for-partially-ordered-sets-part-1#latest

https://forum.azimuthproject.org/discussion/2224/generalized-krein-milman-theorem-for-partially-ordered-sets-part-2#latest

https://forum.azimuthproject.org/discussion/2226/generalized-krein-milman-theorem-for-partially-ordered-sets-part-3#latest

https://forum.azimuthproject.org/discussion/2225/generalized-krein-milman-theorem-for-partially-ordered-sets-part-4#latest

## Comments

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1.

Hmm. Interesting.

Comment Source: Hmm. Interesting.
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2.

Hi, @Christopher Upshaw Here is a problem you may find interesting. Let the partially ordered set be the subsets of $$\mathbb{R}^2$$ with the convex hull operation as the closure system. Create a method for determining whether a convex set is open in the usual topology using only convex sets. You can use the usual set operations. You can know if a convex set is empty or not. Since the intersection of two convex open sets is also convex and open, you should be able to show this for your method. What you can not do is use any of that $$tx + (1 - t)y$$ business. If you want to prove your method works you will need to use the $$tx + (1 - t)y$$ definition of convexity.

This is being done in $$\mathbb{R}^2$$ so that you can figure out the method by scribbling pictures. What I like about this is that any closure system has a topology and, if the closure system is familiar the result is a familiar topology.

This problem is appearing here because it does not seem to have anything to do with category theory.

Comment Source:Hi, @Christopher Upshaw Here is a problem you may find interesting. Let the partially ordered set be the subsets of \$$\mathbb{R}^2 \$$ with the convex hull operation as the closure system. Create a method for determining whether a convex set is open in the usual topology using only convex sets. You can use the usual set operations. You can know if a convex set is empty or not. Since the intersection of two convex open sets is also convex and open, you should be able to show this for your method. What you can not do is use any of that \$$tx + (1 - t)y \$$ business. If you want to prove your method works you will need to use the \$$tx + (1 - t)y \$$ definition of convexity. This is being done in \$$\mathbb{R}^2 \$$ so that you can figure out the method by scribbling pictures. What I like about this is that any closure system has a topology and, if the closure system is familiar the result is a familiar topology. This problem is appearing here because it does not seem to have anything to do with category theory.
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