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# Generalized Krein-Milman Theorem for Partially Ordered Sets. Part 4.

edited June 2018 in Chat

Generalized Krein-Milman Theorem.

Suppose that $$\langle P; \leq, \bot \rangle$$ is a partially ordered set, the set $$\mathsf{C}$$ is a lower semilattice of convex regions, we have $$d \in {\downarrow}\mathsf{C}$$ and $$E \subseteq {\downarrow}d$$ satisfies the lower bound property. Let $$M$$ denote the collection of minimal regions in $$E$$ . Then

• For all anti-chains $$A \subseteq E$$ there exists an anti-chain $$A^{\ast} \subseteq E$$ satisfying $$A \subseteq A^{\ast}$$ and $$d \leq {\mathsf{con}}(A^{\ast})$$ iff $$E$$ satisfies the augmentation property for $$d$$ .

Suppose also that $$E$$ satisfies the augmentation and peripheral properties. Then

• We have $$d \leq {\mathsf{con}}(M)$$ iff $$E$$ satisfies the eventual finite meet property.

Proof.

First suppose that $$E$$ satisfies the augmentation property for $$d$$ . Let $$A \subseteq E^{+}$$ be an anti-chain. Define $$\mathcal{A} = \{ A^{\prime} \subseteq E^{+} \colon A^{\prime} \text{ is an anti-chain and } A \subseteq A^{\prime} \} \text{.}$$ Using Zorn's lemma there exists a maximal anti-chain $$A^{\ast} \in \mathcal{A}$$ with $$A \subseteq A^{\ast}$$. If $$d \nleq {\mathsf{con}}(A^{\ast})$$ the augmentation property implies there exists a larger anti-chain in $$E^{+}$$ , contradicting the maximality of $$A^{\ast}$$.

Now suppose that for all anti-chains $$A \subseteq E$$ there exists an anti-chain $$A^{\ast} \subseteq E$$ satisfying $$A \subseteq A^{\ast}$$ and $$d \leq {\mathsf{con}}(A^{\ast})$$ . If $$d \leq {\mathsf{con}}(A)$$ there is nothing more to prove. For the remainder of this paragraph suppose that $$d \nleq \mathsf{con}(A)$$ . Select an anti-chain $$A^{\ast} \subseteq E$$ satisfying $$A \subseteq A^{\ast}$$ and $$d \leq {\mathsf{con}}(A^{\ast})$$ . Then

• $$\mathsf{con}(A^{\ast}) = \qquad \text{set manipulation}$$
• $$\mathsf{con}((A^{\ast} \smallsetminus A) \cup A) = \qquad$$ distributive property
• $$\mathsf{con}(\cup \lbrace \lbrace e^{\ast} \rbrace \cup A \colon e^{\ast} \in A^{\ast} \smallsetminus A \rbrace \rbrace = \qquad$$ by lemma 2
• $$\mathsf{con}(\cup \lbrace \mathsf{con}(\lbrace e^{\ast} \rbrace \cup A) \colon e^{\ast} \in A^{*} \smallsetminus A \rbrace$$.

If, for all $$e^{\ast} \in A^{\ast} \smallsetminus A$$ , we had $$\mathsf{con}(\lbrace e^{\ast} \rbrace \cup A ) = \mathsf{con}(A)$$ , then we would also have $$\mathsf{con}(A^{\ast}) = \mathsf{con}(A)$$ , contradicting our hypothesis that $$d \leq \mathsf{con}(A^{\ast})$$

For the remainder of this proof we will suppose that $$E$$ satisfies the augmentation and peripheral properties for $$d$$ .

Next suppose that $$d \in \mathsf{con}(M)$$ . Select $$e \in E$$ . Since $$e \in \mathsf{con}(M)$$ the peripheral property implies there exists an $$m \in M$$ with $$e \wedge m \neq \bot$$ . The lower bound property implies there exists an $$e^{\ast} \in E$$ satisfying $$\bot < e^{\ast} \leq { e, m }$$ . Recalling that $$m$$ is minimal we have $$m \leq e$$ . The set $${ m }$$ satisfies the finite meet property.

Finally suppose that $$E$$ satisfies the eventual finite meet property for $$d$$ . The definition of a minimal extreme subregion implies that $$M$$ is an anti-chain. For the purpose of deriving a contradiction, suppose that $$d \nleq \mathsf{con}(M)$$ . The augmentation property implies there is an $$e \in E^{+}$$ for which the set $$M \cup \lbrace e \rbrace$$ is an anti-chain and $$\mathsf{con}(M \cup { e } ) \smallsetminus \mathsf{con}(M)$$ is not empty. Using theorem 1 there is a minimal extreme subregion $$m \leq e$$ . Since $$m \in M$$ the set $$M \cup \lbrace e \rbrace$$ is not an anti-chain. This is the desired contradiction.

The careful reader will note the original Krein-Milman theorem uses the closed convex hull but this version uses the convex hull. If one wishes to use this theorem in the original situation use the closed, convex hull rather than just the convex hull. . The reader may be curious as to how one might find a set $$E$$ with these wonderful properties. We can find extreme subregions by generalizing the definition of an extreme subset in vector spaces.

Definition.

Suppose that $$e, a \in P$$ satisfy $$e \leq a$$ and $${ a } \in \mathsf{Bnd}$$ . We will say that $$e$$ is an extreme subregion of $$a$$ if, for all nonempty $$D \subseteq {\downarrow}a$$ we have $${\downarrow}e \cap \mathsf{con}(D) = {\downarrow}e \cap \mathsf{con}({\downarrow}D \cap {\downarrow}e) \text{.}$$ The reader can check that both $$\bot$$ and $$a$$ are extreme subregions of $$a$$ . Sometimes it proves convenient to allow $$\bot$$ to be an extreme subregion. Defined this way, the collection of extreme subsegions satisfies the peripheral property. If $$e_{0}$$ and $$e_{1}$$ are extreme subregions and $$e_{0} \wedge e_{1}$$ exists then $$e_{0} \wedge e_{1}$$ is also an extreme subregion. If $$\mathsf{con}$$ is algebraic, the set $$E$$ is a nonempty collection of extreme subregions of $$A$$ , and $$\wedge E$$ exists then $$\wedge E$$ is an extreme subregion of $$a$$ .

In a few days I will post a message about extreme subregions.