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# Generalized Krein-Milman Theorem for Partially Ordered Sets. Part 2.

edited June 2018 in Chat

Notation, Definitions and Preliminary Results.

Suppose that $$X$$ is a set. Then $$\mathbf{Pwr} X$$ ( $$\mathbf{Fin} X$$ ) will denote the set of all (finite) subsets of $$X$$ .

In this post we will suppose that $$\langle P; \leq \bot \rangle$$ is a partially ordered set with least element $$\bot$$. We will refer to the elements of $$P$$ as regions. If $$p \leq q$$ we will say that $$p$$ is a subregion of $$q$$. Define $${}^{+} \colon A \mapsto A \smallsetminus \lbrace \bot \rbrace \quad (A \in \mathbf{Pwr} P) \text{.}$$ If $$X$$ is a set then $$\mathbf{Pwr} X$$ is partially ordered by inclusion. We let $${\mathbf{Pwr}}^{+} X$$ ($${\mathbf{Fin}}^{+} X$$) denote the collection of nonempty (finite) subsets of $$X$$. Define

$${\downarrow} \colon A \mapsto \lbrace p \in P \colon p \leq a \text{ for some } a \in A \rbrace \quad (A \in {\mathbf{Pwr}}^{+} P)$$ Collections of elements from a partially ordered set may or may not have a greatest lower bound. For $$A \in {\mathbf{Pwr}}^{+} P$$ and $$p \in P$$ we will write

$$\wedge A = p \quad \text{if } A \text{ has a greatest lower bound, which equals } p$$ $$\wedge A \neq p \quad \text{if } A \text{ does not have a greatest lower bound or the greatest lower bound differs from } p$$ $$\wedge A \text{ exists} \quad \text{if } A \text{ has a greatest lower bound}$$ If $$A = \lbrace a_{0}, a_{1} \rbrace$$ we will write $$a_{0} \wedge a_{1}$$ rather than $$\wedge \lbrace a_{0}, a_{1} \rbrace$$ .

Proposition 1.

Suppose that $$A \in \mathbf{Pwr}^{+} P$$ . If $$\wedge A \neq \bot$$ there exists a $$p \in P^{+}$$ satisfying $$\bot < p \leq A$$ .

Proof.

If $$\wedge A \neq \bot$$ either $$\wedge A = p > \bot$$ or there is a $$p \in P^{+}$$ with $$\bot < p \leq A$$ .

Definitions

Suppose that $$E \in \mathbf{Pwr}^{+} P$$ . We will say that $$E$$ satisfies the finite meet property if, for all nonempty $$A \subseteq E$$ satisfying $$\wedge A = \bot$$ , there exists an $$F \in \mathbf{Fin}^{+} A$$ with $$\wedge F = \bot$$ . We will say that $$E$$ satisfies the eventual finite meet property if, for all $$e_{1} \in E$$ , there exists an $$e_{0} \in E$$ satisfying both $$e_{0} \leq e_{1}$$ and the set $${ e \in E \colon e \leq e_{0} }$$ satisfies the finite meet property.

Suppose that $$E \in \mathbf{Pwr}^{+} P$$ and $$m \in E^{+}$$ . We will say that $$m$$ is a minimal region in $$E$$ if, for all $$e \in E$$ , the relation $$e \leq m$$ implies that $$e = \bot$$ or $$e = m$$ .

Lemma 1.

Suppose that $$E \in \mathbf{Pwr}^{+} P$$ satisfies the finite meet property. Then, for all $$e \in E^{+}$$ , there is a minimal $$m\in E^{+}$$ with $$m \leq e$$ .

Proof.

Select $$e \in E^{+}$$ . Let $$C \subseteq E^{+}$$ be a maximal chain in $$E^{+}$$ with $$e \in C$$ . There are three cases:

• We have $$\wedge C = \bot$$ .
• The set $$C$$ does not have a greatest lower bound.
• We have $$\wedge C > \bot$$ . Suppose the first case holds. Since $$E$$ satisfies the finite meet property there is a finite $$F \subseteq C$$ with $$\wedge F = \bot$$ . Since $$F$$ is finite and totally ordered $$F$$ has a least element $$m$$ and $$m = \wedge F = \bot$$ . Recalling that $$\bot \notin C$$ , we have a contradiction.

Suppose the second case holds. Since $$\wedge C \neq \bot$$ , proposition 1 implies there exists a $$p \in P^{+}$$ with $$\bot < p \leq C$$ . The maximality of $$C$$ implies that $$p \in C$$ . But then $$C$$ has a greatest lower bound, contradicting the hypothesis of this case.

Since the first two cases lead to contradictions, the third case holds. Select $$e^{\ast} \in E$$ satisfying $$e^{\ast} \leq C$$ . If $$e^{\ast} \notin C$$ , the maximality of $$C$$ implies that $$e^{\ast} = \bot$$ . But then $$\wedge C$$ is a minimal element of $$E$$ .

Theorem 1.

Suppose that $$\langle P; \leq, \bot \rangle$$ is a partially ordered set and $$E \in {\mathbf{Pwr}}^{+} P$$ . Then for all $$e \in E^{+}$$ there is a minimal $$m \in E^{+}$$ with $$m\leq e$$ iff and only if $$E$$ satisfies the eventual finite meet property.

Proof.

Suppose that for all $$e \in E^{+}$$ there exists a minimal $$m \in E^{+}$$ satisfying $$m \leq e$$ . Select $$e^{\ast} \in E^{+}$$ and then a minimal $$m \in E^{+}$$ with $$m \leq e^{\ast}$$ . Then $$\lbrace e \in E \colon e \leq m \rbrace$$ is either $$\lbrace m \rbrace$$ or $$\lbrace \bot, m \rbrace$$ . Each of these sets satisfies the finite meet property.

Suppose that $$E$$ satisfies the eventual finite meet property. Select $$e_{0} \in E^{+}$$ . Since $$E$$ satisfies the eventual finite meet property there exists an $$e_{0} \in E$$ satisfying both $$e_{0} \leq e_{1}$$ and the set $${ e \in E \colon e \leq e_{0} }$$ satisfies the finite meet property. Recalling the lemma 1 there is a minimal $$m \in E^{+}$$ with $$m \leq e_{0} \leq e_{1}$$.

The finite meet property is a generalization of the finite intersection property in topology. Suppose that $$X$$ is a set and $$\mathcal{E}$$ is a collection of subsets that has the finite intersection property. Then $${ X \smallsetminus E \colon E \in \mathcal{E} }$$ is a subbase for a compact topology for $$X$$ . In this topology every element of $$\mathcal{E}$$ is closed. Now suppose that $$\mathcal{E}$$ satisfies the eventual finite intersection property. For each element $$E \in \mathcal{E}$$ we are using a different compact topology. Don't be fussy, use a topology that works!