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# Generalized Krein-Milman Theorem for Partially Ordered Sets. Part 3.

edited June 2018 in Chat

Definitions.

Suppose that $$\mathsf{C} \subseteq P$$ , when equipped with the induced partial order is a complete lower semilattice that satisfies both $$\bot \in \mathsf{C}$$ and, for all nonempty $$A \subseteq \mathsf{C}$$ the set $$A$$ has a greatest lower bound in $$P$$ and that greatest lower bound agrees with the greatest lower bound of $$A$$ in $$\mathsf{C}$$ . The elements of $$\mathsf{C}$$ will be said to be \emph{convex} and $$\mathsf{C}$$ will be called a lower semilattice of convex regions. Next we define $$\mathsf{Bnd} = \{ B \in {\mathbf{Pwr}}^{+} \colon B \leq c \text{ for some } c \in \mathsf{C} \} \text{;}$$ $$\mathsf{con} : B \mapsto \wedge \{ c \in \mathsf{C} \colon B \leq c \} \quad (B \in \mathsf{Bnd}) \text{.}$$ The function $$\downarrow$$ will pull double duty. $$\downarrow \colon c \mapsto \{ p \in P \colon p \leq c \} \quad (c \in \mathsf{C}) \text{.}$$ These functions form a Galois connection.The function $$\mathsf{con}$$ is a closure operator.

Lemma 2.

Suppose that $$\mathcal{A} \subseteq {\mathsf{Bnd}}$$ and $$cup \mathcal{A} \in \mathsf{Bnd}$$ . Then $$\mathsf{con}(\cup \mathcal{A} = \mathsf{con}(\cup \lbrace \mathsf{con}(A) \colon A \in \mathcal{A} \rbrace )$$.

Proof.

Since $$\cup \mathcal{A} \in \mathsf{Bnd}$$ the set $$\mathcal{A}$$ is not empty. Then

• $$\mathsf{con}(\cup \mathcal{A}) \subseteq \qquad \text{since } \mathsf{con} \text{ is isotone and extensive}$$
• $$\mathsf{con}(\cup { \mathsf{con}(A) \colon A \in \mathcal{A} } ) \subseteq \qquad \text{since } \mathsf{con} \text{ is isotone}$$
• $$\mathsf{con}(\mathsf{con}(\cup \mathcal{A})) = \qquad \text{since } \mathsf{con} \text{ is idempotent}$$
• $$\mathsf{con}(\cup \mathcal{A}) \text{.}$$

The desired equation holds.

The reader may have noticed that our partially ordered sets have a least element but not necessarily a greatest element. Readers are free to assume that a greatest element exists. Most of what I do uses smaller elements. The idea is that if that the elements are too big maybe we don't know much about them. That need not mean we can't say anything about them. For example If $$p \in P$$ satisfies for all $$B \in \mathsf{Bnd}$$ with $$B \leq p$$ perhaps we have $${\mathsf{con}}(B) \leq p$$. We can interpret this as saying there is no evidence that $$p$$ is not convex. So much for that.

Suppose that $$\mathsf{C}$$ is a lower semilattice of convex regions for $$\langle P; \leq \bot \rangle$$ . Let $$d \in P$$ with $$d \leq c$$ for some $$c \in \mathsf{C}$$ and $$E \subseteq {\downarrow}d$$ .

We will say that $$E$$ satisfies the \emph{lower bound} property if, for all $$e_{0}, e_{1} \in E^{+}$$ that satisfy $$e_{0} \wedge e_{1} \neq \bot$$ \) , there exists an $$e \in E^{+}$$ with $$e \leq { e_{0}, e_{1} }$$ .

We will say that $$E$$ satisfies the \emph{augmentation} property for $$d$$ if, for all anti-chains $$A \subseteq E^{+}$$ either $$d \leq \mathsf{con}(E)$$ or there exists an $$e^{\ast} \in E$$ that satisfies the following requirements:the following conditions are satisfied:

• The set $$A \cup { e^{\ast} }$$ is an anti-chain.
• The set $$\left[ {\downarrow} d \cap {\mathsf{con}}(A \cup { e^{\ast} } ) \right] \smallsetminus \left[ {\downarrow} d \cap {\mathsf{con}}(A) \right]$$ is not empty. The augmentation property is designed to be used with Zorn's lemma.

We will say that $$E$$ satisfies the \emph{peripheral} property for $$d$$ if for all nonempty $$A \subseteq E$$ and all $$e^{\ast} \in E$$ that satisfy $$e^{\ast} \leq {\mathsf{con}}(A)$$ there exists an $$e \in A$$ with $$e^{\ast} \wedge e \neq \bot$$ .