# Are there necessary and sufficient conditions for Krein-Milman type conclusions?

This, the third of three self-answered questions, contains a proof of necessary and sufficient conditions for Krein-Milman type conclusions.

The first question is here.

The second question is here.

The Krein-Milman theorem appears in most if not all functional analysis texts. There is a fancy version:

Webster and Winkler: The Krein-Milman Theorem In Operator Convexity

There are more abstract versions: van de Vel: Theory of Convex Structures

Suppose that $B$ is the closed unit ball in an infinite dimensional Hilbert space. Even though $B$ is not compact it is the convex hull of its extreme points(the collection of unit vectors).

Suppose that $X = S^{1} \times [0, 1]$. Let us say that $C \subseteq X$ is \emph{convex} if and only if for all $c_{0}, c_{1} \in C$ every geodesic segment from $c_{0}$ to $c_{1}$ is also in $C$. For $E \subseteq A \subseteq X$ we will say that $E$ is an \emph{extreme} subset of $A$ if and only if for all $a_{0}, a_{1}$ if there there is a geodesic segment $G$ from $a_{0}$ to $a_{1}$ that intersects $E$ at a point other than an endpoint of the segment then $a_{0}, a_{1} \in E$. Using these definitions the minimal extreme subsets of $X$ are $S^{1} \times \{ 0 \}$ and $S^{1} \times \{ 1 \}$. They are not singletons.

Is there a Krein-Milman type theorem that includes these theorems and examples? Better yet, are there necessary and sufficient conditions for Krein-Milman type conclusions?

1. Definitions. Suppose that $\mathfrak{P} = \langle P, \leq, \bot \rangle$ is a partially ordered set and $\sqsubseteq$ is a quasi-reflexive binary relation on $P$. That is to say, for all $p_{0}, p_{1} \in P$; if $p_{0} \sqsubseteq p_{1}$ then $p_{0} \sqsubseteq p_{0}$ and $p_{1} \sqsubseteq p_{1}$ If $\sqsubseteq$ is quasi-reflexive then we will write $$\mathbf{Dom}(\sqsubseteq) = \{ p \in P \colon p \sqsubseteq p \} \text{.}$$ Abusing notation once again, we will write $${\sqsubseteq}(q) = \{ p \in P \colon p \sqsubseteq q \} \text{.}$$

We will say that $\sqsubseteq$ is a \emph{notion of extremity} for $\mathfrak{P}$ if and only if the following conditions are satisfied:

1. Order Condition. For $p, q \in \mathbf{Dom}(\sqsubseteq)$; if $p \sqsubseteq q$ then $p \leq q$.
2. Least Element Condition. There exists a $\bot_{\sqsubseteq} \in \mathbf{Dom}(\sqsubseteq)$ satisfying $\bot_{\sqsubseteq} \sqsubseteq q$ for all $q \in \mathbf{Dom}(\sqsubseteq)$.
3. Double Meet Condition. For $p_{0}, q_{0}, p_{1}, q_{1} \in \mathbf{Dom}(\sqsubseteq)$; if we have $p_{0} \sqsubseteq q_{0}$ and $p_{1} \sqsubseteq q_{1}$ then we also have $p_{0} \wedge p_{1} \sqsubseteq q_{0} \wedge q_{1}$.

The double meet condition asserts the existence of two greatest lower bounds in $P$. Since $\sqsubseteq$ is quasi-reflexive the double meet condition implies that $\mathbf{Dom}(\sqsubseteq)$ is the underlying set for a lower semilattice. For all $p, q \in \mathbf{Dom}(\sqsubseteq)$; if $p \sqsubseteq q$ then we will say that $p$ is an extreme subelement of $q$. Suppose that $q \in (\mathbf{Dom}({\sqsubseteq})) \smallsetminus \{ \bot_{\sqsubseteq} \}$. Let $M_{q}$ denote the set of minimal elements in the lower semilattice ${\sqsubseteq}(q)$. Define $$V_{q} = \{ p \in {\sqsubseteq}(q) \colon {\sqsubseteq}(q) \cap ({\downarrow}{p}) \text{ satisfies the eventual finite meet condition} \} \text{.}$$

Suppose that $\mathfrak{C} = \langle P, \leq, \bot, \mathsf{con} \rangle$ is a convex structure and $\sqsubseteq$ is a notion of extremity for $\mathfrak{P} = \langle P, \leq, \bot \rangle$. For all $q \in (\mathbf{Dom}(\sqsubseteq)) \smallsetminus \{ \bot_{\sqsubseteq} \}$ we will say that ${\sqsubseteq}(q)$ satisfies the augmentation property if and only if for all nonempty $E \subseteq (\mathbf{Dom}(\sqsubseteq)) \smallsetminus \{ \bot_{\sqsubseteq} \}$; the inequality $$q \wedge \mathsf{con}(E) < q$$ implies there exists an $e^{*} \in {\sqsubseteq}(q)$ satisfying $e \wedge e^{*} < e$ for all $e \in E$ and $$q \wedge \mathsf{con}(E) < q \wedge \mathsf{con}(E \cup \{ e^{*} \} ) \text{.}$$ Equation X ensures that $e^{*} \neq \bot_{\sqsubseteq}$. The inequality $e \wedge e^{*} < e$ is to ensure we do not pick a large element, say $q$, for $e^{*}$.

Before stating theorem 2 we provide some notation and prove a lemma. Suppose that $q \in \mathbf{Dom}(\sqsubseteq)$ and $E \subseteq {\sqsubseteq}(q)$. Define $$\mathcal{G} = \{ G \subseteq {\sqsubseteq}(q) \colon E \preceq G \} \text{.}$$ Recalling proposition 9 in question 2 we see that $\mathcal{G}$ is partially ordered by $\preceq$.

1. Lemma. Suppose that $\mathfrak{C} = \langle P, \leq, \bot, \mathsf{con} \rangle$ is a convex structure with notion of extremity $\sqsubseteq$. For all $q \in \mathbf{Dom}(\sqsubseteq)$ the binary relation $\preceq$ is a partial order on ${\mathcal{G}}_{q}$ with least element $E$.

Proof. The relation $\preceq$ is reflexive because inclusion is reflexive and, in the case where $G_{0} = G_{1}$, both the hull and meet conditions are vacuous. Since inclusion is anti-symmetric the relation $\preceq$ is also anti-symmetric. Suppose that $G_{0}, G_{1}, G_{2} \in \mathcal{G}$ with $G_{0} \preceq G_{1}$ and $G_{1} \preceq G_{2}$. Using the inclusion condition we have $G_{0} \subseteq G_{2}$. Select $g_{0} \in G_{0}$ and $g_{2} \in G_{2}$. Since $G_{0} \subseteq G_{1}$ and $G_{1} \preceq G_{2}$ we see that $g_{0} \wedge g_{2} < g_{0}$. Since $E \in \mathcal{G}$ the inclusion condition ensures that $E$ is the least element.

2. Theorem. Suppose that $\mathfrak{C} = \langle P, \leq, \bot, \mathsf{con} \rangle$ is a convex structure with notion of extremity $\sqsubseteq$. Suppose that $q \in \mathbf{Dom}(\sqsubseteq)$ and ${\sqsubseteq}(q)$ satisfies the augmentation property. Then $q \wedge M_{q} = q$ if and only if for all nonempty $A \subseteq {\sqsubseteq}(q) \smallsetminus \{ \bot_{\sqsubseteq} \}$ and all $p \in {\sqsubseteq}(q)$; if $p$ augments $A$ then there exists an $E \subseteq V_{q}$ satisfying $p \leq \mathsf{con}(A \cup E)$ and, for all $e \in E$, either $e \leq \mathsf{con}(A)$ or $e$ augments $A$.

Proof. In this paragraph we will suppose that $q \wedge M_{q} = q$. Select $p \in {\sqsubseteq}(q)$. Since $p \leq q$ and $M_{q} \subseteq V_{q}$ we have $p \leq \mathsf{con}(V_{q})$. Suppose that $A \subseteq {\sqsubseteq}(q) \smallsetminus \{ \bot_{\sqsubseteq} \}$ is not empty and $p \in {\sqsubseteq}(q)$ augments $A$. Suppose that $m \in M_{q}$ does not augment $A$. Then either $a \leq m$ for some $a \in A$ or $\mathsf{con}(A) = \mathsf{con}(A \cup \{ m \} )$. In either case we have $m \in \mathsf{con}(A)$. Define $$E = \{ m \in M_{q} \colon m \nleq \mathsf{con}(A) \} \text{.}$$ Then $E \subseteq V_{q}$ and every element of $E$ augments $A$. Since $M_{q} \subseteq A \cup E$ and $q \wedge \mathsf{con}(M_{q}) = q$ we see that $p \leq \mathsf{con}(A \cup E)$.

In this paragraph we will suppose that for all nonempty $A \subseteq {\sqsubseteq}(q) \smallsetminus \{ \bot_{\sqsubseteq} \}$ and all $p \in {\sqsubseteq}(q)$; if $p$ augments $A$ then there exists an $E \subseteq V_{q}$ satisfying $p \leq \mathsf{con}(A \cup E)$ and, for all $e \in E$, either $e \leq \mathsf{con}(A)$ or $e$ augments $A$. For the purpose of deriving a contradiction suppose that $q \wedge M_{q} \neq q$. There are two cases: either $\{ q, \mathsf{con}(M_{q}) \}$ does not have a greatest lower bound or $q \wedge \mathsf{con}(M_{q}) < q$. In either case we have $q \nleq \mathsf{con}(M_{q})$. Since the augmentation property holds for ${\sqsubseteq}(q)$ there exists a $p \in {\sqsubseteq}(q)$ that augments $M_{q}$. Recalling the hypothesis in the first sentence of this paragraph there exists an $E \subseteq V_{q}$ satisfying $p \leq \mathsf{con}(M_{q} \cup E)$ and, for all $e \in E$, either $e \leq \mathsf{con}(M_{q})$ or $e$ augments $M_{q}$. Since $p \nleq \mathsf{con}(M_{q})$ there exists an $e \in E$ that augments $M_{q}$. Recalling theorem 2 in question 1 there exists a minimal $m \leq e$. But then $e$ can not augment $M_{q}$. This is the desired contradiction.

Postscript. Readers who have reached this point may observe that while the usual Krein-Milman theorem is about extreme points this result is a point-free version with minimal extreme elements, not singletons.

Suppose the partially ordered set is the power set of some set $X$ with inclusion as the partial order. This result will yield minimal extreme subsets. If we want the minimal extreme subsets to be singletons we need an additional item: If $E_{1}$ is an extreme subset with more than one element then there is a nonempty extreme subset $E_{0} \subseteq E_{1}$ that is a proper subset of $E_{1}$. In the usual Krein-Milman theorem this statement is a consequence of the separation theorem. In van de Vel's work this is a part of the hypotheses for extreme subsets.

It is an easy exercise to check that all the minimal extreme subsets are singletons if and only if the statement in the previous paragraph holds.