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# Generalized Krein-Milman Theorem for Partially Ordered Sets. Part 1.

edited June 2018 in Chat

Introduction.

This series of posts provides a proof of a version of the Krein-Milman theorem for partially ordered sets. We provide necessary and sufficient conditions for our conclusion. Readers who are not familiar with extreme subsets, extreme points, or the Krein-Milman theorem may wish to consult https://en.wikipedia.org/wiki/Extreme_point and https://en.wikipedia.org/wiki/Krein–Milman_theorem . For convenience we provide Wikipedia's statement of the usual Krein-Milman theorem below. Just about any real analysis or functional analysis text will contain this material. For another version of the Krein-Milman theorem for ordered structures see https://arxiv.org/pdf/1301.0760.pdf .

After stating the Krein-Milman theorem we provide some examples that suggest the Krein-Milman theorem is open to generalization, even in vector spaces. We then give some notation and definitions together with some preliminary results.

Krein-Milman Theorem.

This theorem is taken from https://en.wikipedia.org/wiki/Krein–Milman_theorem . Formally, let $$X$$ be a locally convex topological vector space (assumed to be Hausdorff), and let $$K$$ be a compact convex subset of $$X$$. Then, the theorem states that $$K$$ is the closed convex hull of its extreme points.

Examples.

The four examples will use the terminology of vector spaces as given in the Wikipedia link. For completeness of the definitions suppose that $$E \subseteq A$$. Then $$E$$ is an \emph{extreme} subset of $$A$$. if, for all $$a_{0}, a_{1} \in A$$, the existence of a $$t \in (0, 1)$$ satisfying $$ta_{0} + (1 - t)a_{1} \in E$$ implies that $$a_{0}, a_{1} \in E$$ .

Example A.

Let $$B$$ be the closed unit ball in an infinite dimensional Hilbert space. The extreme points of $$B$$ are the unit vectors. Every element of $$B$$ is in the convex hull of the set of unit vectors. But then, even though $$B$$ is not compact in the norm topology, it is the convex hull of its extreme points. This suggests that we may be able to relax the compactness requirement.

Example B.

Let $$A= \lbrace (x,y) \in \mathbb{R}^{2} \colon 0 \leq x \leq 1 \rbrace$$. The minimal extreme subsets of $$A$$ are the lines $$x = 0$$ and $$x = 1$$. This suggests that if we give up compactness we may have to be content with minimal extreme subsets having some internal structure.

Example C.

Let $$I = \lbrace x \in \mathbb{R} \colon 0 \leq x < 1 \rbrace$$. The extreme subsets of $$I$$ are $$\lbrace 0 \rbrace$$ and $$I$$. The set $$\lbrace 0 \rbrace$$ is the only minimal extreme subset. This suggest our theorem will need a condition that says if we have a collection $$\mathcal{E}$$ of extreme subsets whose closed convex hull is smaller than the set in question then there is another extreme subset that, when we add it to the collection and then take the convex hull we get more of the set.

Example D.

Let $$A = \lbrace (x,y) \in \mathbb{R}^{2} \colon 0 \leq x, y \leq 1 \text{ and } xy = 0 \rbrace$$. The set $$A$$ is not convex. The minimal extreme subsets of $$A$$ are $$\lbrace (0,0) \rbrace$$, $$\lbrace (0,1) \rbrace$$, and $$\lbrace (1,0) \rbrace$$. Notice that $$A$$ is contained in the convex hull of its extreme points. Letting $$\mathsf{con}$$ denote the convex hull operator we have

$$A \cap \mathsf{con}(\lbrace (0,0), (0,1), (1,0) \rbrace ) = A \text{.}$$ This suggests the possibility of adapting the Krein-Milman theorem to non-convex sets.