0
$\begingroup$

This is the second of three self-answered questions which will culminate in a proof of necessary and sufficient conditions for Krein-Milman type conclusions.

The first question is here.

The third question is here.

Suppose that $\mathfrak{P} = \langle P, \leq, \bot \rangle$ is a partially ordered set. Let $\mathbf{B}$ be the collection of all nonempty bounded subsets of $P$. That is to say, if $B \in \mathbf{B}$ then there exists a $p \in P$ satisfying $b \leq p$ for all $b \in B$. Let $\mathsf{con} \colon \mathbf{B} \rightarrow P$ satisfy the following conditions:

  1. Increasing Condition. For all $B \in \mathbf{B}$ the element $\mathsf{con}(B)$ is an upper bound for $B$.
  2. Order Condition. For all $B_{0}, B_{1} \in \mathbf{B}$; if for all $b_{0} \in B_{0}$ there exists a $b_{1} \in B_{1}$ with $b_{0} \leq b_{1}$ then $\mathsf{con}(B_{0}) \leq \mathsf{con}(B)$.
  3. Idempotence Condition. For all $B \in \mathbf{B}$ we have $\mathsf{con}(\mathsf{con}(B)) = \mathsf{con}(B)$. Define $\mathbf{C} = \{ \mathsf{con}(B) \colon B \in \mathbf{B} \} $. Elements of $\mathbf{C}$ will be said to be convex. A convex structure will be denoted by $\mathfrak{C} = \langle P, \leq, \bot, \mathsf{con} \rangle$.

Suppose that $q\in P$ and the nonempty set $A \subseteq P \smallsetminus \{ \bot \}$ has $q$ as an upper bound. Are there necessary and sufficient conditions so that for all nonempty $E \subseteq A$ if $q \nleq \mathsf{con}(E)$ then there exists an $E^{*} \subseteq A$ satisfying (i) $E \subseteq E^{*}$, (ii) for all $e \in E$ and all $e^{*} \in E^{*} \smallsetminus E$ we have $e \nleq e^{*}$, and (iii) $q \leq \mathsf{con}(E^{*})$?

$\endgroup$
0
$\begingroup$

1. Definitions. In this collection of definitions we will suppose that $\mathfrak{P} = \langle P, \leq, \bot \rangle$ is a partially ordered set. Recall that $\bot$ is the least element of $P$.

Suppose that $A_{0}, A_{1} \subseteq P$ are not empty. Abusing notation, we will write $A_{0} \leq A_{1}$ if and only if $a_{0} \leq a_{1}$ for all $a_{0} \in A_{0}$ and all $a_{1} \in A_{1}$. We will write $A_{0} \leq_{w} A_{1}$ if and only if for all $a_{0} \in A_{0}$ there exists an $a_{1} \in A_{1}$with $a_{0} \leq a_{1}$.

Suppose that $\mathsf{con} \colon \mathbf{B} \rightarrow P$ satisfies the following conditions:

Increasing Condition. For $B \in \mathbf{B}$ we have $B \leq \mathsf{con}(B)$.

Order Condition. For $B_{0}, B_{1} \in \mathbf{B}$; if $B_{0} \leq_{w} B_{1}$ then we have $\mathsf{con}(B_{0})\leq \mathsf{con}(B_{1})$.

Idempotence Condition. For $B \in \mathbf{B}$ we have $\mathsf{con} \{ \mathsf{con}(B) \} = \mathsf{con}(B)$. Then we will say that $\mathsf{con}$ is a convex hull operator for $\mathfrak{P}$. We will also say that $\mathfrak{C} = \langle P, \leq, \bot, \mathsf{con} \rangle$ is a convex structure. Define $$ \mathbf{C} = \{ \mathsf{con}(B) \colon B \in \mathbf{B} \} \text{.} $$ The elements of $\mathbf{C}$ will be called convex elements. Since $\bot \leq_{w} B$ for any bounded $B$, the order condition implies that $\mathsf{con}(\{ \bot \} )$ is the smallest convex element. We will denote this element by $\bot_{\mathsf{con}}$. It may be the case that $\bot < \bot_{\mathsf{con}}$.

2. Theorem. Suppose that $\mathfrak{C} = \langle P, \leq, \bot, \mathsf{con} \rangle$ is a convex structure. Then $\mathbf{C}$ is the underlying set for a complete lower semilattice whose order is induced by $\leq$.

Proof. Suppose that $C \subseteq \mathbf{C}$ is not empty. Define $$ B = \{ b \in P \colon b \leq C \} \text{.} $$ The set $B$ is bounded. The order and idempotence conditions implies that for all $c \in C$ we have $$ \mathsf{con}(B) \leq \mathsf{con}(\{ c \} ) = c \text{.} $$ Thus $\mathsf{con}(B)$ is a lower bound in $\mathbf{C}$ for $C$. Now suppose that $a \in \mathbf{C}$ is a lower bound for $C$. Since $a \in P$ we see that $a \in B$. The idempotence and order conditions imply that $a = \mathsf{con}(\{ a \}) \leq \mathsf{con} B$. But then $\mathsf{con}(B)$ is the greatest lower bound, in $\mathbf{C}$, for $C$.

3. Corollary. Suppose that $\mathfrak{C} = \langle P, \leq, \bot, \mathsf{con} \rangle$ is a convex structure. For all nonempty $C \subseteq \mathbf{C}$ and all $p \in P$; if $p = \wedge C$ then $p$ is convex.

Proof. Suppose that $C \subseteq \mathbf{C}$ is not empty and $p \in P$ is the greatest lower bound for $C$ in $P$. Using the notation from theorem 1 we see that $p \in B$. But then $p \leq \mathsf{con}( \{ p \} ) \leq \mathsf{con}( B ) \leq p$.

4. Proposition. Suppose that $\mathfrak{C} = \langle P, \leq, \bot, \mathsf{con} \rangle$ is a convex structure, we have $q \in P$, the set $\mathcal{A}$ is a nonempty collection of nonempty subsets of ${\downarrow} q$, and the function $f \colon \mathcal{A} \rightarrow P$ satisfies: for all $A \in \mathcal{A}$ we have $A \leq f(A) \leq \mathsf{con}(A)$. Then $$ \mathsf{con}(\cup \mathcal{A}) = \mathsf{con}(\{ f(A) \colon A \in \mathcal{A} \} ) = \mathsf{con}(\{ \mathsf{con}(A) \colon A \in \mathcal{A} \} ) \text{.} $$

Proof. Using the order condition we have \begin{align} \cup \mathcal{A} &\leq_{w} \{ f(A) \colon A \in \mathcal{A} \} \leq_{w} \{ \mathsf{con}(A) \colon A \in \mathcal{A} \} \text{.} \\ \mathsf{con}(\cup \mathcal{A}) &\leq \mathsf{con}(\{ f(A) \colon A \in \mathcal{A} \}) \leq \mathsf{con}(\{ \mathsf{con}(A) \colon A \in \mathcal{A} \}) \text{.} \end{align} For each $A \in \mathcal{A}$ we have $A \leq_{w} \cup \mathcal{A}$; hence $\mathsf{con}(A) \leq \mathsf{con}(\cup \mathcal{A})$. This means $\{ \mathsf{con}(A) \colon A \in \mathcal{A} \} \leq_{w} \{ \mathsf{con}(\cup \mathcal{A}) \} $. The order and idempotent conditions imply $$ \mathsf{con}(\{ \mathsf{con}(A) \colon A \in \mathcal{A} \} ) \leq \mathsf{con}(\cup \mathcal{A}) \text{.} $$

5. Definitions. Throughout this collection of definitions we will suppose that $\mathfrak{C} = \langle P, \leq, \bot, \mathsf{con} \rangle$ is a convex structure.

For all $E \in \mathbf{B}$ and all $p \in P$ satisfying $E \cup \{ p \} \in \mathbf{B}$ we will say that $p$ augments $E$ if and only if $e \nleq p$ for all $e \in E$ and $\mathsf{con}(E) < \mathsf{con}(E \cup \{ p \} )$.

Suppose that $q \in P$ and the set $A \subseteq {\downarrow} q$ is not empty. We will say that $A$ satisfies the augmentation property if and only if for all nonempty $E \subseteq A$ if $q \nleq \mathsf{con}(E)$ then there exists a $p \in A$ that augments $E$. For all nonempty $E_{0}, E_{1} \subseteq A$ we will write $$ E_{0} \preceq E_{1} $$ if and only if $E_{0} \subseteq E_{1}$ and, for all $e_{1} \in E_{1} \smallsetminus E_{0}$, the element $e_{1}$ augments $E_{0}$.

6. Lemma. Suppose that $\mathfrak{C} = \langle P, \leq, \bot, \mathsf{con} \rangle$ is a convex structure. Suppose also that $A_{0}, A_{1} \in \mathsf{B}$ and $p \in P$ satisfy: 1. We have $A_{0} \subseteq A_{1}$. 2. The set $A_{1}o \cup \{ p \} $ is bounded. 3. We have $\mathsf{con}(A_{1}) < \mathsf{con}(A_{1} \cup \{ p \} )$. Then $\mathsf{con}(A_{0}) < \mathsf{con}(A_{0} \cup \{ p \} )$.

Proof. The order condition implies $\mathsf{con}(A_{0}) \leq \mathsf{con}(A_{0} \cup \{ p \} )$. For the purpose of deriving a contradiction suppose that $\mathsf{con}(A_{0}) = \mathsf{con}(A_{0} \cup \{ p \} )$. Since $$ A_{1} \cup \{ p \} = A_{1} \cup (A_{0} \cup \{ p \} ) $$ proposition 4 and the properties of a convex hull operator imply \begin{align} \mathsf{con}(A_{1} \cup \{ p \} ) &= \mathsf{con}(A_{1} \cup (A_{0} \cup \{ p \} )) \\ &= \mathsf{con}(\{ \mathsf{con}(A_{1}), \mathsf{con}(A_{0} \cup \{ p \} ) \} ) \\ &= \mathsf{con}(\{ \mathsf{con}(A_{1}), \mathsf{con}(A_{0}) \} ) \\ &= \mathsf{con}( \{ \mathsf{con}(A_{1}) \} ) \\ &= \mathsf{con}(A_{1}) \text{.} \end{align} This is the desired contradiction.

7. Lemma. Suppose that $\mathfrak{C} = \langle P, \leq, \bot, \mathsf{con} \rangle$ is a convex structure. The relation $\preceq$ is a partial order on $\mathbf{B}$.

Proof. Select $E \in \mathbf{B}$. Since inclusion is reflexive and $E \smallsetminus E$ is empty the relation $\preceq$ is reflexive. Since inclusion is anti-symmetric so is $\preceq$. Select $E_{0}, E_{1}, E_{2} e\in \mathbf{B}$ and suppose we have both $E_{0} \preceq E_{1}$ and $E_{1} \preceq E_{2}$. If $E_{0} = E_{1}$ or $E_{1}= E_{2}$ then $E_{0} \preceq E_{2}$. For the remainder of this proof we will suppose that neither $E_{1} \smallsetminus E_{0}$ nor $E_{2} \smallsetminus E_{1}$ is empty. Select $e_{2} \in E_{2}$. If $e_{2} \in E_{1}$ then, since $E_{0} \preceq E_{1}$ the element $e_{2}$ augments $E_{0}$. Suppose that $e_{2} \in E_{2} \smallsetminus E_{1}$. Since $E_{0} \subseteq E_{1}$ and $E_{1} \preceq E_{2}$ we have $\mathsf{con}(E_{1}) < \mathsf{con}(E_{1} \cup \{ e_{2} \} )$. Recalling lemma 6 we see that $\mathsf{con}(E_{0}) < \mathsf{con}(E_{0} \cup \{ p \} )$. But then $E_{0} \preceq E_{2}$.

8. Theorem. Suppose that $\mathfrak{C} = \langle P, \leq, \bot, \mathsf{con} \rangle$ is a convex structure. Suppose also that $q \in P$ and $A \subseteq {\downarrow} q$ is not empty. Then $A$ satisfies the augmentation property if and only if for all nonempty $E \subseteq A$ there exists an $E^{*} \subseteq A$ satisfying both $E \preceq E^{*}$ and $q \leq \mathsf{con}(E^{*})$.

Proof. In this paragraph we will suppose that $A$ satisfies the augmentation property. Choose a nonempty $E \subseteq A$. Define $$ \mathcal{G} = \{ G \subseteq A \colon E \preceq G \} \text{.} $$ Since $E \in \mathcal{G}$ the set $\mathcal{G}$ is not empty. The relation $\preceq$ partially orders $\mathcal{G}$. Suppose that $\mathcal{T} \subseteq \mathcal{G}$ is not empty and is totally ordered by inclusion. We have $E \subseteq \cup \mathcal{T}$. Select $g \in \cup \mathcal{T}$. For some $G \in \mathcal{T}$ we have $g \in G$. Since $E\preceq G$ the element $g$ augments $E$. But then $\cup \mathcal{T} \in \mathcal{G}$. Applying Zorn's lemma we see that $\mathcal{G}$ has a maximal element $E^{*}$ For the purpose of deriving a contradiction suppose that $q \nleq \mathsf{con}(E^{*})$. Since $A$ satisfies the augmentation property there exists a $p \in A$ that augments $E^{*}$. But then $E^{*} \prec E^{*} \cup \{ p \} $ which contradicts the maximality of $E^{*}$.

In this paragraph we will suppose that for all nonempty $E \subseteq A$ there exists an $E^{*} \subseteq A$ satisfying both $E \preceq E^{*}$ and $q \leq \mathsf{con}(E^{*})$. Select a nonempty $E \subseteq A$. If $q \leq \mathsf{con}(E)$ there is nothing to prove. Suppose that $q \nleq \mathsf{con}(E)$. Choose $E^{*} \subseteq A$ satisfying both $E \preceq E^{*}$ and $q \leq E^{*}$. Now choose $e^{*} \in E^{*} \smallsetminus E$ and note that $e^{*}$ augments $E$.

9. Proposition. Suppose that $\mathfrak{C} = \langle P, \leq, \bot, \mathsf{con} \rangle$ is a convex structure. The relation $\preceq$ is a partial order on $\mathbf{B}$.

Proof. Inclusion is reflexive. In the case that $E_{1} = E_{0}$ the meet and hull conditions are vacuous. But then $\preceq$ is reflexive. Since inclusion is anti-symmetric so is $\preceq$. Select $E_{0}, E_{1}, E_{2} \in \mathbf{B}$ and suppose that $E_{0} \preceq E_{1}$ and $E_{1} \preceq E_{2}$. If $E_{0} = E_{}$ or $E_{1} = E_{2}$ then $E_{0} \preceq E_{2}$. For the remainder of the proof we will suppose that neither $E_{1} \smallsetminus E_{0}$ nor $E_{2} \smallsetminus E_{1}$ is empty. Select $e_{0} \in E_{0}$ and $e_{2} \in E_{2} \smallsetminus E_{0}$. If $e_{2} \in E_{1}$ then, since $E_{0} \preceq E_{1}$, it is not the case that $e_{0} \leq e_{2}$. If $e_{2} \in E_{2} \smallsetminus E_{1}$ then, since $e_{0} \in E_{1}$ and $E_{1} \preceq E_{2}$, it is not the case that $e_{0} \leq e_{2}$.Thus the meet condition holds. Select $e_{2} \in E_{2}$. Lemma 6 implies that $\mathsf{con}(E_{0}) < \mathsf{con}(E_{0} \cup \{ e_{2} \} )$. But then the hull condition holds.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.