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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!

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