Characterization of Prime Filter by Finite Suprema

Theorem
Let $L = \struct {S, \vee, \preceq}$ be a join semilattice.

Let $F$ be a filter in $L$.

Then
 * $F$ is a prime filter


 * for all non-empty finite subset $A$ of $S: \paren {\sup A \in F \implies \exists a \in A: a \in F}$
 * for all non-empty finite subset $A$ of $S: \paren {\sup A \in F \implies \exists a \in A: a \in F}$

Sufficient Condition
Let $F$ be a prime ideal.

Define $\map \PP X: \equiv X \ne \O \land \sup X \in F \implies \exists x \in X: x \in F$

where $X$ is subset of $S$.

Let $A$ be a non-empty finite subset of $S$.

By definition of empty set:
 * $\map \PP \O$

We will prove that:
 * $\forall x \in A, B \subseteq A: \map \PP B \implies \map \PP {B \cup \set x}$

Let $x \in A, B \subseteq A$ such that:
 * $\map \PP B$ (Induction Hypothesis)

Assume that:
 * $B \cup \set x \ne \O$ and $\map \sup {B \cup \set x} \in F$

Case $B = \O$:

By Union with Empty Set:
 * $B \cup \set x = \set x$

By Supremum of Singleton:
 * $\sup \set x = x$

By definition of singleton:
 * $x \in \set x$

Thus
 * $\exists a \in B \cup \set x: a \in F$

Case $B \ne \O$:

By Subset of Finite Set is Finite:
 * $B$ is finite.

By Existence of Non-Empty Finite Suprema in Join Semilattice:
 * $B$ admits a supremum.

By Supremum of Singleton:
 * $\set x$ admits a supremum.

By definition of prime filter:
 * $\sup B \in F$ or $x \in F$

Case $\sup B \in F$:

By Induction Hypothesis:
 * $\exists a \in B: a \in F$

By definition of union:
 * $a \in B \cup \set x$

Thus:
 * $\exists a \in B \cup \set x: a \in F$

Case $x \in F$:

By definition of union:
 * $x \in B \cup \set x$

Thus:
 * $\exists a \in B \cup \set x: a \in F$

By Induction of Finite Set:
 * $\map \PP A$

Thus the result.

Necessary Condition
Suppose that
 * for all non-empty finite subset $A$ of $S: \paren {\sup A \in F \implies \exists a \in A: a \in F}$

Let $x, y \in S$ such that
 * $x \vee y \in F$

By Unordered Pair is Finite:
 * $\set {x, y}$ is a finite set.

By definition of unordered tuple:
 * $x \in \set {x, y}$

By definition of non-empty set:
 * $\set {x, y}$ is a non-empty set.

By definition of join:
 * $\sup \set {x, y} = x \vee y$

By assumption:
 * $\exists a \in \set {x, y}: a \in F$

Thus by definition of unordered tuple:
 * $x \in F$ or $y \in F$

Hence $I$ is prime filter.