# 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 $S: \paren {\sup A \in F \implies \exists a \in A: a \in F}$

## Proof

### 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 $B = \O$:

$B \cup \set x = \set x$
$\sup \set x = x$

By definition of singleton:

$x \in \set x$

Thus

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

$\Box$

Case $B \ne \O$:

$B$ is finite.
$B$ admits a supremum.
$\set x$ admits a supremum.
 $\displaystyle \map \sup {B \cup \set x}$ $=$ $\displaystyle \map \sup {\bigcup \set {B, \set x} }$ Definition of Set Union $\displaystyle$ $=$ $\displaystyle \sup \set {\sup B, x}$ Supremum of Suprema: $\displaystyle$ $=$ $\displaystyle \paren {\sup B} \vee x$ Definition of Join (Order Theory)

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:

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

$\Box$

Case $x \in F$:

By definition of union:

$x \in B \cup \set x$

Thus:

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

$\Box$

$\map \PP A$

Thus the result.

$\Box$

### 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

$x \vee y \in F$
$\set {x, y}$ 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.

$\blacksquare$