# 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 $\mathcal P\left({X}\right) :\equiv X \ne \varnothing \land \sup X \in F \implies \exists x \in X: x \in F$

where $X$ is subset of $S$.

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

By definition of empty set:

$\mathcal P\left({\varnothing}\right)$

We will prove that

$\forall x \in A, B \subseteq A: \mathcal P\left({B}\right) \implies \mathcal P\left({B \cup \left\{ {x}\right\} }\right)$

Let $x \in A, B \subseteq A$ such that

$\mathcal P\left({B}\right)$ (Induction Hypothesis)

Assume that

$B \cup \left\{ {x}\right\} \ne \varnothing$ and $\sup \left({B \cup \left\{ {x}\right\} }\right) \in F$

Case $B = \varnothing$:

$B \cup \left\{ {x}\right\} = \left\{ {x}\right\}$
$\sup \left\{ {x}\right\} = x$

By definition of singleton:

$x \in \left\{ {x}\right\}$

Thus

$\exists a \in B \cup \left\{ {x}\right\}: a \in F$

$\Box$

Case $B \ne \varnothing$:

$B$ is finite.
$B$ admits an supremum.
$\left\{ {x}\right\}$ admits an supremum.
 $\displaystyle \sup\left({B \cup \left\{ {x}\right\} }\right)$ $=$ $\displaystyle \sup\left({\bigcup\left\{ {B, \left\{ {x}\right\} }\right\} }\right)$ definition of union $\displaystyle$ $=$ $\displaystyle \sup \left\{ {\sup B, x}\right\}$ Supremum of Suprema: $\displaystyle$ $=$ $\displaystyle \left({\sup B}\right) \vee x$ definition of join

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:

$a \in B \cup \left\{ {x}\right\}$

Thus

$\exists a \in B \cup \left\{ {x}\right\}: a \in F$

$\Box$

Case $x \in F$:

By definition of union:

$x \in B \cup \left\{ {x}\right\}$

Thus

$\exists a \in B \cup \left\{ {x}\right\}: a \in F$

$\Box$

$\mathcal P\left({A}\right)$

Thus the result.

$\Box$

### Necessary Condition

Suppose that

for all non-empty finite subset $A$ of $S: \left({ \sup A \in F \implies \exists a \in A: a \in F}\right)$

Let $x, y \in S$ such that

$x \vee y \in F$
$\left\{ {x, y}\right\}$ is a finite set.

By definition of unordered tuple:

$x \in \left\{ {x, y}\right\}$

By definition of non-empty set:

$\left\{ {x, y}\right\}$ is a non-empty set.

By definition of join:

$\sup \left\{ {x, y}\right\} = x \vee y$

By assumption:

$\exists a \in \left\{ {x, y}\right\}: a \in F$

Thus by definition of unordered tuple:

$x \in F$ or $y \in F$

Hence $I$ is prime filter.

$\blacksquare$