# Characterization of Prime Ideal by Finite Infima

## Theorem

Let $L = \left({S, \wedge, \preceq}\right)$ be a meet semilattice.

Let $I$ be an ideal in $L$.

Then

$I$ is a prime ideal
for all non-empty finite subset $A$ of $S: \left({ \inf A \in I \implies \exists a \in A: a \in I}\right)$

## Proof

### Sufficient Condition

Let $I$ be a prime ideal.

Define $\mathcal P\left({X}\right) :\equiv X \ne \varnothing \land \inf X \in I \implies \exists x \in X: x \in I$

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 $\inf \left({B \cup \left\{ {x}\right\} }\right) \in I$

Case $B = \varnothing$:

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

By definition of singleton:

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

Thus

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

$\Box$

Case $B \ne \varnothing$:

$B$ is finite.
$B$ admits an infimum.
$\left\{ {x}\right\}$ admits an infimum.
 $\ds \inf\left({B \cup \left\{ {x}\right\} }\right)$ $=$ $\ds \inf\left({\bigcup\left\{ {B, \left\{ {x}\right\} }\right\} }\right)$ definition of union $\ds$ $=$ $\ds \inf \left\{ {\inf B, x}\right\}$ Infimum of Infima: $\ds$ $=$ $\ds \left({\inf B}\right) \wedge x$ definition of meet
$\inf B \in I$ or $x \in I$

Case $\inf B \in I$:

By Induction Hypothesis:

$\exists a \in B: a \in I$

By definition of union:

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

Thus

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

$\Box$

Case $x \in I$:

By definition of union:

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

Thus

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

$\Box$

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

Thus the result.

$\Box$

### Necessary Condition

Suppose that

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

Let $x, y \in S$ such that

$x \wedge y \in I$
$\left\{ {x, y}\right\}$ 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 meet:

$\inf \left\{ {x, y}\right\} = x \wedge y$

By assumption:

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

Thus by definition of unordered tuple:

$x \in I$ or $y \in I$

Hence by Characterization of Prime Ideal:

$I$ is prime ideal.

$\blacksquare$