Characterization of Prime Ideal by Finite Infima

Theorem
Let $L = \struct {S, \wedge, \preceq}$ 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 $S: \paren {\inf A \in I \implies \exists a \in A: a \in I}$

Sufficient Condition
Let $I$ be a prime ideal.

Define $\map P X: \equiv X \ne \O \land \inf X \in I \implies \exists x \in X: x \in I$

where $X$ is subset of $S$.

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

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

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

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

Assume that
 * $B \cup \set x \ne \O$ and $\map \inf {B \cup \set x} \in I$

Case $B = \O$:

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

By Infimum of Singleton:
 * $\inf \set x = x$

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

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

Case $B \ne \O$:

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

By Existence of Non-Empty Finite Infima in Meet Semilattice:
 * $B$ admits an infimum.

By Infimum of Singleton:
 * $\set x$ admits an infimum.

By Characterization of Prime Ideal:
 * $\inf B \in I$ or $x \in I$

Case $\inf B \in I$:

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

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

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

Case $x \in I$:

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

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

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

Thus the result.

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

Let $x, y \in S$ such that
 * $x \wedge y \in I$

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 meet:
 * $\inf \set {x, y} = x \wedge y$

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

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

Hence by Characterization of Prime Ideal:
 * $I$ is prime ideal.