Characterization of Prime Ideal by Finite Infima

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Theorem

Let $L = \struct {S, \wedge, \preceq}$ be a meet semilattice.

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

Then

$I$ is a prime ideal

if and only if

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


Proof

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:

$\map P B$

This will be used as an 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

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

$\Box$


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.
\(\ds \map \inf {B \cup \set x}\) \(=\) \(\ds \map \inf {\bigcup \set {B, \set x} }\) Definition of Set Union
\(\ds \) \(=\) \(\ds \inf \set {\inf B, x}\) Infimum of Infima
\(\ds \) \(=\) \(\ds \paren {\inf B} \wedge x\) Definition of Meet

By Characterization of Prime Ideal:

$\inf B \in I$ or $x \in I$


Case $\inf B \in I$

By the induction hypothesis:

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

By definition of union:

$a \in B \cup \set x$

Thus:

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

$\Box$


Case $x \in I$

By definition of union:

$x \in B \cup \set x$

Thus

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

$\Box$


By Induction of Finite Set:

$\map P A$

Thus the result.

$\Box$


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.

$\blacksquare$


Sources

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