Characterization of Prime Element in Meet Semilattice

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Theorem

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

Let $p \in S$,


Then:

$p$ is prime element

if and only if:

for all non-empty finite subsets $A$ of $S$:
if $\inf A \preceq p$, then there exists element $x$ of $A$ such that $x \preceq p$.


Proof

Sufficient Condition

Let $p$ be prime element.

Let $A$ be non-empty finite subsets of $S$.

Define

$\map P X : \equiv X \ne \O \land \inf X \preceq p \implies \exists x \in X: x \preceq p$

where $X \subseteq S$.

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

$\map P B$

and

$B \cup \set x \ne \O$ and $\map \inf {B \cup \set x} \preceq p$

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

$\exists z \in B \cup \set x: z \preceq p$

$\Box$


Case $B \ne \O$.

By Subset of Finite Set is Finite:

$B$ is finite.

By Existence of Non-Empty Finite Infima in Meet Semilattice:

$B, \set x$ admit infima.

By Infimum of Infima:

$\map \inf {B \cup \set x} = \inf B \wedge \inf \set x$

By definition of prime element:

$\inf B \preceq p$ or $x \preceq p$

Case $\inf B \preceq p$.

By assumption:

$\exists z \in B: z \preceq p$

By definition of union:

$z \in B \cup \set x$

Thus

$\exists z \in B \cup \set x: z \preceq p$

$\Box$


Case $x \preceq p$.

By definitions of union and singleton:

$x \in B \cup \set x$

Thus

$\exists z \in B \cup \set x: z \preceq p$

$\Box$


By definition of empty set:

$\map P \O$

By Induction of Finite Set:

$\map P A$

Thus

if $\inf A \preceq p$, then there exists element $x$ of $A$ such that $x \preceq p$.

$\Box$


Necessary Condition

Suppose that

for all non-empty finite subsets $A$ of $S$:
if $\inf A \preceq p$, then there exists element $x$ of $A$ such that $x \preceq p$.

Let $x, y \in S$ such that

$x \wedge y \preceq p$

Define $A := \set {x, y}$

By definitions of non-empty set and finite set:

$A$ is non-empty finite subset of $S$.

By definition of meet:

$\inf A = x \wedge y$

By assumption:

$\exists z \in A: z \preceq p$

Thus by definition of unordered tuple:

$x \preceq p$ or $y \preceq p$

$\blacksquare$


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