Prime Element iff There Exists Way Below Open Filter which Complement has Maximum

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Theorem

Let $L = \struct {S, \vee, \wedge, \preceq}$ be a bounded below continuous distributive lattice.

Let $p \in S$ such that:

$p \ne \top$

where $\top$ denotes the top of $L$.


Then

$p$ is a prime element

if and only if:

there exists a way below open filter $F$ in $L$: $p = \map \max {\relcomp S F}$


Proof

Sufficient Condition

Let $p$ be a prime element.

By definition of continuous:

$\forall x \in S: x^\ll$ is directed

We will prove that:

$\forall x \in S: \paren {x \in \relcomp S {p^\preceq} \implies \exists y \in S: y \in \relcomp S {p^\preceq} \land y \ll x}$

Let $x \in S$ such that:

$x \in \relcomp S {p^\preceq}$

By definition of relative complement:

$x \notin p^\preceq$

By definition of lower closure of element:

$x \npreceq p$

By Axiom of Approximation in Up-Complete Semilattice:

$\exists y \in S: y \ll x \land y \npreceq p$

By definition of lower closure of element:

$y \notin p^\preceq$

By definition of relative complement:

$y \in \relcomp S {p^\preceq}$

Thus:

$\exists y \in S: y \in \relcomp S {p^\preceq} \land y \ll x$

$\Box$


By Prime Element iff Complement of Lower Closure is Filter:

$F := \relcomp S {p^\preceq}$ is way below open filter in $L$.

By definitions of antisymmetry and lower closure of element:

$\lnot \exists y \in S: y \in F \land y \prec p$

By Relative Complement of Relative Complement::

$\relcomp S F = p^\preceq$

By definitions of reflexivity and lower closure of element:

$p \in \relcomp S F$

Thus by definition of greatest element:

$p = \map \max {\relcomp S F}$

$\Box$


Necessary Condition

Let:

there exists a way below open filter $F$ in $L$: $p = \map \max {\relcomp S F}$

By Maximal Element of Complement of Filter is Meet Irreducible:

$p$ is meet irreducible.

Thus by Prime Element iff Meet Irreducible in Distributive Lattice:

$p$ is a prime element.

$\blacksquare$


Sources

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