Continuous iff Way Below Closure is Ideal and Element Precedes Supremum

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Theorem

Let $L = \struct {S, \wedge, \preceq}$ be an up-complete meet semilattice.

Then

$L$ is continuous

if and only if

$\forall x \in S: x^\ll$ is an ideal in $L$ and $x \preceq \map \sup {x^\ll}$ and:
for every ideal $I$ in $L$: $x \preceq \sup I \implies x^\ll \subseteq I$

where $x^\ll$ denotes the way below closure of $x$.


Proof

Sufficient Condition

Let $L$ be continuous.

Let $x \in S$.

By definition of continuous:

$x^\ll$ is directed.

By Way Below Closure is Lower Section:

$x^\ll$ is a lower section.

Thus by definition:

$x^\ll$ is an ideal in $L$.

By definition of continuous:

$L$ satisfies the axiom of approximation.

By the axiom of approximation:

$x = \map \sup {x^\ll}$

Thus by definition of reflexivity:

$x \preceq \map \sup {x^\ll}$

Let $I$ be an ideal in $L$ such that

$x \preceq \sup I$

We will prove that

$x^\ll \subseteq I$

Let $y \in x^\ll$

By definition of way below closure:

$y \ll x$

Thus by Way Below iff Second Operand Preceding Supremum of Ideal implies First Operand is Element of Ideal:

$y \in I$

$\Box$


Necessary Condition

Assume that:

$\forall x \in S: x^\ll$ is an ideal in $L$ and $x \preceq \map \sup {x^\ll}$ and:
for every ideal $I$ in $L$: $x \preceq \sup I \implies x^\ll \subseteq I$

Let $x \in S$.

By assumption:

$x \preceq \map \sup {x^\ll}$

By Way Below implies Preceding:

$x$ is upper bound for $x^\ll$

By definition of supremum:

$\map \sup {x^\ll} \preceq x$

Thus by definition of antisymmetry:

$x = \map \sup {x^\ll}$

Thus by definition:

$L$ satisfies the axiom of approximation.

Thus by assumption and definition of ideal:

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

Hence $L$ is continuous.

$\blacksquare$


Sources

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