Continuous iff For Every Element There Exists Ideal 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

for every element $x$ of $S$ there exists ideal $I$ in $L$:
$x \preceq \sup I$ and for every ideal $J$ in $L: x \preceq \sup J \implies I \subseteq J$


Proof

Sufficient Condition

Let $L$ be continuous.

By Continuous iff Way Below Closure is Ideal and Element Precedes Supremum:

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

Thus

for every element $x$ of $S$ there exists ideal $I$ in $L$:
$x \preceq \sup I$ and for every ideal $J$ in $L: x \preceq \sup J \implies I \subseteq J$

$\Box$


Necessary Condition

Assume that

for every element $x$ of $S$ there exists ideal $I$ in $L$:
$x \preceq \sup I$ and for every ideal $J$ in $L: x \preceq \sup J \implies I \subseteq J$

Let $x \in S$.

There exists ideal $I$ in $L$:

$x \preceq \sup I$ and for every ideal $J$ in $L: x \preceq \sup J \implies I \subseteq J$

We will prove that

$I \subseteq x^\ll$

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

Let $y \in I$.

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

$x \preceq \sup J$

Then $I \subseteq J$

Thus by definition of subset:

$y \in J$

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

$y \ll x$

Thus by definition of way below closure:

$y \in x^\ll$

$\Box$


By Way Below iff Second Operand Preceding Supremum of Ideal implies First Operand is Element of Ideal

$\forall y \in x^\ll: y \in I$

By definition of subset:

$x^\ll \subseteq I$

Thus by definition of set equality:

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

Hence by Continuous iff Way Below Closure is Ideal and Element Precedes Supremum:

$L$ is continuous

$\blacksquare$


Sources

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