Characterization of Locale/Statement 5 Implies Statement 3

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Theorem

Let $L = \struct{S, \preceq}$ be a complete Brouwerian lattice.

Then:

$L$ satisfies the infinite join distributive law

Proof

Let $A \subseteq S$.

Let $a \in S$.

Lemma 1

$\sup \set{a \wedge b : b \in A} \preceq a \wedge \sup A$

Proof of Lemma 1

By definition of meet::

$\forall b \in A : a \wedge b \preceq a, a \wedge b \preceq b$

By definition of upper bound:

$a$ is an upper bound for $\set{a \wedge b : b \in A}$

By definition of supremum:

$\sup \set{a \wedge b : b \in A} \preceq a$

From Finer Supremum Precedes Supremum:

$\sup \set{a \wedge b : b \in A} \preceq \sup A$

By definition of meet:

$\sup \set{a \wedge b : b \in A} \preceq a \wedge \sup A$

$\Box$

Lemma 2

$a \wedge \sup A \preceq \sup \set{a \wedge b : b \in A}$

Proof of Lemma 2

By definition of supremum:

$\forall b \in A : a \wedge b \preceq \sup \set{a \wedge b : b \in A}$

From Relative Pseudocomplement Preserves Order:

$\forall b \in A : a \to a \wedge b \preceq a \to \sup \set{a \wedge b : b \in A}$

By definition of supremum:

$\sup \set{a \to a \wedge b : b \in A} \preceq a \to \sup \set{a \wedge b : b \in A}$

From Inequality with Meet Operation is Equivalent to Inequality with Relative Pseudocomplement in Brouwerian Lattice:

$\forall b \in A : b \preceq a \to a \wedge b$

From Finer Supremum Precedes Supremum:

$\sup A \preceq \sup \set{a \to a \wedge b : b \in A}$

By Ordering Axiom $(3)$: Antisymmetry:

$\sup A \preceq a \to \sup \set{a \wedge b : b \in A}$

From Inequality with Meet Operation is Equivalent to Inequality with Relative Pseudocomplement in Brouwerian Lattice:

$a \wedge \sup A \preceq \sup \set{a \wedge b : b \in A}$

$\Box$

By Ordering Axiom $(1)$: Reflexivity:

$a \wedge \sup A = \sup \set{a \wedge b : b \in A}$

Since $A$ and $a$ were arbitrary:

$\forall A \subseteq S, a \in S : a \wedge \sup A = \sup \set{a \wedge b : b \in A}$

It follows that $L$ satisfies the infinite join distributive law by definition.

$\blacksquare$