Proper Filter is Included in Ultrafilter in Boolean Lattice

Theorem

Let $L = \struct {S, \vee, \wedge, \preceq}$ be a Boolean lattice.

Let $F$ be a proper filter on $L$.

Then there exists ultrafilter $G$ on $L$: $F \subseteq G$

Proof

By Singleton of Bottom is Ideal:

$I := \set \bot$ is an ideal in $L$.

where $\bot$ denotes the bottom of $L$.

We will prove that

$F \cap I = \O$

Let $x \in I$.

By definition of singleton:

$x = \bot$

Thus by Bottom not in Proper Filter:

$x \notin F$

$\Box$

By If Ideal and Filter are Disjoint then There Exists Prime Filter Including Filter and Disjoint from Ideal:

there exists a prime filter $G$ on $L$: $F \subseteq G$ and $G \cap I = \O$

By definition of singleton:

$\bot \in I$

By definitions of empty set and intersection:

$\bot \notin G$

By Bottom not in Proper Filter:

$G$ is proper.

Thus by Proper and Prime iff Ultrafilter in Boolean Lattice:

$G$ is ultrafilter.

Thus $F \subseteq G$

$\blacksquare$

Sources

• 1980: G. Gierz, K.H. Hofmann, K. Keimel, J.D. Lawson, M.W. Mislove and D.S. Scott: A Compendium of Continuous Lattices

• Mizar article WAYBEL_7:26