Filter Containing Complements is Not Proper

Theorem

Let $L = \struct {S, \lor, \land, \preceq}$ be a bounded lattice.

Let $F \subseteq S$ be a filter on $L$.

Suppose there exist $a, b \in F$ such that:

$b$ is a complement of $a$.

Then:

$F = S$

Proof

By filter axiom $\paren 2$:

$\exists c \in F: c \preceq a \land c \preceq b$

By definition of complement:

$b \land a = \bot$

Thus, by definition of meet:

$c \preceq \bot$

Therefore:

\(\ds \forall x \in S: \, \)

\(\ds x\)

\(\succeq\)

\(\ds \bot\)

Definition of Bottom of Lattice

\(\ds \)

\(\succeq\)

\(\ds c\)

\(\ds \leadsto \ \ \)

\(\ds \forall x \in S: \, \)

\(\ds x\)

\(\in\)

\(\ds F\)

Filter Axiom $\paren 3$

\(\ds \leadsto \ \ \)

\(\ds S\)

\(\subseteq\)

\(\ds F\)

Hence, by definition of set equality:

$F = S$

$\blacksquare$