Filter Containing Complements is Not Proper
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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
- $\exists c \in F: c \preceq a \land c \preceq b$
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:
- $F = S$
$\blacksquare$