Meet Semilattice Filter iff Ordered Set Filter
Theorem
Let $\struct {S, \wedge, \preceq}$ be a meet semilattice.
Let $F \subseteq S$ be a non-empty subset of $S$.
Then:
- $F$ is a meet semilattice filter of $\struct {S, \wedge, \preceq}$ if and only if $F$ is an ordered set filter of $\struct {S, \preceq}$.
Proof
Necessary Condition
Let $F$ be a meet semilattice filter of $\struct {S, \wedge, \preceq}$.
To show that $F$ is an ordered set filter of $\struct {S, \preceq}$ it is sufficient to show:
\(\ds \forall x, y \in F: \exists z \in F:\) | \(\ds z \preceq x \text{ and } z \preceq y \) |
Let $x, y \in F$.
Let $z = x \wedge y$.
By definition of meet semilattice filter, $F$ is a subsemilattice, so:
- $z \in F$
By definition of Meet:
- $z \preceq x \text{ and } z \preceq y$
The result follows.
$\Box$
Sufficient Condition
Let $F$ be an ordered set filter of $\struct {S, \preceq}$.
To show that $F$ is a meet semilattice filter of $\struct {S, \wedge, \preceq}$ it is sufficient to show:
$F$ is a subsemilattice of $S$: | \(\ds \forall x, y \in F:\) | \(\ds x \wedge y \in F \) |
Let $x, y \in F$.
By definition of ordered set filter:
- $\exists z \in F : z \preceq x \text { and } z \preceq y$
By definition of meet:
- $z \preceq x \wedge y$
By definition of ordered set filter:
- $x \wedge y \in F$
The result follows.
$\blacksquare$