Filters form Complete Lattice

Theorem

Let $L = \left({S, \wedge, \preceq}\right)$ be a bounded above meet semilattice.

Let $F = \left({\mathit{Filt}\left({L}\right), \subseteq}\right)$ be an inclusion ordered set,

where $\mathit{Filt}\left({L}\right)$ denotes set of all filters on $L$.

Then $F$ is a complete lattice.

Proof

Define $L^{-1} := \left({S, \succeq}\right)$ be a dual ordered set of $L$.

By Dual Pairs (Order Theory):

$L^{-1}$ is a bounded below meet semilattice.

By Filters equal Ideals in Dual Ordered Set:

$\mathit{Filt}\left({L}\right) = \mathit{Ids}\left({L^{-1} }\right)$

where $\mathit{Ids}\left({L^{-1} }\right)$ denotes the set of all ideals in $L^{-1}$.

Thus by Ideals form Complete Lattice:

$F$ is a complete lattice.

$\blacksquare$

Sources

• Mizar article WAYBEL16:18