Filters of Lattice of Power Set form Bounded Below Ordered Set

From ProofWiki
Jump to navigation Jump to search

Theorem

Let $X$ be a set.

Let $L = \struct {\powerset X, \cup, \cap, \subseteq}$ be an inclusion lattice of power set of $X$.

Let $F = \struct {\map {\operatorname {Filt} } L, \subseteq}$ be an inclusion ordered set,

where $\map {\operatorname {Filt} } L$ denotes the set of all filters on $L$.


Then $F$ is bounded below and $\bot_F = \set X$

where $\bot_F$ denotes the smallest element of $F$.


Proof

By Singleton of Set is Filter in Lattice of Power Set:

$\set X$ is a filter on $L$.

Let $A \in \map {\operatorname {Filt} } L$.

By definition of non-empty set:

$\exists x: x \in A$

By definition of power set:

$x \subseteq X$

By definition of upper section:

$X \in A$

Thus by definitions of singleton and subset:

$\set X \subseteq A$

Thus by definitions:

$F$ is bounded below and $\bot_F = \set X$

$\blacksquare$


Sources