Filters of Lattice of Power Set form Bounded Below Ordered Set

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 $\bot_F = \set X$