Filters of Lattice of Power Set form Bounded Below Ordered Set

Theorem
Let $X$ be a set.

Let $L = \left({\mathcal P\left({X}\right), \cup, \cap, \subseteq}\right)$ be an inclusion lattice of power set of $X$.

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

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

Then $F$ is bounded below and $\bot_F = \left\{{X}\right\}$

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

Proof
By Singleton of Set is Filter in Lattice of Power Set:
 * $\left\{{X}\right\}$ is a filter on $L$.

Let $A \in \mathit{Filt}\left({L}\right)$.

By definition of non-empty set:
 * $\exists x: x \in A$

By definition of power set:
 * $x \subseteq X$

By definition of upper set:
 * $X \in A$

Thus by definitions of singleton and subset:
 * $\left\{{X}\right\} \subseteq A$

Thus by definitions:
 * $F$ is bounded below and $\bot_F = \left\{{X}\right\}$