Filters of Lattice of Power Set form Bounded Above 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 above and $\top_F = \mathcal P\left({X}\right)$

where $\top_F$ denotes the greatest element of $F$.

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

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

Thus by definition of filter:
 * $A \subseteq \mathcal P\left({X}\right)$

Thus by definitions:
 * $F$ is bounded above and $\top_F = \mathcal P\left({X}\right)$