Filters of Lattice of Power Set form Bounded Above 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 above and $\top_F = \powerset X$

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

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

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

Thus by definition of filter:

$A \subseteq \powerset X$

Thus by definitions:

$F$ is bounded above and $\top_F = \powerset X$

$\blacksquare$