Top in Filter

Theorem
Let $\struct {S, \preceq}$ be a bounded above ordered set.

Let $F$ be a filter on $S$.

Then $\top \in F$

where $\top$ denotes the greatest element of $S$.

Proof
By definition of filter in ordered set:
 * $F$ is non-empty and upper.

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

By definition of greatest element:
 * $x \preceq \top$

Thus by definition of upper section:
 * $\top \in F$