Upper Closure of Element is Filter

Theorem
Let $\left({S, \preceq}\right)$ ne an ordered set.

Let $s$ be an element of $S$.

Then $s^\succeq$ is filter in $\left({S, \preceq}\right)$

where $s^\succeq$ denotes the upper closure of $s$.

Proof
By Singleton is Directed and Filtered Subset
 * $\left\{ {s}\right\}$ is a filtered subset of $S$

By Filtered iff Upper Closure Filtered:
 * $\left\{ {s}\right\}^\succeq$ is a filtered subset of $S$

By Upper Closure is Upper Set:
 * $\left\{ {s}\right\}^\succeq$ is a upper set in $S$

By Upper Closure of Singleton
 * $\left\{ {s}\right\}^\succeq = s^\succeq$

By definition of reflexivity:
 * $s \preceq s$

By definition of upper closure of element:
 * $s \in s^\succeq$

Thus by definition:
 * $s^\succeq$ is non-empty filtered and upper.

Thus by definition:
 * $s^\succeq$ is a filter in $\left({S, \preceq}\right)$