Lower Closure of Element is Ideal

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

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

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

where $s^\preceq$ denotes the lower closure of $s$.

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

By Directed iff Lower Closure Directed:
 * $\left\{ {s}\right\}^\preceq$ is a directed subset of $S$

By Lower Closure is Lower Set:
 * $\left\{ {s}\right\}^\preceq$ is a lower set in $S$

By Lower Closure of Singleton
 * $\left\{ {s}\right\}^\preceq = s^\preceq$

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

By definition of lower closure of element:
 * $s \in s^\preceq$

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

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