Lower Closure is Closure Operator

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

Then
 * lower closure of set is a closure operator.

Inflationary
Let $X$ be a subset of $S$.

Let $x \in X$.

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

Thus by definition of lower closure of set:
 * $x \in X^\preceq$

Thus by definition of subset:
 * $X \subseteq X^\preceq$

Increasing
Let $X, Y$ be subsets of $S$ such that
 * $X \subseteq Y$

Let $x \in X^\preceq$.

By definition of lower closure of set:
 * $\exists y \in X: x \preceq y$

By definition of subset:
 * $y \in Y$

Thus by definition of lower closure of set:
 * $x \in Y^\preceq$

Thus by definition of subset:
 * $X^\preceq \subseteq Y^\preceq$

Idempotent
Let $X$ be a subset of $S$.

Let $x \in \left({X^\preceq}\right)^\preceq$.

By definition of lower closure of set:
 * $\exists y \in X^\preceq: x \preceq y$

By definition of lower closure of set:
 * $\exists z \in X: y \preceq z$

By definition of transitivity:
 * $x \preceq z$

Thus by definition of lower closure of set:
 * $x \in X^\preceq$

Thus by definition of subset:
 * $\left({X^\preceq}\right)^\preceq \subseteq X^\preceq$

By definition of inflationary:
 * $X \subseteq X^\preceq$

By definition of increasing:
 * $X^\preceq \subseteq \left({X^\preceq}\right)^\preceq$

Thus definition of set equality;
 * $\left({X^\preceq}\right)^\preceq = X^\preceq$

Thus by definition:
 * lower closure of set is a closure operator.