Union of Upper Sections is Upper

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

Let $A$ be a set of subsets of $S$ such that
 * $\forall X \in A: X$ is an upper set.

Then:
 * $\bigcup A$ is also an upper set.

Proof
Let $x \in \bigcup A, y \in S$ such that
 * $x \preceq y$

By definition of union:
 * $\exists Y \in A: x \in Y$

By assumption:
 * $Y$ is an upper set.

By definition of upper set:
 * $y \in Y$

Thus by definition of union:
 * $y \in \bigcup A$

Hence
 * $\bigcup A$ is an upper set.