Upper Closure of Coarser Subset is Subset of Upper Closure

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

Let $A, B$ be subsets of $S$ such that
 * $A$ is coarser than $B$.

Then $A^\succeq \subseteq B^\succeq$

Proof
Let $x \in A^\succeq$

By definition of upper closure of subset:
 * $\exists y \in A: y \preceq x$

By definition of coarser subset:
 * $\exists z \in B: z \preceq y$

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

Thus by definition of upper closure of subset:
 * $x \in B^\succeq$