Upper Bound is Upper Bound for Subset

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

Let $A, B$ be subsets of $S$ such that
 * $B \subseteq A$

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

Let $U$ be an upper bound for $A$.

Then $U$ is an upper bound for $B$.

Proof
Assume that:
 * $U$ is upper bound for $A$.

By definition of upper bound:
 * $\forall x \in A: x \preceq U$

By definition of subset:
 * $\forall x \in B: x \in A$

Hence:
 * $\forall x \in B: x \preceq U$

Thus by definition
 * $U$ is sn upper bound for $B$.