Upper Bound for Subset

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

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

Let $\left({T, \preceq}\right)$ be a subset of $\left({S, \preceq}\right)$.

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

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

But as $\forall y \in T: y \in S$ by definition of subset, it follows that:
 * $\forall y \in T: y \preceq U$.

Hence the result, again by definition of upper bound.

Also see

 * Lower Bound for Subset