Greatest Element is Upper Bound

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

Let $T \subseteq S$.

Let $T$ have a greatest element $M \in T$.

Then $M$ is an upper bound of $T$.

It follows by definition that $T$ is bounded above.

Proof
Let $M \in T$ be a greatest element of $T$.

By definition:
 * $\forall y \in T: y \preceq M$

But as $T \subseteq S$, it follows that $M \in S$.

Hence:
 * $\exists M \in S: \forall y \in T: y \preceq M$

Thus $T$ is bounded above by the upper bound $M$.

Also see

 * Smallest Element is Lower Bound