Greatest Element is Supremum

Theorem
Let $\struct {S, \preceq}$ be an ordered set.

Let $T \subseteq S$.

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

Then $M$ is the supremum of $T$ in $S$.

Proof
Let $M$ be the greatest element of $T$.

Then by definition:
 * $\forall x \in T: x \preceq M$

By definition of supremum, it is necessary to show that:


 * $(1): \quad M$ is an upper bound of $T$ in $S$
 * $(2): \quad M \preceq U$ for all upper bounds $U$ of $T$ in $S$.

By Greatest Element is Upper Bound, $M$ is an upper bound of $T$ in $S$.

It remains to be shown that:
 * $M \preceq U$ for all upper bounds $U$ of $T$ in $S$.

Let $U \in S$ be an upper bound of $T$ in $S$.

By definition of upper bound:


 * $\forall t \in T: t \preceq U$

We have that $M \in T$.

Therefore:
 * $M \preceq U$

Hence the result.

Also see

 * Smallest Element is Infimum