Greatest Element is Upper Bound

Theorem

Let $\struct {S, \preceq}$ 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 $M$.

$\blacksquare$

Also see

• Smallest Element is Lower Bound

Sources

• 1977: K.G. Binmore: Mathematical Analysis: A Straightforward Approach ... (previous) ... (next): $\S 2$: Continuum Property: $\S 2.7$: Maximum and Minimum