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.

$\blacksquare$

Examples

Example 1

Let $S$ be the subset of the real numbers $\R$ defined as:

$S = \set {1, 2, 3}$

Then the greatest element of $S$ is $3$.

From Greatest Element is Supremum it follows that:

$\sup S = 3$

Example 2

Let $V$ be the subset of the real numbers $\R$ defined as:

$V := \set {x \in \R: x > 0}$

From Supremum of Subset of Real Numbers: Example 3, $V$ has no supremum.

It follows from Greatest Element is Supremum that $V$ has no greatest element either.