Ordered Set with Multiple Maximal Elements has no Greatest Element

Theorem

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

Let $\struct {S, \preccurlyeq}$ have more than one maximal element.

Then $\struct {S, \preccurlyeq}$ has no greatest element.

Proof

Let $s$ and $t$ both be maximal elements of $\struct {S, \preccurlyeq}$ such that $s \ne t$.

Then by definition:

$\forall x \in S: s \preccurlyeq x \implies s = x$

and:

$\forall x \in S: t \preccurlyeq x \implies t = x$

Aiming for a contradiction, suppose $S$ has a greatest element $m$.

Then by definition:

$\forall y \in S: y \preccurlyeq m$

Hence:

$s \preccurlyeq m$

and also:

$y \preccurlyeq m$

But because $s$ and $t$ are both maximal elements of $\struct {S, \preccurlyeq}$:

$s = m$

and:

$t = m$

This contradicts the fact that $s \ne t$.

Hence by Proof by Contradiction there can be no greatest element of $\struct {S, \preccurlyeq}$.

$\blacksquare$