Maximal Element need not be Greatest Element

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

Let $M \in S$ be a maximal element of $S$.

Then $M$ is not necessarily the greatest element of $S$.

Proof
Proof by Counterexample:

Let $S = \set {a, b, c}$.

Let $\preccurlyeq$ be defined as:
 * $x \preccurlyeq y \iff \tuple {x, y} \in \set {\tuple {a, a}, \tuple {b, b}, \tuple {c, c}, \tuple {a, b}, \tuple {a, c} }$

A straightforward but laborious process determines that $\preccurlyeq$ is a partial ordering on $S$.

We have that:
 * $c \preccurlyeq x \implies c = x$

and:
 * $b \preccurlyeq x \implies b = x$

and so by definition, both $b$ and $c$ are maximal elements of $S$.

Suppose $b$ is the greatest element of $S$.

Then from Greatest Element is Unique it follows that $c$ can not be the greatest element of $S$.

Hence the result.

In fact, from the definition of the greatest element of $S$:
 * $x$ is the greatest element of $S$ $\forall y \in S: y \preccurlyeq x$

it can be seen directly that neither $b$ nor $c$ is the greatest element of $S$.

Also see

 * Maximal Element in Toset is Unique and Greatest