Maximal Element need not be Greatest Element

Theorem
Let $\left({S, \preccurlyeq}\right)$ be an ordered set.

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

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

Proof
Proof by Counterexample:

Let $S = \left\{{a, b, c}\right\}$ and let $\preccurlyeq$ be defined as:
 * $x \preccurlyeq y \iff \left({x, y}\right) \in \left\{{\left({a, a}\right), \left({b, b}\right), \left({c, c}\right), \left({a, b}\right), \left({a, c}\right)}\right\}$

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 $S$ iff $\forall y \in S: y \preccurlyeq x$

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