Ordered Set may not have Maximal Element

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

It may be the case that $S$ has no maximal elements.

Proof
Consider the set $S$ defined as:
 * $S = \N \setminus \set 0$

That is, $S$ is the set of natural numbers with $0$ removed.

Let $\preccurlyeq$ be the ordering on $S$ defined as:
 * $\forall a, b \in S: a \preccurlyeq b \iff a \divides b$

where $a \divides b$ denotes that $a$ is a divisor of $b$.

From Divisor Relation on Positive Integers is Partial Ordering, $\struct {S, \preccurlyeq}$ is a partially ordered set.

$S$ has a maximal element $m$.

Consider the natural number $n = 2 m$.

Then:
 * $m \divides n$

but:
 * $m \ne n$

and so by definition $m$ is not a maximal element of $S$.

Hence by Proof by Contradiction it follows that there exists no such $m$.

Hence the result.