Maximal Element need not be Unique
Theorem
Let $\struct {S, \preccurlyeq}$ be an ordered set.
It is possible for $S$ to have more than one maximal element.
Proof
Consider the set $T$ defined as:
- $T = \set {0, 1}$
Let $S$ be defined as:
- $S := \powerset T \setminus T$
where $\powerset T$ denotes the power set of $T$.
That is:
- $S = \set {\O, \set 0, \set 1}$
Let $\preccurlyeq$ be the relation defined on $S$ as:
- $\forall a, b \in S: a \preccurlyeq b \iff a \subseteq b$
That is, $\preccurlyeq$ is the subset relation on $S$.
From Subset Relation is Ordering, $\struct {S, \preccurlyeq}$ is an ordered set.
Let $a \in S$ such that $\set 1 \preccurlyeq a$.
Then by inspection it is apparent that:
- $a = \set 1$
That is, $\set 1$ is a maximal element of $\struct {S, \preccurlyeq}$.
Similarly, let $a \in S$ such that $\set 0 \preccurlyeq a$.
Then by inspection it is apparent that:
- $a = \set 0$
That is, $\set 0$ is also a maximal element of $\struct {S, \preccurlyeq}$.
Hence $S$ has more than one maximal element.
$\blacksquare$
Also see
Sources
- 1996: Winfried Just and Martin Weese: Discovering Modern Set Theory. I: The Basics ... (previous) ... (next): Part $1$: Not Entirely Naive Set Theory: Chapter $2$: Partial Order Relations