Greatest Element is Unique

Theorem

Let $\struct {S, \preceq}$ be a ordered set.

If $S$ has a greatest element, then it can have only one.

That is, if $a$ and $b$ are both greatest elements of $S$, then $a = b$.

Let $V$ be a basic universe.

Let $\RR \subseteq V \times V$ be an ordering.

Let $A$ be a subclass of the field of $\RR$.

Suppose $A$ has a greatest element $g$ with respect to $\RR$.

Then $g$ is unique.

Let $a$ and $b$ both be greatest elements of $S$.

Then by definition:

$\forall y \in S: y \preceq a$

$\forall y \in S: y \preceq b$

Thus it follows that:

$b \preceq a$

$a \preceq b$

But as $\preceq$ is an ordering, it is antisymmetric.

Hence, by definition of antisymmetric relation, $a = b$.

$\blacksquare$

1960: Paul R. Halmos: Naive Set Theory ... (previous) ... (next): $\S 14$: Order
1965: Seth Warner: Modern Algebra ... (previous) ... (next): Chapter $\text {IV}$: Rings and Fields: $23$. The Field of Rational Numbers
1982: P.M. Cohn: Algebra Volume 1 (2nd ed.) ... (previous) ... (next): Chapter $1$: Sets and mappings: $\S 1.5$: Ordered Sets
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: Exercise $5$
2000: James R. Munkres: Topology (2nd ed.) ... (previous) ... (next): $1$: Set Theory and Logic: $\S 3$: Relations
2000: James R. Munkres: Topology (2nd ed.) ... (previous) ... (next): $1$: Set Theory and Logic: $\S 3$: Relations: Exercise $3.11$