Supremum of Doubleton in Totally Ordered Set

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

Let $D = \set {a, b} \subseteq S$ be an arbitrary doubleton subset of $S$.

Then:
 * $\map \sup D \in D$

where $\sup D$ denotes the supremum of $D$.

Proof
Let $D = \set {a, b} \subseteq S$ as asserted.

As $S$ is a totally ordered set, either $a \preccurlyeq b$ or $b \preccurlyeq a$.

let $a \preccurlyeq b$.

We have by definition of supremum that:
 * $\forall x \in D: x \preccurlyeq \map \sup D$
 * $\forall y \in S:$ if $y$ is an upper bound of $D$, then $\map \sup D \preccurlyeq y$.

Now we have:
 * $\forall x \in D: x \preccurlyeq b$

Let $y \in S$ be an upper bound of $D$.

Then:
 * $\forall x \in D: x \preccurlyeq y$

In particular:
 * $b \preccurlyeq y$

Hence by definition:
 * $\map \sup D = b$

Similarly, if $b \preccurlyeq a$, then :
 * $\map \sup D = a$

In either case:


 * $\map \sup D \in D$