Supremum of Doubleton in Totally Ordered Set

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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 $b \preccurlyeq a$.

Without loss of generality 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 mutatis mutandis:

$\map \sup D = a$

In either case:

$\map \sup D \in D$

$\blacksquare$