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:
- $\map \sup D \in D$
$\blacksquare$