Supremum of Singleton
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Theorem
Let $\struct {S, \preceq}$ be an ordered set.
Then for all $a \in S$:
- $\sup \set a = a$
where $\sup$ denotes supremum.
Proof
Since $a \preceq a$, $a$ is an upper bound of $\set a$.
Let $b$ be another upper bound of $\set a$.
Then necessarily $a \preceq b$.
It follows that indeed:
- $\sup \set a = a$
as desired.
$\blacksquare$