Supremum of Singleton

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.

$\blacksquare$

Also see

• Infimum of Singleton