Supremum of Singleton

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Let $\struct {S, \preceq}$ be an ordered set.

Then for all $a \in S$:

$\sup \set a = a$

where $\sup$ denotes supremum.


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.


Also see