Infimum of Singleton

From ProofWiki
Jump to navigation Jump to search


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

Then for all $a \in S$:

$\inf \set a = a$

where $\inf$ denotes infimum.


Since $a \preceq a$, $a$ is a lower bound for $\set a$.

Let $b$ be another lower bound for $\set a$.

Then necessarily $b \preceq a$.

It follows that indeed:

$\inf \set a = a$

as desired.


Also see