Supremum is Dual to Infimum

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Theorem

Let $\left({S, \preceq}\right)$ be an ordered set.

Let $a \in S$ and $T \subseteq S$.


The following are dual statements:

$a$ is a supremum for $T$
$a$ is an infimum for $T$


Proof

By definition, $a$ is a supremum for $T$ iff:

$a$ is an upper bound for $T$
$a \preceq b$ for all upper bounds $b$ of $T$

The dual of this statement is:

$a$ is a lower bound for $T$
$b \preceq a$ for all lower bounds $b$ of $T$

by Dual Pairs (Order Theory).


By definition, this means $a$ is an infimum for $T$.


The converse follows from Dual of Dual Statement (Order Theory).

$\blacksquare$


Also see