1.3 - Supremum and Infimum

1.3.1: Infimum Definitions

a

Infimum

A real number $g$ is the greatest lower bound, or the infimum, for a set $A \subseteq R$ if it meets the criteria that:

$g$ is a lower bound for $A$
If $g^{'}$ is any lower bound for $A$ then $g \geq g^{'}$ .

b

We prove the following lemma:

Infimum Lemma

Suppose $g \in R$ is a lower bound for a set $A \subseteq R$ . Then $g = inf (A)$ iff for all $ε > 0$ there is some $a \in A$ where $g + ε > a$ .

Proof
( $\Rightarrow$ ). Suppose $g = inf (A)$ , so then $g$ is both a lower bound and the greatest one. Let $ε > 0$ be arbitrary. Notice that then $g + ε > g$ , then sing $g$ is the greatest lower bound then $g + ε$ must not be a lower bound for $A$ , so then there is some $a \in A$ where $a < g + ε$ , per our requirements.

( $\Leftarrow$ ). Suppose for all $ε > 0$ there is some $a \in A$ where $g + ε > a$ . We need to show that $g$ is the infimum for $A$ . We already know that $g$ is a lower bound, satisfying (i). For (ii), let $g^{'}$ be any other lower bound for $A$ . We need to show that $g \geq g^{'}$ , so then for all $a \in A$ then $g^{'} \leq a$ . Choose $ε = g^{'} - g > 0$ since we've assumed that $g^{'}$ isn't the greatest lower bound. Thus, then $g + (g^{'} - g) = g^{'} > a$ is a contradiction as $g^{'}$ is a lower bound. Hence, $g^{'}$ must be the some other lower bound, so then $g^{'} - g < 0$ and thus $g^{'} < g$ as required.
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