Completeness property for Infima: If $E \subseteq \mathbb{R}$ is non-empty and bounded below then $E$ has a finite infimum.The book explicitly states to use the reflection principle to prove that inf(E) exists and is finite
Thoughts: 1) By the reflection property a non-empty E $\subseteq \mathbb{R}$ has an infimum IFF $-E$ has a supremum.
2) Can use Supremum of $-E$ which is finite by the completeness axiom to be the greatest finite lower bound of $E$
I am fairly confident with my answer
Proof:
$\exists m \in E : m \leq x \space \space \forall x \in E$
$-m \geq -x$
$-m$ is an upperbound of $-E$
So $-E$ is non-empty and bounded above. Which means $Sup(-E)$ exists and is finite.
$Sup(-E) \leq -m \space \forall$ -m of $-E$
$-Sup(-E) \geq m$
$-Sup(-E)$ is the greatest lower bound of $E$
$-Sup(-E)= inf(E) \geq x \space \space \forall x \space \in E$
Thus $inf(E)$ exists and is finite.