Infimum of Subset of Extended Real Numbers is Arbitrarily Close

Theorem
Let $A \subseteq \overline \R$ be a subset of the extended real numbers.

Let $b$ be an infimum (in $\R$) of $A$.

Let $\epsilon \in \R_{>0}$.

Then:
 * $\exists x \in A \cap \R: x - b < \epsilon$

Proof
We have that:
 * $A$ is a a set of extended real numbers
 * $A$ is bounded below (in $\R$) as the real number $b$ is a lower bound for $A$.

From this follows by Infimum of Real Subset:
 * $\map \inf {A \cap \R} \in \R$ as $\inf A \in \R$

Let $\epsilon \in \R_{>0}$.

Then: