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:

\(\ds \map \inf {A \cap \R}\)

\(\in\)

\(\ds \R\)

\(\ds \leadsto \ \ \)

\(\ds \exists x \in A \cap \R: \, \)

\(\ds x - b\)

\(<\)

\(\ds \epsilon\)

Infimum of Subset of Real Numbers is Arbitrarily Close

$\blacksquare$