Characterizing Property of Infimum of Subset of Real Numbers

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Theorem

Let $S \subset \R$ be a non-empty subset of the real numbers.

Let $S$ be bounded below.

Let $\alpha \in \R$.


The following are equivalent:

$(1): \quad \alpha$ is the infimum of $S$
$(2): \quad \alpha$ is a lower bound for $S$
and:
$\forall \epsilon \in \R_{> 0}$ there exists $x \in S$ with $x < \alpha + \epsilon$


Proof

$(1)$ implies $(2)$

Let $\alpha$ be the infimum of $S$.

Then by definition, $\alpha$ is a lower bound for $S$.

Let $\epsilon>0$.

Because $\alpha+\epsilon>\alpha$, it is not a lower bound for $S$.

Thus there exists $x\in S$ with $x < \alpha + \epsilon$.

$\Box$


$(2)$ implies $(1)$

Let $\alpha$ be a lower bound of $S$ such that $\forall \epsilon > 0$ there exists $x \in S$ with $x < \alpha + \epsilon$.

Let $d \in \R$ be a lower bound of $S$.

We have to prove that $d \le \alpha$.

Aiming for a contradiction, suppose $d > \alpha$.

Let $\epsilon = d - \alpha > 0$.

Then there exists $x \in S$ such that $x < \alpha + \left({d - \alpha}\right) = d$.

But this contradicts our assumption that $d$ is a lower bound of $S$.

$\blacksquare$


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