Infima of two Real Sets

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Theorem

Let $S$ and $T$ be sets of real numbers.

Let $S$ and $T$ admit infima.


Then:

$\inf S \ge \inf T \iff \forall \epsilon \in \R_{>0}: \forall s \in S: \exists t \in T: s + \epsilon > t$


Proof

Let:

$-S = \left\{{-s: s \in S}\right\}$
$-T = \left\{{-t: t \in T}\right\}$

Observe that:

$s \in S \iff -s \in -S$
$t \in T \iff -t \in -T$


We know that $\inf S$ and $\inf T$ exist.

The expression $\inf S \ge \inf T$ exists as $\inf S$ and $\inf T$ exist.

In other words, for fixed sets $S$ and $T$, $\inf S \ge \inf T$ is either true or false.


We find:

\(\displaystyle \inf S\) \(\ge\) \(\displaystyle \inf T\)
\(\displaystyle \iff \ \ \) \(\displaystyle -\sup -S\) \(\ge\) \(\displaystyle -\sup -T\) by the lemma
\(\displaystyle \iff \ \ \) \(\displaystyle \sup -S\) \(\le\) \(\displaystyle \sup -T\)
\(\displaystyle \iff \ \ \) \(\displaystyle \forall \epsilon \in \R_{>0}: \forall x \in -S: \exists y \in -T: x\) \(<\) \(\displaystyle y + \epsilon\) Suprema of two Real Sets
\(\displaystyle \iff \ \ \) \(\displaystyle \forall \epsilon \in \R_{>0}: \forall s \in S: \left({x: = -s}\right): \exists t \in T: \left({y: = -t}\right): x\) \(<\) \(\displaystyle y + \epsilon\) as $s \in S \iff -s \in −S$ and $t \in T \iff -t \in −T$
\(\displaystyle \iff \ \ \) \(\displaystyle \forall \epsilon \in \R_{>0}: \forall s \in S: \exists t \in T: -s\) \(<\) \(\displaystyle -t + \epsilon\)
\(\displaystyle \iff \ \ \) \(\displaystyle \forall \epsilon \in \R_{>0}: \forall s \in S: \exists t \in T: s + \epsilon\) \(>\) \(\displaystyle t\)


Lemma

Let $X$ be a set of real numbers.

Let $X$ admit an infimum.

Let $-X = \left\{{-x: x \in X}\right\}$.


Then:

$ \sup -X = -\inf X$


Proof

Because $X$ admits an infimum, it follows that it is not empty.

The result follows by Negative of Infimum is Supremum of Negatives.

$\Box$

$\blacksquare$


Also see