# Infima of two Real Sets

## 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.

$\Box$

$\blacksquare$