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:
- $-S = \set {-s: s \in S}$
- $-T = \set {-t: t \in T}$
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:
| \(\ds \inf S\) | \(\ge\) | \(\ds \inf T\) | ||||||||||||
| \(\ds \leadstoandfrom \ \ \) | \(\ds -\sup -S\) | \(\ge\) | \(\ds -\sup -T\) | by the lemma | ||||||||||
| \(\ds \leadstoandfrom \ \ \) | \(\ds \sup -S\) | \(\le\) | \(\ds \sup -T\) | |||||||||||
| \(\ds \leadstoandfrom \ \ \) | \(\ds \forall \epsilon \in \R_{>0}: \forall x \in -S: \exists y \in -T: \, \) | \(\ds x\) | \(<\) | \(\ds y + \epsilon\) | Suprema of two Real Sets | |||||||||
| \(\ds \leadstoandfrom \ \ \) | \(\ds \forall \epsilon \in \R_{>0}: \forall s \in S: \set {x: = -s}: \exists t \in T: \paren {y: = -t}: \, \) | \(\ds x\) | \(<\) | \(\ds y + \epsilon\) | as $s \in S \iff -s \in −S$ and $t \in T \iff -t \in −T$ | |||||||||
| \(\ds \leadstoandfrom \ \ \) | \(\ds \forall \epsilon \in \R_{>0}: \forall s \in S: \exists t \in T: \, \) | \(\ds -s\) | \(<\) | \(\ds -t + \epsilon\) | ||||||||||
| \(\ds \leadstoandfrom \ \ \) | \(\ds \forall \epsilon \in \R_{>0}: \forall s \in S: \exists t \in T: \, \) | \(\ds s + \epsilon\) | \(>\) | \(\ds t\) |
Lemma
Let $X$ be a set of real numbers.
Let $X$ admit an infimum.
Let $-X = \set {-x: x \in X}$.
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.
$\Box$
$\blacksquare$