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:

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.

Also see

 * Suprema of two Real Sets