# Suprema of two Real Sets

## Theorem

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

Let $S$ and $T$ admit suprema.

Then:

$\sup S \le \sup T \iff \forall \epsilon \in \R_{>0}: \forall s \in S: \exists t \in T: s < t + \epsilon$

## Proof

### Necessary Condition

Let $\sup S \le \sup T$.

The aim is to establish that:

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

Let $\epsilon \in \R_{>0}$.

Let:

$S' = \set {s \in S : \exists t \in T : s \le t}$

There are two cases for an $s \in S$:

$s \in S'$, in other words: $\exists t \in T : s \le t$

and:

$s \in S \setminus S'$, in other words: $\nexists t \in T : s \le t$

First, consider the case that $s \in S'$.

Let $s \in S'$.

Then $t \in T$ exists such that:

 $\ds s$ $\le$ $\ds t$ $\ds \leadsto \ \$ $\ds s$ $\le$ $\ds t < t + \epsilon$ as $t < t + \epsilon$ is true $\ds \leadsto \ \$ $\ds s$ $<$ $\ds t + \epsilon$

We have found:

$\forall \epsilon \in \R_{>0}: \forall s \in S': \exists t \in T: s < t + \epsilon$

Next, consider the second case, that $s \in S \setminus S'$.

Let $s \in S \setminus S'$.

Then:

 $\ds \nexists t \in T : s$ $\le$ $\ds t$ $\ds \leadsto \ \$ $\ds \forall t \in T : s$ $>$ $\ds t$ so, $s$ is an upper bound for $T$ $\ds \leadsto \ \$ $\ds s$ $\ge$ $\ds \sup T$ as (a) $s$ is an upper bound for $T$ and (b) $\sup T$ is the least upper bound of $T$ $\ds \leadsto \ \$ $\ds \sup S$ $\ge$ $\ds s \ge \sup T$ as $\sup S \ge s$ is true since (a) $\sup S$ is an upper bound for $S$ and (b) $s \in S$ $\ds \leadsto \ \$ $\ds \sup T$ $\ge$ $\ds \sup S \ge s \ge \sup T$ as $\sup S \le \sup T$ is true $\ds \leadsto \ \$ $\ds \sup T$ $\ge$ $\ds s \ge \sup T$ $\ds \leadsto \ \$ $\ds s$ $=$ $\ds \sup T$

So, $s = \sup T$ is the only element of $S \setminus S'$.

A $t \in T$ exists such that:

 $\ds \sup T - t$ $<$ $\ds \epsilon$ by Supremum of Subset of Real Numbers is Arbitrarily Close $\ds \leadsto \ \$ $\ds \sup T$ $<$ $\ds t + \epsilon$ $\ds \leadsto \ \$ $\ds s$ $<$ $\ds t + \epsilon$ as $s = \sup T$

We have found:

$\forall \epsilon \in \R_{>0}: \forall s \in S \setminus S': \exists t \in T: s < t + \epsilon$

We conclude by combining the results for the two cases:

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

$\Box$

### Sufficient Condition

Let $\forall \epsilon \in \R_{>0}: \forall s \in S: \exists t \in T: s < t + \epsilon$.

The aim is to establish that $\sup S \le \sup T$.

Let $\epsilon \in \R_{>0}$.

An $s \in S$ exists such that:

 $\ds \sup S - s$ $<$ $\ds \epsilon$ by Supremum of Subset of Real Numbers is Arbitrarily Close $\ds \iff \ \$ $\ds s$ $>$ $\ds \sup S - \epsilon$

Let $s$ be such an element of $S$.

A $t \in T$ exists such that:

 $\ds s$ $<$ $\ds t + \epsilon$ as $\forall \epsilon' \in \R_{>0}: \forall s' \in S: \exists t \in T: s' < t + \epsilon'$ $\ds \iff \ \$ $\ds t$ $>$ $\ds s - \epsilon$

Let $t$ be such an element of $T$.

Then:

 $\ds t$ $>$ $\ds s - \epsilon$ $\ds \leadsto \ \$ $\ds t$ $>$ $\ds s - \epsilon \text { and } s > \sup S - \epsilon$ as $s > \sup S - \epsilon$ is true $\ds \leadsto \ \$ $\ds t$ $>$ $\ds s - \epsilon \text{ and } s - \epsilon > \sup S - 2 \epsilon$ $\ds \leadsto \ \$ $\ds t$ $>$ $\ds s - \epsilon > \sup S - 2 \epsilon$ $\ds \leadsto \ \$ $\ds t$ $>$ $\ds \sup S - 2 \epsilon$ $\ds \leadsto \ \$ $\ds \sup T$ $\ge$ $\ds t > \sup S - 2 \epsilon$ as $\sup T \ge t$ is true since (a) $\sup T$ is an upper bound for $T$ and (b) $t \in T$ $\ds \leadsto \ \$ $\ds \sup T$ $>$ $\ds \sup S - 2 \epsilon$ $\ds \leadsto \ \$ $\ds \sup T$ $\ge$ $\ds \sup S$ as $\epsilon$ is an arbitrary element of $\R_{>0}$ $\ds \leadsto \ \$ $\ds \sup S$ $\le$ $\ds \sup T$

$\blacksquare$