# 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:

\(\displaystyle s\) | \(\le\) | \(\displaystyle t\) | |||||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle s\) | \(\le\) | \(\displaystyle t < t + \epsilon\) | as $t < t + \epsilon$ is true | |||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle s\) | \(<\) | \(\displaystyle t + \epsilon\) |

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:

\(\displaystyle \nexists t \in T : s\) | \(\le\) | \(\displaystyle t\) | |||||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle \forall t \in T : s\) | \(>\) | \(\displaystyle t\) | so, $s$ is an upper bound for $T$ | |||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle s\) | \(\ge\) | \(\displaystyle \sup T\) | as (a) $s$ is an upper bound for $T$ and (b) $\sup T$ is the least upper bound of $T$ | |||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle \sup S\) | \(\ge\) | \(\displaystyle s \ge \sup T\) | as $\sup S \ge s$ is true since (a) $\sup S$ is an upper bound for $S$ and (b) $s \in S$ | |||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle \sup T\) | \(\ge\) | \(\displaystyle \sup S \ge s \ge \sup T\) | as $\sup S \le \sup T$ is true | |||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle \sup T\) | \(\ge\) | \(\displaystyle s \ge \sup T\) | ||||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle s\) | \(=\) | \(\displaystyle \sup T\) |

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

A $t \in T$ exists such that:

\(\displaystyle \sup T - t\) | \(<\) | \(\displaystyle \epsilon\) | by Supremum of Subset of Real Numbers is Arbitrarily Close | ||||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle \sup T\) | \(<\) | \(\displaystyle t + \epsilon\) | ||||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle s\) | \(<\) | \(\displaystyle t + \epsilon\) | 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:

\(\displaystyle \sup S - s\) | \(<\) | \(\displaystyle \epsilon\) | by Supremum of Subset of Real Numbers is Arbitrarily Close | ||||||||||

\(\displaystyle \iff \ \ \) | \(\displaystyle s\) | \(>\) | \(\displaystyle \sup S - \epsilon\) |

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

A $t \in T$ exists such that:

\(\displaystyle s\) | \(<\) | \(\displaystyle t + \epsilon\) | as $\forall \epsilon' \in \R_{>0}: \forall s' \in S: \exists t \in T: s' < t + \epsilon'$ | ||||||||||

\(\displaystyle \iff \ \ \) | \(\displaystyle t\) | \(>\) | \(\displaystyle s - \epsilon\) |

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

Then:

\(\displaystyle t\) | \(>\) | \(\displaystyle s - \epsilon\) | |||||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle t\) | \(>\) | \(\displaystyle s - \epsilon \text { and } s > \sup S - \epsilon\) | as $s > \sup S - \epsilon$ is true | |||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle t\) | \(>\) | \(\displaystyle s - \epsilon \text{ and } s - \epsilon > \sup S - 2 \epsilon\) | ||||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle t\) | \(>\) | \(\displaystyle s - \epsilon > \sup S - 2 \epsilon\) | ||||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle t\) | \(>\) | \(\displaystyle \sup S - 2 \epsilon\) | ||||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle \sup T\) | \(\ge\) | \(\displaystyle t > \sup S - 2 \epsilon\) | as $\sup T \ge t$ is true since (a) $\sup T$ is an upper bound for $T$ and (b) $t \in T$ | |||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle \sup T\) | \(>\) | \(\displaystyle \sup S - 2 \epsilon\) | ||||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle \sup T\) | \(\ge\) | \(\displaystyle \sup S\) | as $\epsilon$ is an arbitrary element of $\R_{>0}$ | |||||||||

\(\displaystyle \leadsto \ \ \) | \(\displaystyle \sup S\) | \(\le\) | \(\displaystyle \sup T\) |

$\blacksquare$