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:
| \(\ds \nexists t \in T: \, \) | \(\ds s\) | \(\le\) | \(\ds t\) | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \forall t \in T: \, \) | \(\ds s\) | \(>\) | \(\ds t\) | 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 | |||||||||||
| \(\ds \leadstoandfrom \ \ \) | \(\ds s\) | \(>\) | \(\ds \sup S - \epsilon\) |
Let $s$ be such an element of $S$.
A $t \in T$ exists such that:
| \(\ds s\) | \(<\) | \(\ds t + \epsilon\) | as $\forall \epsilon' \in \R_{>0}: \forall s' \in S: \exists t \in T: s' < t + \epsilon'$ | |||||||||||
| \(\ds \leadstoandfrom \ \ \) | \(\ds t\) | \(>\) | \(\ds s - \epsilon\) |
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 $\R_{>0}$ | ||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \sup S\) | \(\le\) | \(\ds \sup T\) |
$\blacksquare$