Suprema of two Real Sets

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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 $\R_{>0}$
\(\displaystyle \leadsto \ \ \) \(\displaystyle \sup S\) \(\le\) \(\displaystyle \sup T\)

$\blacksquare$


Also see