Condition for Supremum of Subset to equal Supremum of Set
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Lemma
Let $S$ be a real set.
Let $T$ be a subset of $S$.
Let $S$ and $T$ admit suprema.
Then:
- $\sup S = \sup T \iff \forall \epsilon \in \R_{>0}: \forall s \in S: \exists t \in T: s < t + \epsilon$
Proof
Necessary Condition
Let $\sup S = \sup T$.
The aim is to establish that $\forall \epsilon \in \R_{>0}: \forall s \in S: \exists t \in T: s < t + \epsilon$.
We have:
\(\ds \sup S\) | \(=\) | \(\ds \sup T\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \sup S\) | \(\le\) | \(\ds \sup T\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \forall \epsilon \in \R_{>0}: \forall s \in S: \exists t \in T: s\) | \(<\) | \(\ds t + \epsilon\) | Suprema of two Real Sets |
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 $\sup S = \sup T$.
We have:
\(\ds \forall \epsilon \in \R_{>0}: \forall s \in S: \exists t \in T: s\) | \(<\) | \(\ds t + \epsilon\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \sup S\) | \(\le\) | \(\ds \sup T\) | Suprema of two Real Sets | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds \sup S\) | \(\le\) | \(\ds \sup T \le \sup S\) | as $\sup T \le \sup S$ is true by Supremum of Subset | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds \sup S\) | \(=\) | \(\ds \sup T\) |
$\blacksquare$