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:

\(\displaystyle \sup S\) \(=\) \(\displaystyle \sup T\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle \sup S\) \(\le\) \(\displaystyle \sup T\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle \forall \epsilon \in \R_{>0}: \forall s \in S: \exists t \in T: s\) \(<\) \(\displaystyle t + \epsilon\) 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:

\(\displaystyle \forall \epsilon \in \R_{>0}: \forall s \in S: \exists t \in T: s\) \(<\) \(\displaystyle t + \epsilon\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle \sup S\) \(\le\) \(\displaystyle \sup T\) Suprema of two Real Sets
\(\displaystyle \leadsto \ \ \) \(\displaystyle \sup S\) \(\le\) \(\displaystyle \sup T \le \sup S\) as $\sup T \le \sup S$ is true by Supremum of Subset
\(\displaystyle \leadsto \ \ \) \(\displaystyle \sup S\) \(=\) \(\displaystyle \sup T\)

$\blacksquare$