Condition for Supremum of Subset to equal Supremum of Set

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

\(\ds \leadsto \ \ \)

\(\ds \sup S\)

\(=\)

\(\ds \sup T\)

$\blacksquare$