# 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:

 $\displaystyle \sup S$ $=$ $\displaystyle \sup T$ $\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 \ \$ $\displaystyle \sup S$ $=$ $\displaystyle \sup T$

$\blacksquare$