# Condition for Infimum of Subset to equal Infimum of Set

## Lemma

Let $S$ be a real set.

Let $T$ be a subset of $S$.

Let $S$ and $T$ admit infima.

Then:

$\inf S = \inf T \iff \forall \epsilon \in \R_{>0}: \forall s \in S: \exists t \in T: s + \epsilon > t$

## Proof

### Necessary Condition

Let $\inf S = \inf T$.

The aim is to establish that $\forall \epsilon \in \R_{>0}: \forall s \in S: \exists t \in T: s + \epsilon > t$.

We have:

 $\displaystyle \inf S$ $=$ $\displaystyle \inf T$ $\displaystyle \implies \ \$ $\displaystyle \inf S$ $\ge$ $\displaystyle \inf T$ $\displaystyle \iff \ \$ $\displaystyle \forall \epsilon \in \R_{>0}: \forall s \in S: \exists t \in T: s + \epsilon$ $>$ $\displaystyle t$ Infima of two Real Sets

### Sufficient Condition

Let $\forall \epsilon \in \R_{>0}: \forall s \in S: \exists t \in T: s + \epsilon > t$.

The aim is to establish that $\inf S = \inf T$.

We have:

 $\displaystyle \forall \epsilon \in \R_{>0}: \forall s \in S: \exists t \in T: s + \epsilon$ $>$ $\displaystyle t$ $\displaystyle \iff \ \$ $\displaystyle \inf S$ $\ge$ $\displaystyle \inf T$ Infima of two Real Sets $\displaystyle \iff \ \$ $\displaystyle \inf S$ $\ge$ $\displaystyle \inf T \ge \inf S$ as $\inf T \ge \inf S$ is true by Infimum of Subset $\displaystyle \iff \ \$ $\displaystyle \inf S$ $=$ $\displaystyle \inf T$

$\blacksquare$