Condition for Infimum of Subset to equal Infimum 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 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
\(\displaystyle \iff \ \ \) \(\displaystyle \inf S\) \(=\) \(\displaystyle \inf T\)

$\blacksquare$