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:
\(\ds \inf S\) | \(=\) | \(\ds \inf T\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \inf S\) | \(\ge\) | \(\ds \inf T\) | |||||||||||
\(\ds \leadstoandfrom \ \ \) | \(\ds \forall \epsilon \in \R_{>0}: \forall s \in S: \exists t \in T: \, \) | \(\ds s + \epsilon\) | \(>\) | \(\ds t\) | 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:
\(\ds \forall \epsilon \in \R_{>0}: \forall s \in S: \exists t \in T: \, \) | \(\ds s + \epsilon\) | \(>\) | \(\ds t\) | |||||||||||
\(\ds \leadstoandfrom \ \ \) | \(\ds \inf S\) | \(\ge\) | \(\ds \inf T\) | Infima of two Real Sets | ||||||||||
\(\ds \leadstoandfrom \ \ \) | \(\ds \inf S\) | \(\ge\) | \(\ds \inf T \ge \inf S\) | as $\inf T \ge \inf S$ is true by Infimum of Subset | ||||||||||
\(\ds \leadstoandfrom \ \ \) | \(\ds \inf S\) | \(=\) | \(\ds \inf T\) |
$\blacksquare$