Infimum of Real Subset

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Lemma

Let $S$ be a set of extended real numbers.

Let $S$ be bounded below in $\R$.

Let $T = S \cap \R$.


Then:

$S$ admits an infimum in $\R$ if and only if $T$ admits an infimum in $\R$

and, if $\inf S$ and $\inf T$ exist as real numbers:

$\inf S = \inf T$


Proof

We observe that $T$ constitutes the real numbers of $S$.

Since there is a real number that is a lower bound for $S$, $-\infty$ is not an element of $S$.

Accordingly, $\infty$ is the only possible element of $S \setminus T$.

Therefore:

$S$ is a subset of $T \cup \set \infty$


First, we show that $S$ and $T$ have the same set of lower bounds.


Let $b$ be a lower bound in $\R$ for $S$.

Then $b$ is a lower bound for $T$ as $T$ is a subset of $S$.

Therefore:

the set of lower bounds for $S$ is a subset of the set of lower bounds for $T$


Assume that $c$ is a lower bound in $\R$ for $T$.

Then $c$ is a lower bound for $T \cup \set \infty$ as well since $c < \infty$.

Accordingly, $c$ is a lower bound for $S$ since $S$ is a subset of $T \cup \set \infty$.

Therefore:

the set of lower bounds for $T$ is a subset of the set of lower bounds for $S$


We have:

the set of lower bounds for $T$ is a subset of the set of lower bounds for $S$
the set of lower bounds for $S$ is a subset of the set of lower bounds for $T$

Therefore:

the set of lower bounds for $T$ equals the set of lower bounds for $S$ by definition


Next, we show that $S$ and $T$ have the same infima.


We have that $S$ and $T$ have the same set of lower bounds.

Therefore, $S$ and $T$ have the same greatest lower bound in $\overline \R$.

Accordingly, as a corollary, if one of the sets $S$ and $T$ admits an infimum in $\R$, so does the other.

Furthermore, these infima are equal.

$\blacksquare$