Infimum of Real Subset

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 $\R$ $T$ 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.