Infimum of Union of Bounded Below Sets of Real Numbers

Theorem

Let $A$ and $B$ be sets of real numbers.

Let $A$ and $B$ both be bounded below.

Then:

$\map \inf {A \cup B} = \min \set {\inf A, \inf B}$

where $\inf$ denotes the infimum.

Let $A$ and $B$ both be bounded below.

By the Continuum Property, $A$ and $B$ both admit an infimum.

Let $x \in A \cup B$.

Then either $x \ge \inf A$ or $x \ge \inf B$ by definition of infimum.

Hence:

$x \ge \min \set {\inf A, \inf B}$

and so $\min \set {\inf A, \inf B}$ is certainly a lower bound of $A \cup B$.

It remains to be shown that $\min \set {\inf A, \inf B}$ is the greatest lower bound.

Aiming for a contradiction, suppose there exists $m \in \R$ such that:

$m > \min \set {\inf A, \inf B}$

and:

$\forall x \in A \cup B: x \ge m$

Then:

$\min \set {\inf A, \inf B} = \sup A$

and so:

$m > \sup A$

But then by definition of infimum:

$\exists a \in A: a < m$

and so $m$ is not a lower bound of $A \cup B$.

This contradicts our assumption that $m$ is an infimum of $A \cup B$.

It follows by Proof by Contradiction that $\min \set {\inf A, \inf B}$ is the infimum of $A \cup B$.

The same argument shows, mutatis mutandis, that if $\inf A \le \inf B$, $\min \set {\inf A, \inf B}$ is the infimum of $A \cup B$.

$\blacksquare$