Union of Bounded Below Real Subsets is Bounded Below

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

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

Then $A \cup B$ is also bounded below.

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

Then by definition $A$ and $B$ both have a lower bound $L_A$ and $L_B$ respectively.

Suppose $L_A \ge L_B$.

Then:
 * $\forall a \in A: a \ge L_B$

and also, by definition:
 * $\forall b \in B: b \ge L_B$

and so $L_B$ is a lower bound for $A$.

Otherwise, suppose $L_A < L_B$.

Then:
 * $\forall b \in B: b \ge L_A$

and also, by definition:
 * $\forall a \in A: a \ge L_A$

Let $x \in A \cup B$.

Then from the above, either $x \ge L_A$ or $x \ge L_B$.

So either $L_A$ or $L_B$ is a lower bound for $A \cup B$.

Hence, by definition, $A \cup B$ is bounded below by either $L_A$ or $L_B$.