Union of Bounded Below Real Subsets is Bounded Below
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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 $L_B$.
$\blacksquare$
Sources
- 1975: W.A. Sutherland: Introduction to Metric and Topological Spaces ... (previous) ... (next): $1$: Review of some real analysis: Exercise $1.5: 3$