Union of Non-Disjoint Bounded Subsets of Metric Space is Bounded

Theorem
Let $M = \struct {A, d}$ be a metric space.

Let $B$ and $C$ be bounded subsets of $M$ such that $B \cap C \ne \O$.

Let $\map \diam B$ and $\map \diam C$ denote the diameters of $B$ and $C$.

Then $B \cup C$ is a bounded subset of $M$ such that:
 * $\map \diam {B \cup C} \le \map \diam B + \map \diam C$

Proof
there exists $x, y \in B \cup C$ such that:
 * $\map d {x, y} > \map \diam B + \map \diam C$

$x$ and $y$ cannot both be in $B$ or $C$, otherwise either $\map d {x, y} \le \map \diam B$ or $\map d {x, y} \le \map \diam C$.

, let $x \in B$ and $y \in C$.

Let $z \in B \cap C$.

This is possible because $B \cap C \ne \O$ by hypothesis.

Then:

This contradicts our supposition that $\map d {x, y} > \map \diam B + \map \diam C$.

Hence: