Finite Union of Bounded Subsets

Theorem
Let $$M = \left({A, d}\right)$$ be a metric space.

Then the union of any number of bounded subsets of $$M$$ is itself bounded.

Proof
It is sufficient to prove this for two subsets, as the general result follows by induction.

Suppose $$S_1$$ and $$S_2$$ are bounded subsets of $$M = \left({A, d}\right)$$.

Let $$a_1, a_2 \in A$$.

Let $$K_1, K_2 \in \reals$$ such that:
 * $$\forall x \in S_1: d \left({x, a_1}\right) \le K_1$$;
 * $$\forall x \in S_2: d \left({x, a_2}\right) \le K_2$$.

WLOG let $$a = a_1$$ and $$K = \max \left\{{K_1, K_2 + d \left({a_1, a_2}\right)}\right\}$$.

Then $$\forall x \in S_1 \cup S_2$$:
 * Either $$x \in S_1$$ and so $$d \left({x, a}\right) \le K_1 \le K$$;
 * or $$x \in S_2$$ and so $$d \left({x, a}\right) = d \left({x, a_1}\right) \le d \left({x, a_2}\right) + d \left({a_2, a_1}\right) \le K_2 + d \left({a_2, a_1}\right) \le K$$.

Hence the result.