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 $K_1, K_2 \in \R$ 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.

$\blacksquare$

Sources

• W.A. Sutherland: Introduction to Metric and Topological Spaces (1975): Proposition $2.2.15$