# Finite Union of Bounded Subsets

## Theorem

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

Then the union of any finite 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:

$(1): \quad \forall x \in S_1: d \left({x, a_1}\right) \le K_1$
$(2): \quad \forall x \in S_2: d \left({x, a_2}\right) \le K_2$

Without loss of generality, 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.

$\blacksquare$