Finite Union of Bounded Subsets

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

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


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 $M = \struct {A, d}$.

Let $a_1, a_2 \in A$.

Let $K_1, K_2 \in \R$ such that:

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

Without loss of generality, let $a = a_1$ and $K = \max \set {K_1, K_2 + \map d {a_1, a_2} }$.

Then $\forall x \in S_1 \cup S_2$:


$x \in S_1$ and so $\map d {x, a} \le K_1 \le K$


$x \in S_2$ and so $\map d {x, a} = \map d {x, a_1} \le \map d {x, a_2} + \map d {a_2, a_1} \le K_2 + \map d {a_2, a_1} \le K$

Hence the result.