Compact Subspace of Metric Space is Bounded

Theorem

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

Let $C$ be a subspace of $M$.

If $C$ is compact, then it is bounded.

Let $a \in M$.

Let $n \in \N_{>0}$.

Let $\map {B_n} a$ be the open $n$-ball of $a$.

By definition of open $n$-ball, $\forall x \in C: \map d {x, a} < n$, there exists $n \in \N$ such that:

$\ds C \subseteq \bigcup_{n \mathop = 1}^\infty \map {B_n} a$

Thus the collection:

$\set {\map {B_n} a: n \in \N}$

forms an open cover of $C$.

Because $C$ is compact, it has a finite subcover, say:

$\set {\map {B_{n_1} } a, \map {B_{n_2} } a, \ldots, \map {B_{n_r} } a}$

Let $n = \max \set {n_1, n_2, \ldots, n_r}$.

Then:

$\ds C \subseteq \bigcup_{n \mathop = 1}^r \map {B_{n_r} } a = \map {B_n} a$

The result follows by definition of bounded.

$\blacksquare$

We are given that $C$ is compact.

So $C$ is a priori bounded.

Hence the result.

$\blacksquare$

1975: W.A. Sutherland: Introduction to Metric and Topological Spaces ... (previous) ... (next): $5$: Compact spaces: $5.4$: Properties of compact spaces: Proposition $5.4.1$