Compact Subspace of Metric Space is Bounded

From ProofWiki
Jump to navigation Jump to search

Theorem

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

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


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


Proof

Let $a \in M$.

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

Let $B_n \left({a}\right)$ be the open $n$-ball of $a$.

Then $\displaystyle C \subseteq \bigcup_{n \mathop = 1}^\infty B_n \left({a}\right)$ because $\forall x \in C: d \left({x, a}\right) < n$ for some $n \in \N$.

Thus the collection $\left\{{B_n \left({a}\right): n \in \N}\right\}$ forms an open cover of $C$.

Because $C$ is compact, it has a finite subcover, say: $\left\{{B_{n_1} \left({a}\right), B_{n_2} \left({a}\right), \ldots, B_{n_r} \left({a}\right)}\right\}$.

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

Then:

$\displaystyle C \subseteq \bigcup_{n \mathop = 1}^r B_{n_r} \left({a}\right) = B_n \left({a}\right)$

The result follows by definition of bounded.

$\blacksquare$

Sources