Sequentially Compact Metric Space has Finite Net

Theorem

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

Then, for every $\epsilon > 0$, there exists a finite $\epsilon$-net of $M$.

Proof

Suppose there is no finite $\epsilon$-net for $M$.

We aim to obtain a contradiction by constructing a sequence $\left \langle {x_n} \right \rangle$ in $M$ with no convergent subsequence.

Let $x_1 \in M$.

Suppose (inductively) that $x_1, x_2, \ldots, x_r$ have been chosen such that $d \left({x_i, x_j}\right) \ge \epsilon$ for all $i, j \le r: i \ne j$.

By hypothesis, $\left\{{x_1, x_2, \ldots, x_n}\right\}$ is not an $\epsilon$-net.

So there must exist $x_{r+1} \in M$ such that $d \left({x_{r+1}, x_i}\right) \ge \epsilon$ for all $1 \le i \le r$.

This completes the inductive step in the construction of $\left \langle {x_n} \right \rangle$ such that $d \left({x_m, x_n}\right) \ge \epsilon$ for all $m \ne n$.

Thus $\left \langle {x_n} \right \rangle$ has no Cauchy subsequence.

So by Convergent Sequence is Cauchy Sequence, it has no convergent subsequence either.

Thus by definition, $M$ is not sequentially compact.

The result follows from this contradiction.

$\blacksquare$