A sequentially compact metric space is totally bounded

Let \((X, d)\) be a sequentially compact metric space. We want to prove that \(X\) is totally bounded: for every \(\varepsilon > 0\), \(X\) can be covered by finitely many open balls of radius \(\varepsilon\).

Suppose towards a contradiction that \(X\) is not totally bounded for some \(\varepsilon_0 > 0\). We construct a sequence \((x_n)_{n=1}^\infty\) inductively as follows: 1. Choose an arbitrary point \(x_1 \in X\). 2. Since \(B(x_1, \varepsilon_0)\) does not cover \(X\), choose \(x_2 \in X \setminus B(x_1, \varepsilon_0)\). Then \(d(x_2, x_1) \geq \varepsilon_0\). 3. Inductively, given \(x_1, x_2, \ldots, x_k\) such that \(d(x_i, x_j) \geq \varepsilon_0\) for all \(1 \leq i < j \leq k\): Since \(\bigcup_{i=1}^k B(x_i, \varepsilon_0)\) is a finite union of \(\varepsilon_0\)-balls and \(X\) is not covered by any finite collection of \(\varepsilon_0\)-balls, there exists: \[ x_{k+1} \in X \setminus \bigcup_{i=1}^k B(x_i, \varepsilon_0). \] By construction, \(d(x_{k+1}, x_i) \geq \varepsilon_0\) for all \(1 \leq i \leq k\).

In this way, we obtain an infinite sequence \((x_n)_{n=1}^\infty\) in \(X\) such that: \[ d(x_n, x_m) \geq \varepsilon_0 \quad \text{for all } n \neq m. \]

Since \(X\) is sequentially compact, the sequence \((x_n)\) must contain a convergent subsequence \((x_{n_k})_{k=1}^\infty\). Every convergent sequence in a metric space is Cauchy, so there exists \(N \in \mathbb{N}\) such that for all \(j, k \geq N\): \[ d(x_{n_j}, x_{n_k}) < \varepsilon_0. \] However, taking \(j \neq k\), our sequence satisfies \(d(x_{n_j}, x_{n_k}) \geq \varepsilon_0\), a direct contradiction.

Therefore, \(X\) must be totally bounded.