- Show that if a subset of a metric space is complete and totally bounded, then it is compact.
Solution
In a metric space, compactness is equivalent to sequential compactness. We will show that every sequence in a complete, totally bounded metric space \((X, d)\) has a convergent subsequence.
Let \((x_n)_{n=1}^\infty\) be an arbitrary sequence in \(X\).
Constructing nested subsequences by total boundedness:
For \(\varepsilon = 1\): Since \(X\) is totally bounded, \(X\) can be covered by finitely many balls of radius \(1\). By the Pigeonhole Principle, at least one of these balls contains infinitely many terms of \((x_n)\). Let \(B(y_1, 1)\) be such a ball, and let \((x_{1, k})_{k=1}^\infty\) be a subsequence of \((x_n)\) entirely contained in \(B(y_1, 1)\).
For \(\varepsilon = 1/2\): The ball \(B(y_1, 1)\) (or \(X\)) is covered by finitely many balls of radius \(1/2\). At least one ball \(B(y_2, 1/2)\) contains infinitely many terms of the subsequence \((x_{1, k})\). Let \((x_{2, k})_{k=1}^\infty\) be a subsequence of \((x_{1, k})\) contained in \(B(y_2, 1/2)\).
Inductively, for each \(m \geq 1\): Having chosen the subsequence \((x_{m-1, k})\), covered by finitely many balls of radius \(1/m\), we extract a subsequence \((x_{m, k})_{k=1}^\infty\) of \((x_{m-1, k})\) that is entirely contained in some ball \(B(y_m, 1/m)\).
Diagonal subsequence: Consider the diagonal sequence \((z_j)_{j=1}^\infty\) defined by \(z_j = x_{j, j}\).
For each \(m \geq 1\), the tail \((z_j)_{j=m}^\infty\) is a subsequence of \((x_{m, k})_{k=1}^\infty\), so for all \(j \geq m\), \(z_j \in B(y_m, 1/m)\).
Therefore, for any \(j, \ell \geq m\): \[ d(z_j, z_\ell) \leq d(z_j, y_m) + d(y_m, z_\ell) < \frac{1}{m} + \frac{1}{m} = \frac{2}{m}. \]
Cauchy sequence and convergence: Given any \(\varepsilon > 0\), choose \(m \in \mathbb{N}\) such that \(\frac{2}{m} < \varepsilon\). Then for all \(j, \ell \geq m\), \(d(z_j, z_\ell) < \varepsilon\). Thus \((z_j)_{j=1}^\infty\) is a Cauchy sequence in \(X\).
Since \(X\) is complete, the Cauchy sequence \((z_j)\) converges to some \(x^* \in X\).
Conclusion: \((z_j)\) is a subsequence of the original sequence \((x_n)\) that converges in \(X\). Thus, every sequence in \(X\) has a convergent subsequence, which proves that \(X\) is sequentially compact, and therefore compact.