Show that every compact subspace of a metric space is bounded in that metric and is closed. Find a metric space in which not every closed bounded subspace is compact.
Solution
Goal: Prove that compact subspaces of metric spaces are closed and bounded, and provide a metric space containing a closed, bounded subspace that fails to be compact.
1. Compact subspaces are closed:Let \(K\) be a compact subspace of a metric space \((X, d)\). Then \(K\) is closed in \(X\).Proof:
1.1. Every metric space is Hausdorff (\(T_2\)).
1.2. Let \(x \in X \setminus K\). For each \(y \in K\), let \(r_y = d(x, y) > 0\).
1.3. Define \(U_y = B(y, \frac{r_y}{2})\) and \(V_y = B(x, \frac{r_y}{2})\). Then \(U_y \cap V_y = \emptyset\).
1.4. The family \(\{U_y\}_{y \in K}\) is an open cover of \(K\). Since \(K\) is compact, there exists a finite subcover \(\{U_{y_1}, \dots, U_{y_k}\}\).
1.5. The set \(V = \bigcap_{i=1}^k V_{y_i}\) is an open neighborhood of \(x\).
1.6. Since \(V \cap U_{y_i} = \emptyset\) for each \(i\), \(V \cap K \subseteq V \cap (\bigcup_{i=1}^k U_{y_i}) = \emptyset\).
1.7. Thus \(X \setminus K\) is open, so \(K\) is closed in \(X\).
2. Compact subspaces are bounded:Let \(K\) be a compact subspace of \((X, d)\). Then \(\operatorname{diam}(K) < \infty\).Proof:
2.1. If \(K = \emptyset\), \(\operatorname{diam}(K) = 0 < \infty\).
2.2. If \(K \neq \emptyset\), fix an arbitrary basepoint \(x_0 \in K\).
2.3. For every \(x \in K\), \(d(x_0, x) < n\) for some integer \(n \in \mathbb{Z}_+\), so \(K \subseteq \bigcup_{n=1}^\infty B(x_0, n)\).
2.4. Since \(\{B(x_0, n)\}_{n=1}^\infty\) is an open cover of the compact space \(K\), there exists a finite subcover \(\{B(x_0, n_1), \dots, B(x_0, n_k)\}\).
2.5. Setting \(N = \max\{n_1, \dots, n_k\}\), we have \(K \subseteq B(x_0, N)\).
2.6. By the triangle inequality, for all \(x, y \in K\):\[d(x, y) \le d(x, x_0) + d(x_0, y) < N + N = 2N.\]
2.7. Thus \(\operatorname{diam}(K) \le 2N < \infty\), so \(K\) is bounded.
3. Counterexample (closed + bounded \(\not\implies\) compact):Proof:
3.1. Consider an infinite set \(X\) (e.g. \(X = \mathbb{Z}_+\)) equipped with the discrete metric \(d(x, y) = 1\) for all \(x \neq y\) and \(d(x, x) = 0\).
3.2. The entire space \(X\) is closed in itself and is bounded since \(\operatorname{diam}(X) = 1 \le 1\).
3.3. The open cover \(\mathcal{U} = \{B(x, 1/2)\}_{x \in X} = \{\{x\}\}_{x \in X}\) consists of singleton open sets.
3.4. Since \(X\) is infinite, no finite subcollection of \(\mathcal{U}\) can cover \(X\).
3.5. Thus the closed bounded metric space \((X, d)\) is not compact. Q.E.D.