A collection \(\mathcal{A}\) of subsets of \(X\) is said to be locally discrete if each point of \(X\) has a neighborhood that intersects at most one element of \(\mathcal{A}\). A collection \(\mathcal{B}\) is countably locally discrete (or “\(\sigma\)-locally discrete”) if it equals a countable union of locally discrete collections. Prove the following.
Theorem (Bing metrization theorem). A space \(X\) is metrizable if and only if it is regular and has a basis that is countably locally discrete.
Solution
1. Sufficiency (\(\impliedby\)): Regular spaces with \(\sigma\)-locally discrete bases are metrizable.Proof:
1.1. A collection \(\mathcal{A}\) of subsets of \(X\) is locally finite if every point of \(X\) has an open neighborhood intersecting only finitely many elements of \(\mathcal{A}\).
1.2. If \(\mathcal{A}\) is locally discrete, every point has a neighborhood intersecting at most one element of \(\mathcal{A}\), so \(\mathcal{A}\) is locally finite.
1.3. If \(\mathcal{B} = \bigcup_{n=1}^\infty \mathcal{B}_n\) is a countably locally discrete basis, each \(\mathcal{B}_n\) is locally discrete, hence locally finite. Thus \(\mathcal{B}\) is a countably locally finite (\(\sigma\)-locally finite) basis.
1.4. By the Nagata-Smirnov Metrization Theorem, any regular space with a countably locally finite basis is metrizable.
1.5. Therefore \(X\) is metrizable.
2. Necessity (\(\implies\)): Metrizable spaces possess a \(\sigma\)-locally discrete basis.Proof:
2.1. Let \((X, d)\) be a metric space. Metric spaces are \(T_4\), hence regular.
2.2. Every open cover \(\mathcal{A}\) of \(X\) has an open refinement \(\mathcal{E} = \bigcup_{n=1}^\infty \mathcal{E}_n\) that covers \(X\), with each \(\mathcal{E}_n\) locally discrete.Proof: Well-order \(\mathcal{A}\) by \(<\). For \(U \in \mathcal{A}\) and \(n \in \mathbb{Z}_+\) put\[S_n(U) = \{x \in X : B_d(x, \tfrac1n) \subseteq U\}, \qquad T_n(U) = S_n(U) \setminus \bigcup_{V < U} V,\]and let \(E_n(U) = \bigcup_{x \in T_n(U)} B_d(x, \frac{1}{3n})\), an open set. Then \(E_n(U) \subseteq U\), because a point within \(\frac{1}{3n}\) of a point \(x \in S_n(U)\) lies in \(B_d(x, \frac1n) \subseteq U\).If \(V < W\) in \(\mathcal{A}\), \(x \in T_n(V)\) and \(y \in T_n(W)\), then \(B_d(x, \frac1n) \subseteq V\) and \(y \notin V\), so \(d(x, y) \ge \frac1n\). Hence points of \(E_n(V)\) and \(E_n(W)\) are at distance at least \(\frac1n - \frac{2}{3n} = \frac{1}{3n}\), and a ball of radius \(\frac{1}{6n}\) meets at most one member of \(\mathcal{E}_n = \{E_n(U) : U \in \mathcal{A}\}\). So \(\mathcal{E}_n\) is locally discrete.For \(x \in X\), let \(U\) be the first element of \(\mathcal{A}\) containing \(x\), and choose \(n\) with \(B_d(x, \frac1n) \subseteq U\). Then \(x \in T_n(U) \subseteq E_n(U)\). So \(\mathcal{E}\) covers \(X\).
2.3. For \(m \in \mathbb{Z}_+\), apply step 2 to the cover \(\mathcal{A}_m = \{B_d(x, \frac1m) : x \in X\}\) to get an open refinement \(\mathcal{C}_m\) that covers \(X\) and is countably locally discrete. Each member of \(\mathcal{C}_m\) lies in a ball of radius \(\frac1m\), so it has diameter at most \(\frac2m\).
2.4. \(\mathcal{B} = \bigcup_{m=1}^\infty \mathcal{C}_m\) is countably locally discrete, as a countable union of countably locally discrete collections.
2.5. \(\mathcal{B}\) is a basis. For an open set \(U \subseteq X\) and \(x \in U\), choose \(m\) with \(B_d(x, \frac{3}{m}) \subseteq U\). Some \(C \in \mathcal{C}_m\) contains \(x\), and \(\operatorname{diam} C \le \frac2m\), so \(x \in C \subseteq B_d(x, \frac3m) \subseteq U\).
2.6. Q.E.D. Steps 1, 4, and 5 show that \(X\) is regular and has a countably locally discrete basis.