Proof Sketch of Nagata-Smirnov Metrization

From a Base to a Metric

The Nagata–Smirnov criterion identifies a useful bridge between a combinatorial description of open sets and a metric. Its sufficiency direction starts with a regular \(T_1\) space having a base that is a countable union of locally finite families. The key is not merely that the base is a countable union: local finiteness lets us assemble continuous functions without losing control of continuity. We trace the argument from refinements to functions and then to a metric.

Recall that a family \(\mathcal{A}\) of subsets of \(X\) is locally finite if every point has an open neighborhood meeting only finitely many members of \(\mathcal{A}\). A base \(\mathcal{B}\) is sigma-locally-finite if \(\mathcal{B}=\bigcup_{n\in\mathbb{N}}\mathcal{B}_n\), where each \(\mathcal{B}_n\) is locally finite. Regularity is used in the convention that for a closed set \(C\) and a point \(x\notin C\), there are disjoint open sets containing \(x\) and \(C\), respectively. The \(T_1\) condition says that singletons are closed.

The argument uses a standard refinement lemma, stated here to make the technical input explicit. Its detailed cover construction is the refinement part of the standard proof of the criterion: it applies the locally finite open-refinement theorem successively to the layers of a sigma-locally-finite base, using regularity to shrink neighborhoods before refining. The resulting sequence of covers is often called a development.

The important point in this lemma is the star condition, which is stronger than merely finding one small cover member at \(x\). It controls every member of the cover that contains \(x\). At each stage of the standard construction, regularity permits choosing neighborhoods whose closures stay inside prescribed open sets; locally finite refinements keep the choices locally finite. The sigma-locally-finite base supplies countably many layers to process. We use the lemma in precisely the form stated, rather than assuming that an individual layer of the base covers \(X\).

From Locally Finite Covers to Functions

For each cover \(\mathcal{V}_n\), take a locally finite partition of unity subordinate to it. Concretely, this gives a family \(\mathcal{F}_n\) of continuous functions \(f:X\to[0,1]\) such that the family of supports is locally finite, \(\sum_{f\in\mathcal{F}_n}f(x)=1\) for every \(x\), and each \(f\) has support contained in some member of \(\mathcal{V}_n\). A partition of unity is locally finite, so the sum at each point has only finitely many nonzero terms. This standard consequence of regularity and locally finite refinements is obtained by shrinking a locally finite cover and applying the continuous separation of a closed set from a disjoint closed set.

These functions detect the topology. Fix \(x\in U\), with \(U\) open, and choose \(n\) such that \(\operatorname{St}(x,\mathcal{V}_n)\subseteq U\). Since the partition sums to \(1\) at \(x\), some \(f\in\mathcal{F}_n\) has \(f(x)>0\). Its support lies in a member \(V\in\mathcal{V}_n\). As \(f(x)>0\), that member contains \(x\), so \(V\subseteq\operatorname{St}(x,\mathcal{V}_n)\subseteq U\). Thus \(f\) is positive at \(x\) and vanishes outside \(U\).

Assembling the Metric

For each \(n\), compare points by the largest difference among the functions in \(\mathcal{F}_n\). The functions take values in \([0,1]\), so the comparison is bounded. To cover the possibility that a family is empty, define its comparison to be zero; equivalently, take the supremum in the ordered set \([0,1]\), where the supremum of the empty set is \(0\).

Proof. Each \(\rho_n\) is nonnegative and symmetric, and \(\rho_n(x,x)=0\). For any \(x,y,z\) and any \(f\in\mathcal{F}_n\), the real triangle inequality gives \(|f(x)-f(z)|\le |f(x)-f(y)|+|f(y)-f(z)|\le\rho_n(x,y)+\rho_n(y,z)\). Taking the supremum gives the triangle inequality for \(\rho_n\); it also holds when \(\mathcal{F}_n\) is empty, since then all three values are zero. Since \(0\le\rho_n(x,y)\le1\), the series defining \(d\) converges and is finite. Summing the triangle inequalities with the positive weights \(2^{-n}\) proves the triangle inequality for \(d\).

To show that distinct points have positive distance, let \(x\ne y\). Since \(X\) is \(T_1\), \(U=X\setminus\{y\}\) is open and contains \(x\). The function-detection argument gives an \(n\) and an \(f\in\mathcal{F}_n\) with \(f(x)>0\) and support contained in \(U\). Thus \(f(y)=0\), so \(\rho_n(x,y)\ge |f(x)-f(y)|=f(x)>0\), and \(d(x,y)\ge 2^{-n}f(x)>0\). If \(X\) has at most one point, the same formula satisfies the metric axioms: there are no distinct points requiring separation.

It remains to compare the topologies. First fix \(x\) and \(n\). We verify that \(y\mapsto\rho_n(x,y)\) is continuous; this is where local finiteness is essential. At a point \(y_0\), choose an open neighborhood \(V\) meeting only finitely many supports from \(\mathcal{F}_n\). Also, only finitely many supports contain \(x\), because the family of supports is locally finite at \(x\). Outside the union of these two finite collections, a function vanishes throughout \(V\) and at \(x\). Its difference \(|f(x)-f(y)|\) is therefore zero for \(y\in V\). For each of the finitely many remaining functions, \(y\mapsto|f(x)-f(y)|\) is continuous. On \(V\), \(\rho_n(x,y)\) is the maximum of zero and these finitely many continuous functions, hence is continuous. This argument includes the case that there are no remaining functions.

For each \(N\), the tail of the metric series is bounded uniformly in \(x,y\) by \(\sum_{n>N}2^{-n}=2^{-N}\), since \(0\le\rho_n(x,y)\le1\). Hence \(y\mapsto d(x,y)\) is a uniform limit of continuous partial sums and is continuous in the original topology. Therefore every metric ball \(B_d(x,r)=\{y:d(x,y)<r\}\) is open in the original topology.

Conversely, let \(U\) be open and \(x\in U\). Choose \(n\) with \(\operatorname{St}(x,\mathcal{V}_n)\subseteq U\), and choose \(f\in\mathcal{F}_n\) with \(f(x)>0\). Its support is contained in a cover member containing \(x\), hence in \(U\). If \(y\notin U\), then \(f(y)=0\), so \(d(x,y)\ge2^{-n}\rho_n(x,y)\ge2^{-n}|f(x)-f(y)|=2^{-n}f(x)\). Consequently \(B_d(x,2^{-n}f(x))\subseteq U\). Every original open neighborhood contains a metric ball about each of its points. Together with the previous inclusion, this proves equality of the topologies. \(\square\)

Two Checks on the Construction

Common pitfall. One cannot omit the empty-family convention and write \(\rho_n=\sup\{|f(x)-f(y)|:f\in\mathcal{F}_n\}\) for every \(n\): when \(\mathcal{F}_n=\varnothing\), the supremum is not a real number under the usual real-valued convention. Set \(\rho_n=0\) in that case (or take the supremum in \([0,1]\)). A second error is to say that functions whose supports miss a neighborhood contribute zero to \(|f(x)-f(y)|\). They vanish there, but \(f(x)\) may be nonzero. The continuity proof above separately includes the finitely many supports containing \(x\); their contributions are continuous functions on the chosen neighborhood.

The construction has a useful division of labor. The covers make neighborhoods small in the star sense; partitions of unity turn those covers into continuous measurements; and the weighted sum combines the measurements while keeping the series finite and continuous. The weights \(2^{-n}\) are not special: any positive summable sequence would control the tail. What matters is that the functions detect every point-neighborhood pair, and that each level is locally finite.