Nagata-Smirnov Metrization Criterion

From Local Finiteness to a Metrization Criterion

A metric gives a space a precise notion of distance, but a topology records only which sets are open. A metrizability criterion identifies topological conditions that guarantee some metric induces the given topology. The Nagata–Smirnov theorem uses a base organized into locally finite families. This organization is weaker than requiring the whole base to be locally finite, but strong enough—together with regularity and the \(T_1\) condition—to characterize metrizable spaces.

Let \((X,\mathcal{T})\) be a topological space. A family \(\mathcal{F}\) of subsets of \(X\) is locally finite if, for every \(x\in X\), there is an open neighborhood \(U\) of \(x\) that intersects only finitely many members of \(\mathcal{F}\). A family \(\mathcal{B}\) is a base if every open set is a union of members of \(\mathcal{B}\); equivalently, whenever \(x\in U\in\mathcal{T}\), some \(B\in\mathcal{B}\) satisfies \(x\in B\subseteq U\).

Definition: A base \(\mathcal{B}\) is sigma-locally-finite if it can be written as \(\mathcal{B}=\bigcup_{n\in\mathbb{N}}\mathcal{B}_n\), where each family \(\mathcal{B}_n\) is locally finite. The whole base need not be locally finite.

The separation assumptions in the criterion are as follows. The space is \(T_1\) if every singleton \(\{x\}\) is closed. We use “regular” for the condition that whenever \(x\notin C\) and \(C\) is closed, there are disjoint open sets containing \(x\) and \(C\), respectively. Stating \(T_1\) separately makes explicit both requirements in the theorem.

Theorem (Nagata–Smirnov Metrization Criterion): A topological space is metrizable if and only if it is regular, \(T_1\), and has a sigma-locally-finite base. The proof of this characterization is developed in the later proof tutorial. In particular, its sufficient condition is that a regular \(T_1\) space possess a sigma-locally-finite base; that condition guarantees a metric inducing its topology.

The statement is an equivalence, not just a test that gives a sufficient condition. It says that every metrizable space has the specified kind of base, and that regularity, \(T_1\), and such a base together imply metrizability. The metric-space direction is useful to understand before using the criterion in reverse.

Why Metric Spaces Have Such Bases

We use the standard metric-space refinement theorem: every open cover of a metric space has a locally finite open refinement. A refinement is a cover each of whose sets is contained in some member of the original cover. This result is also known as the paracompactness theorem for metric spaces; its proof is a standard result beyond this tutorial. The construction below shows how it yields a sigma-locally-finite base.

Let \((X,d)\) be a metric space. For each \(n\in\mathbb{N}\), consider the open cover \(\mathcal{U}_n=\{B_d(x,1/n):x\in X\}\). Choose a locally finite open refinement \(\mathcal{V}_n\) of \(\mathcal{U}_n\). Since \(\mathcal{V}_n\) covers \(X\), every \(x\in X\) belongs to some \(V\in\mathcal{V}_n\). For that \(V\), there is a \(y\in X\) such that \(V\subseteq B_d(y,1/n)\). Because \(x\in V\), we have \(d(x,y)<1/n\).

Now let \(U\) be open with \(x\in U\). By the definition of the metric topology, there is an \(r>0\) such that \(B_d(x,r)\subseteq U\). Choose \(n\) with \(2/n<r\), and take \(V\in\mathcal{V}_n\) containing \(x\). For every \(z\in V\), both \(d(z,y)<1/n\) and \(d(x,y)<1/n\), so the triangle inequality gives \(d(z,x)<2/n<r\). Thus \(x\in V\subseteq U\). This proves that \(\bigcup_{n\in\mathbb{N}}\mathcal{V}_n\) is a base, and each family \(\mathcal{V}_n\) is locally finite by construction. Hence this base is sigma-locally-finite.

Worked Example: A Sigma-Locally-Finite Base for the Real Line

There is also a direct construction for \(\mathbb{R}\) with its standard topology. For \(n\in\mathbb{N}\), set \(h_n=2^{-n}\) and define \(\mathcal{B}_n=\{(k h_n-2h_n,k h_n+2h_n):k\in\mathbb{Z}\}\). Each interval is open.

Local finiteness at a fixed level. Fix \(n\), and let \(x\in\mathbb{R}\). The neighborhood \((x-1,x+1)\) can meet an interval centered at \(k h_n\) only if \(k h_n\in(x-1-2h_n,x+1+2h_n)\). This is a bounded interval, so it contains only finitely many integer multiples of the positive number \(h_n\). Thus \(\mathcal{B}_n\) is locally finite.

The base property. Suppose \(x\in U\) and \(U\) is open. Choose \(r>0\) with \((x-r,x+r)\subseteq U\). Choose \(n\) so large that \(5h_n/2<r\), and choose an integer \(k\) with \(|x-kh_n|\le h_n/2\). Such an integer exists by rounding \(x/h_n\) to a nearest integer. Then \(x\) belongs to the interval centered at \(kh_n\), since \(h_n/2<2h_n\). If \(z\) belongs to that interval, then \[ |z-x|\le |z-kh_n|+|kh_n-x|<2h_n+h_n/2=5h_n/2<r. \] So the interval contains \(x\) and is contained in \(U\). The union of the families \(\mathcal{B}_n\) is therefore a base.

Check: At \(n=2\), \(h_2=2^{-2}=1/4\), so the interval indexed by \(k=4\) is \((4/4-2/4,4/4+2/4)=(1/2,3/2)\). It contains \(1\), and its endpoints are at distance \(1/2\) from \(1\), as the formula predicts.

Using the Criterion

Worked Example: Applying the Criterion to the Unit Interval

Give \([0,1]\) the subspace topology inherited from \(\mathbb{R}\). For each \(n\in\mathbb{N}\), let \(h_n=2^{-n}\) and take the finite family \[ \mathcal{C}_n=\{[0,1]\cap(kh_n-2h_n,kh_n+2h_n): k=0,1,\ldots,2^n\}. \] Every member is open in the subspace topology, and every \(\mathcal{C}_n\) is locally finite because it is finite.

To check that their union is a base, let \(x\in U\), where \(U\) is relatively open in \([0,1]\). There is an \(r>0\) such that \([0,1]\cap(x-r,x+r)\subseteq U\). Choose \(n\) with \(5h_n/2<r\). Since \(0\le x\le1\), a nearest integer \(k\) to \(x/h_n\) can be chosen from \(0,1,\ldots,2^n\), and \(|x-kh_n|\le h_n/2\). The interval in \(\mathcal{C}_n\) indexed by \(k\) contains \(x\). Each of its points \(z\) satisfies \(|z-x|<2h_n+h_n/2=5h_n/2<r\). Its intersection with \([0,1]\) is therefore contained in \(U\), proving the base property.

The usual metric on \([0,1]\) induces its subspace topology, so this space is \(T_1\) and regular. The construction gives it a sigma-locally-finite base, and the Nagata–Smirnov criterion applies. Here it recovers metrizability from topological properties and a base condition.

Proposition: Every space with a countable base has a sigma-locally-finite base.

Proof. Let \(\mathcal{B}=\{B_n:n\in\mathbb{N}\}\) be a countable base; if the base is finite, repeat members in this list. For each \(n\), the one-member family \(\{B_n\}\) is locally finite: every point has the neighborhood \(X\), which meets at most that one member. The given base is the union of these locally finite families, so it is sigma-locally-finite. \(\square\)

This implication does not say that every sigma-locally-finite base is countable. For example, if \(X\) is any set with the discrete topology, then \(\{\{x\}:x\in X\}\) is a base and is itself locally finite: the neighborhood \(\{x\}\) meets only one member. If \(X\) is uncountable, this base is uncountable. Thus sigma-local finiteness allows a base to have many members, provided they are arranged in countably many locally finite families.

Hypotheses and a Common Pitfall

Worked Example: A Space the Criterion Cannot Metrize

Let \(X=\{a,b\}\) have the indiscrete topology \(\mathcal{T}=\{\varnothing,X\}\). This space is not \(T_1\): the only open set containing \(b\) is \(X\), so \(X\setminus\{a\}=\{b\}\) is not open, and \(\{a\}\) is not closed. It is not metrizable either. In a metric topology, for distinct points \(a,b\), the ball \(B_d(a,d(a,b)/2)\) contains \(a\) and excludes \(b\), so the topology is \(T_1\). The indiscrete topology on these two distinct points cannot be induced by a metric.

The space does have a sigma-locally-finite base: \(\{X\}\) is a one-member base and is locally finite. The example shows why the base condition alone is not the criterion. The required separation hypotheses matter.

Common pitfall. It is incorrect to infer that the union of locally finite families is itself locally finite. In the real-line construction, for every \(n\), the interval indexed by \(k=0\) is \((-2h_n,2h_n)\), and all these intervals contain \(0\). Every open neighborhood of \(0\) therefore meets infinitely many distinct members of the full union: for sufficiently large \(n\), \(2h_n\) is smaller than the neighborhood's radius. So the whole base is not locally finite at \(0\), even though each individual \(\mathcal{B}_n\) is locally finite. Sigma-local finiteness asserts local finiteness separately at each level, not for the union.

The criterion is useful because it replaces the task of constructing a metric with two topological checks: verify regularity and \(T_1\), then organize a base into countably many locally finite families. The metric-space refinement theorem explains why the condition is necessary; the Nagata–Smirnov theorem supplies the converse. The theorem is a characterization, while in an application the forward implication gives the practical sufficient condition for metrizability.

Check Your Understanding

For each question, distinguish local finiteness of an individual family from sigma-local finiteness of a base.

State the definition of a locally finite family and the definition of a sigma-locally-finite base.
Which three hypotheses in the Nagata–Smirnov criterion together imply metrizability?
Why does a countable base automatically give a sigma-locally-finite base?
In the real-line construction, why does choosing \(5h_n/2<r\) ensure the selected interval is contained in \((x-r,x+r)\)?
Why does the indiscrete topology on \(\{a,b\}\) fail the \(T_1\) hypothesis and fail to be metrizable?