From Bases to a Metrization Criterion
A base describes a topology by listing open sets small enough to fit inside every open neighborhood. Bing’s metrization criterion gives a condition on how such a base is distributed: it must be a countable union of discrete families. This is different from asking for a countable base. A sigma-discrete base may be uncountable, which allows the criterion to apply to spaces that are not second countable.
Throughout, \((X,\mathcal{T})\) is a topological space. A family \(\mathcal{B}\) of open subsets of \(X\) is a base if for every open set \(U\) and every \(x\in U\), there is some \(B\in\mathcal{B}\) such that \(x\in B\subseteq U\). A family \(\mathcal{D}\) of subsets of \(X\) is discrete if each \(x\in X\) has an open neighborhood that intersects at most one member of \(\mathcal{D}\). This condition concerns how the sets meet neighborhoods; it does not require the sets in the family to be open.
For the separation hypotheses, \(X\) is \(T_1\) if every singleton \(\{x\}\) is closed. Here, regular means that whenever \(F\) is closed and \(x\notin F\), there are disjoint open sets containing \(x\) and \(F\), respectively. We state \(T_1\) separately so the role of each hypothesis is explicit.
Bing’s theorem is a standard metrization theorem. Its implication from a sigma-discrete base to a compatible metric requires a substantial construction, so that construction is not proved here; it is treated in the proof sketch devoted to Bing metrization. We will prove directly several consequences and check the hypotheses in the examples. The examples also show why sigma-discreteness should not be confused with countability.
Countable Bases and Discrete Spaces
A basic observation supplies a useful class of examples. Every family consisting of one set is discrete: at every point, the open neighborhood \(X\) meets at most that one member. Consequently, any countable family can be expressed as a countable union of discrete families, by placing one member in each family. Thus every countable base is sigma-discrete. This observation does not assert that a sigma-discrete family is countable: its discrete subfamilies can themselves have arbitrarily many members.
Worked Example: An Arbitrary Discrete Space
Let \(X\) be any set with the discrete topology \(\mathcal{T}=\mathcal{P}(X)\). We verify the criterion, including when \(X\) is empty.
If \(X\ne\varnothing\), take \(\mathcal{B}=\{\{x\}:x\in X\}\). Every open set \(U\subseteq X\) is a union of its singleton subsets, so \(\mathcal{B}\) is a base. It is itself a discrete family: for each \(x\in X\), the open neighborhood \(\{x\}\) meets the member \(\{x\}\) and meets no other singleton. Hence \(\mathcal{B}\) is sigma-discrete, for example by taking \(\mathcal{B}_1=\mathcal{B}\) and \(\mathcal{B}_n=\varnothing\) for \(n\ge2\).
The space is \(T_1\), since \(X\setminus\{x\}\) is open for every \(x\). It is regular: if \(F\) is closed and \(x\notin F\), the open sets \(\{x\}\) and \(F\) are disjoint and contain \(x\) and \(F\). Finally, the discrete metric \(d(x,y)=0\) when \(x=y\), and \(d(x,y)=1\) when \(x\ne y\), induces the discrete topology: the ball \(B(x,1/2)\) is \(\{x\}\), and every subset is a union of such balls. This verifies metrizability directly as well as through the criterion.
If \(X=\varnothing\), its topology and the base \(\mathcal{B}=\varnothing\) are both empty. The base condition has no points to check, and the empty family is discrete. The unique function \(d:\varnothing\times\varnothing\to\mathbb{R}\) satisfies the metric axioms vacuously and induces the unique topology. Thus the conclusion also holds in the empty case.
In the nonempty example, the base has one member for each point. If \(X\) is uncountable, this base is uncountable, yet it is discrete rather than merely a countable union of singleton families. In fact the discrete topology on an uncountable set is not second countable: if \(\mathcal{C}\) were a countable base, then for each \(x\), the open set \(\{x\}\) would have to contain a basis member containing \(x\), which forces that member to equal \(\{x\}\). The base would therefore need to contain uncountably many distinct singletons.
Worked Example: The Standard Topology on the Real Line
Let \(\mathbb{R}\) have its standard topology. Consider the family \(\mathcal{B}\) of all intervals \((p,q)\) with \(p,q\in\mathbb{Q}\) and \(p<q\). This is a countable family because \(\mathbb{Q}\times\mathbb{Q}\) is countable. We check that it is a base.
Suppose \(U\) is open and \(x\in U\). By the definition of the standard topology, there is \(\varepsilon>0\) such that \((x-\varepsilon,x+\varepsilon)\subseteq U\). The density of \(\mathbb{Q}\) in \(\mathbb{R}\) gives rationals \(p,q\) with \(x-\varepsilon<p<x<q<x+\varepsilon\). Then \(x\in(p,q)\subseteq U\), as required. For a concrete instance, if \(U=(-1/3,7/4)\) and \(x=1/2\), the rational endpoints \(p=0\) and \(q=1\) satisfy \(-1/3<0<1/2<1<7/4\), so \(x\in(0,1)\subseteq U\).
Since \(\mathcal{B}\) is countable, enumerate its members as \(B_1,B_2,\ldots\), and set \(\mathcal{B}_n=\{B_n\}\). Each \(\mathcal{B}_n\) is discrete by the one-member observation, so \(\mathcal{B}\) is sigma-discrete. To check the other hypotheses directly, distinct real numbers \(x,y\) have disjoint open neighborhoods of radius \(|x-y|/3\), so the space is \(T_1\). If \(F\) is closed and \(x\notin F\), the open complement \(\mathbb{R}\setminus F\) contains some interval \((x-r,x+r)\) with \(r>0\). The open sets \((x-r/3,x+r/3)\) and \(\bigcup_{y\in F}(y-r/3,y+r/3)\) are disjoint: a point in both would imply \(|x-y|<2r/3<r\) for some \(y\in F\), contradicting \((x-r,x+r)\cap F=\varnothing\). Thus \(\mathbb{R}\) is regular as well as \(T_1\), and Bing’s criterion applies.
Why the Separation Hypotheses Matter
The base condition alone does not guarantee metrizability. A quick way to see the role of \(T_1\) is to find a space with a sigma-discrete base that is not \(T_1\). Since every metric space is \(T_1\), such a space cannot be metrizable.
Worked Example: A Sigma-Discrete Base Without \(T_1\)
Let \(X=\{a,b\}\) with topology \(\mathcal{T}=\{\varnothing,\{a\},X\}\). The family \(\mathcal{B}=\{\{a\},X\}\) is a base: the only nonempty open sets are \(\{a\}\) and \(X\), and each is a member of \(\mathcal{B}\). Write \(\mathcal{B}_1=\{\{a\}\}\) and \(\mathcal{B}_2=\{X\}\). Each is a one-member discrete family, so \(\mathcal{B}\) is sigma-discrete.
However, \(\{a\}\) is not closed, because its complement \(\{b\}\) is not open. Thus this space is not \(T_1\). It cannot be metrizable: in any metric space, for distinct points \(a,b\), the ball \(B(a,d(a,b)/2)\) excludes \(b\), and the corresponding separation by open balls implies that singletons are closed. Notice also that \(\mathcal{B}\) itself is not discrete: every open neighborhood of \(b\) is \(X\), which meets both members. It is the decomposition into two discrete families, not the whole base as one discrete family, that makes it sigma-discrete.
This example does not contradict Bing’s theorem, because the theorem requires both regularity and \(T_1\). The hypotheses should be checked rather than inferred from the base. In metric spaces, the separation assumptions are automatic; here is the direct verification.
Proof. Let \(d\) be a metric inducing the topology. If \(x\ne y\), then \(d(x,y)>0\), and \(B(x,d(x,y)/2)\) contains \(x\) but not \(y\). Therefore \(X\setminus\{y\}\) is open: each of its points \(x\) has such a ball contained in \(X\setminus\{y\}\). Hence \(\{y\}\) is closed, proving \(T_1\).
For regularity, let \(F\) be closed and \(x\notin F\). If \(F=\varnothing\), the disjoint open sets \(X\) and \(\varnothing\) suffice. Otherwise, since \(\mathbb{R}\setminus F\) is open, choose \(r>0\) such that \(B(x,r)\cap F=\varnothing\). Set \(U=B(x,r/3)\) and \(V=\bigcup_{y\in F}B(y,r/3)\). Both sets are open, \(x\in U\), and \(F\subseteq V\). If \(z\in U\cap V\), then for some \(y\in F\), \(d(x,z)<r/3\) and \(d(z,y)<r/3\). The triangle inequality gives \(d(x,y)<2r/3<r\), contradicting \(B(x,r)\cap F=\varnothing\). Thus \(U\cap V=\varnothing\), which proves regularity. \(\square\)
Using the Criterion Carefully
The practical advantage of Bing’s criterion is that it replaces a search for a metric by a topological check on a base, together with the separation conditions. It also separates two questions that are easy to conflate: does a space have a countable base, and does it have a sigma-discrete base? A countable base always answers the second question positively, but the discrete-space example shows that the converse need not hold.
Common pitfall. One might argue, “The base is a union of discrete families, so it is countable.” That inference is false: discreteness places no countability restriction on the number of members in a family. For example, in the uncountable discrete space above, the single family of all singleton sets is discrete and uncountable. Another error is to conclude that any sigma-discrete base guarantees metrizability. The two-point example has such a base but fails \(T_1\); Bing’s criterion cannot be applied.
Verify that each open set is locally represented by members of the proposed family: every \(x\in U\) must lie in some basis member contained in \(U\).
Give an explicit countable decomposition into discrete subfamilies. If the base is countable, singleton subfamilies provide such a decomposition.
Verify regularity and \(T_1\), unless they are already known from a metric or another stated hypothesis.
Only after these hypotheses are established does the criterion conclude that the space is metrizable.
Check Your Understanding
For each question, identify the base or separation condition being tested.
- Why is every countable base sigma-discrete? State the discrete subfamilies explicitly.
- In an uncountable discrete space, explain why the singleton base is discrete but the space is not second countable.
- For the topology \(\{\varnothing,\{a\},\{a,b\}\}\), identify which singleton is not closed and explain why the space is not metrizable.
- State all three conditions in Bing’s metrization criterion.
- In the regularity proof for a metric space, explain why a point in \(B(x,r/3)\cap B(y,r/3)\) would contradict \(B(x,r)\cap F=\varnothing\) when \(y\in F\).