Choosing a Metrization Criterion
A metrization criterion turns checkable topological properties into a conclusion about a metric: the given topology is induced by some metric. The criteria in this tutorial use bases and separation properties. Their hypotheses are not interchangeable. In particular, second countability is a convenient sufficient condition, not a necessary condition for metrizability.
Recall that a base \(\mathcal{B}\) for a topology on \(X\) is a family of open sets such that for every open \(U\) and every \(x\in U\), some \(B\in\mathcal{B}\) satisfies \(x\in B\subseteq U\). A space is second-countable if it has a countable base. A space is \(T_1\) if each singleton is closed. Here, regularity means \(T_1\) together with the condition that whenever \(x\notin C\) and \(C\) is closed, there are disjoint open sets containing \(x\) and \(C\), respectively.
Every discrete family is locally finite: a neighborhood meeting at most one member meets only finitely many. The converse need not hold. Consequently, the sigma-discrete hypothesis is the more restrictive-looking of these two base conditions. The criteria below nevertheless each give a characterization of metrizability when their separation hypothesis is included.
The Three Tests
The two “if and only if” criteria have both directions: they assert that the stated hypotheses imply metrizability and that every metrizable space has a base of the specified kind. The Urysohn theorem is commonly used as a direct sufficient test. If its hypotheses hold, there is no need to construct a sigma-locally-finite or sigma-discrete base separately.
One useful connection can be checked directly. If \(\mathcal{B}=\{B_1,B_2,\ldots\}\) is a countable base, then each one-member family \(\{B_n\}\) is discrete: every neighborhood of any point meets at most that one member. It is therefore locally finite as well. Thus a countable base is both sigma-discrete and sigma-locally-finite. This observation lets a regular second-countable space meet all three tests, although Urysohn’s theorem is often the shortest route.
Proof. The union of the one-member families \(\{B_n\}\), for \(n\in\mathbb{N}\), is exactly \(\mathcal{B}\). For each \(n\), a family containing only \(B_n\) cannot have a neighborhood meeting two distinct members, because it has no two distinct members. It is discrete, hence locally finite. Therefore \(\mathcal{B}\) is sigma-discrete and sigma-locally-finite. \(\square\)
Worked Tests
Worked Example: Euclidean Space
Test \(\mathbb{R}^n\), with its Euclidean topology, for metrizability. Here \(n\in\mathbb{N}\), and the metric is \(d(x,y)=\|x-y\|\).
Step 1: \(T_1\) and regularity. For distinct \(x,y\in\mathbb{R}^n\), the open ball \(B(x,\|x-y\|/2)\) misses \(y\). This shows singletons are closed: for each \(x\ne y\), there is an open neighborhood of \(x\) missing \(y\), so \(\mathbb{R}^n\setminus\{y\}\) is open. Now let \(C\) be closed and \(x\notin C\). The open set \(\mathbb{R}^n\setminus C\) contains \(x\), so there is \(r>0\) with \(B(x,r)\subseteq\mathbb{R}^n\setminus C\). The open sets \(B(x,r/2)\) and \(\mathbb{R}^n\setminus\overline{B(x,r/2)}\) are disjoint. The second contains \(C\), because every point of \(C\) is outside \(B(x,r)\), and hence outside the closed ball of radius \(r/2\). This verifies regularity.
Step 2: a countable base. Take all balls \(B(q,s)\) whose centers \(q\in\mathbb{Q}^n\) have rational coordinates and whose radii \(s\) are positive rationals. There are countably many such balls. To check the base property, let \(U\) be open, \(x\in U\), and choose \(\varepsilon>0\) such that \(B(x,\varepsilon)\subseteq U\). Choose \(q\in\mathbb{Q}^n\) with \(\|q-x\|<\varepsilon/4\). Density of the rationals allows this choice. Choose a positive rational \(s\) with \(\varepsilon/4<s<\varepsilon/2\). Then \(\|q-x\|<s\), so \(x\in B(q,s)\). If \(y\in B(q,s)\), the triangle inequality gives \(\|y-x\|\leq\|y-q\|+\|q-x\|<s+\varepsilon/4<3\varepsilon/4<\varepsilon\). Thus \(B(q,s)\subseteq U\), as required.
Conclusion. The hypotheses of Urysohn’s theorem hold, so \(\mathbb{R}^n\) is metrizable. The same countable base is also sigma-discrete and sigma-locally-finite by the one-member-family argument, so it meets the base hypotheses of the other two criteria as well. In this example the Urysohn test is the most direct.
Check: the Euclidean metric used to define the topology is already a metric witness. The criterion verifies that the topological hypotheses give the same conclusion, rather than supplying a different topology.
Worked Example: An Uncountable Discrete Space
Let \(X\) be an uncountable set with the discrete topology, in which every subset is open. The Urysohn test cannot be applied using second countability, but the Bing and Nagata–Smirnov tests can.
Step 1: separation. The discrete metric \(d(x,y)=0\) when \(x=y\), and \(d(x,y)=1\) when \(x\ne y\), induces the discrete topology. In particular, the space is \(T_1\). It is also regular directly: if \(x\notin C\), then the open sets \(\{x\}\) and \(X\setminus\{x\}\) are disjoint, contain \(x\) and \(C\), respectively.
Step 2: check the base conditions. The family \(\mathcal{B}=\{\{x\}:x\in X\}\) is a base, since for any open \(U\) and \(x\in U\), the base member \(\{x\}\) satisfies \(x\in\{x\}\subseteq U\). It is discrete: the neighborhood \(\{x\}\) of \(x\) meets exactly one member of \(\mathcal{B}\), namely \(\{x\}\). Thus \(\mathcal{B}\) is itself a discrete family, and therefore also locally finite. It is a sigma-discrete and sigma-locally-finite base (one family suffices).
Conclusion. The hypotheses of both Bing and Nagata–Smirnov hold, so either criterion establishes metrizability. The discrete metric above is an explicit witness.
Why Urysohn’s test does not apply: every base must contain \(\{x\}\) for each \(x\in X\). Indeed, apply the base property to the open set \(\{x\}\) and its point \(x\); the only set \(B\) satisfying \(x\in B\subseteq\{x\}\) is \(\{x\}\). Hence every base is uncountable. The space is not second-countable, although it is metrizable.
Worked Example: The Rational Line
Let \(X=\mathbb{Q}\) with the subspace topology inherited from \(\mathbb{R}\). This is a concrete regular second-countable space, so the Urysohn criterion applies.
Step 1: second countability. The sets \(\mathbb{Q}\cap(a,b)\), for rational \(a<b\), form a countable base. They are open in the subspace topology. If \(q\in U\) and \(U\) is open in \(\mathbb{Q}\), there is a real open set \(O\) with \(U=O\cap\mathbb{Q}\). Since \(q\in O\), some interval around \(q\) lies in \(O\). Density of \(\mathbb{Q}\) lets us choose rational \(a<q<b\) close enough that \((a,b)\subseteq O\). Then \(q\in\mathbb{Q}\cap(a,b)\subseteq U\).
Step 2: \(T_1\) and regularity. Distinct points of \(\mathbb{Q}\) have disjoint sufficiently small metric neighborhoods, so singletons are closed. Let \(F\) be closed in \(\mathbb{Q}\) and \(q\notin F\). Because \(\mathbb{Q}\setminus F\) is open, choose \(\varepsilon>0\) such that \(B_{\mathbb{Q}}(q,\varepsilon)\subseteq\mathbb{Q}\setminus F\). The sets \(U=B_{\mathbb{Q}}(q,\varepsilon/3)\) and \(V=\mathbb{Q}\setminus\overline{B_{\mathbb{Q}}(q,\varepsilon/2)}\) are open and disjoint. Every point of \(F\) has distance at least \(\varepsilon\) from \(q\), so \(F\subseteq V\). This verifies regularity.
Conclusion. Urysohn’s theorem gives metrizability. In fact, the subspace metric \(d(p,q)=|p-q|\) induces the given topology. The countable base also gives sigma-discrete and sigma-locally-finite bases, so the other two criteria are available, but they require no additional construction here.
A Common Mistake and a Practical Choice
Common mistake. “An uncountable space cannot be metrizable, because a metrizable space has a countable base.” The implication is false. The uncountable discrete space above is metrizable, and its singleton base is uncountable. The correct implication is that a second-countable metrizable space has a countable base; metrizability by itself does not impose second countability.
For a new space, first check \(T_1\) and regularity, then ask which base condition is actually available. If a countable base is already known, Urysohn is usually the shortest test. If the space is not second-countable, look instead for a sigma-locally-finite base or a sigma-discrete base and use the corresponding criterion. These tests establish metrizability only when their hypotheses have been verified; the name of a criterion is not a substitute for checking them.
| Criterion | Hypotheses to verify | Useful situation |
|---|---|---|
| Urysohn | Regular, \(T_1\), second-countable | A countable base is readily available |
| Nagata–Smirnov | Regular, \(T_1\), sigma-locally-finite base | The base is organized into locally finite families |
| Bing | Regular, \(T_1\), sigma-discrete base | The base is organized into discrete families |
Check Your Understanding
For each question, identify the criterion and verify every hypothesis it requires.
- Why does a countable base automatically give both a sigma-discrete base and a sigma-locally-finite base?
- For an uncountable discrete space, which hypothesis of the Urysohn test fails, and which base criterion still applies?
- State the separation and base hypotheses in the Nagata–Smirnov criterion.
- Can a metrizable space fail to be second-countable? Give the example from this tutorial and identify a metric for it.
- If a space is regular, \(T_1\), and second-countable, which criterion gives the most direct conclusion, and what is that conclusion?