A Sufficient Test for Metrizability
A metric does more than assign distances: its open balls determine a topology. The Urysohn metrization theorem gives a condition stated entirely in topological terms that guarantees such a metric exists. Its hypotheses play different roles. Regularity lets us shrink neighborhoods without losing control of their closures; second countability lets us organize the needed separations in a countable list.
We use the convention that a space is regular if it is \(T_1\) and, whenever \(x\notin C\) with \(C\) closed, there are disjoint open sets containing \(x\) and \(C\), respectively. A \(T_1\) space is one in which every singleton is closed. A space is second countable if it has a countable basis: a countable collection \(\mathcal B\) of open sets such that for every open \(U\) and every \(x\in U\), some \(B\in\mathcal B\) satisfies \(x\in B\subseteq U\). Some texts build \(T_1\) into “regular”; others do not. Here the theorem’s \(T_1\) requirement is explicit.
The conclusion is topological: the metric need not be the metric one might initially expect, and it is not unique. The theorem asserts existence of a metric that gives exactly the original open sets, not merely a metric that separates distinct points.
What Regularity and Countability Contribute
Regularity has a useful shrinking consequence. If \(x\in U\) and \(U\) is open, apply regularity to \(x\) and the closed set \(X\setminus U\). It gives disjoint open sets \(V,W\) with \(x\in V\) and \(X\setminus U\subseteq W\). Since \(V\cap W=\varnothing\), \(V\subseteq X\setminus W\). The set \(X\setminus W\) is closed and contained in \(U\), so \(\overline V\subseteq X\setminus W\subseteq U\). Thus \(\ x\in V\subseteq\overline V\subseteq U\).
Second countability makes it possible to use only countably many such shrinkings. In the proof, one considers pairs of basis elements \(B_i,B_j\) with \(\overline{B_i}\subseteq B_j\). There are at most countably many pairs. Each suitable pair gives a continuous function that is \(0\) on \(\overline{B_i}\) and \(1\) outside \(B_j\). These functions detect the topology: if \(x\in U\), a suitable pair can be chosen with \(x\in B_i\) and \(B_j\subseteq U\). The coordinate belonging to this pair is zero at \(x\), and equals one outside \(U\).
Two standard results of normal-space theory are used in this construction. A regular second-countable space is normal, and Urysohn’s lemma says that in a normal space, disjoint closed sets can be separated by a continuous function into \([0,1]\): the function can be prescribed to be \(0\) on one closed set and \(1\) on the other. Their proofs are part of the general theory of normality and Urysohn’s lemma. Second countability implies the Lindelöf property, and the standard theorem that regular Lindelöf spaces are normal supplies the first result.
How the Functions Give a Metric
Here is the metric step of the argument. List the separating functions as \(f_1,f_2,\ldots\), with each \(f_n:X\to[0,1]\) continuous. If the list is finite, append constant zero functions. The construction above ensures both that distinct points are separated by some function and that, for each \(x\in U\) open, some coordinate is \(0\) at \(x\) and \(1\) on \(X\setminus U\). Define
The series converges because \(0\leq |f_n(x)-f_n(y)|\leq1\) and \(\sum_{n=1}^{\infty}2^{-n}=1\). The triangle inequality for absolute value, applied term by term and then summed, gives \(d(x,z)\leq d(x,y)+d(y,z)\). Symmetry and nonnegativity follow from absolute value. If \(x\ne y\), a separating coordinate has \(f_n(x)\ne f_n(y)\), so its positive contribution makes \(d(x,y)>0\). Also \(d(x,x)=0\). Thus \(d\) is a metric.
To check that its topology is the original one, first take an original open neighborhood \(U\) of \(x\). Choose a coordinate \(f_n\) with \(f_n(x)=0\) and \(f_n(y)=1\) for \(y\notin U\). For every \(y\notin U\), the \(n\)-th summand gives \(d(x,y)\geq 2^{-n}\). Hence \(B_d(x,2^{-n})\subseteq U\).
In the other direction, fix \(\varepsilon>0\). Choose \(N\) so that \(\sum_{n>N}2^{-n}<\varepsilon/2\). By continuity of the finitely many functions \(f_1,\ldots,f_N\), there is an original open neighborhood \(V\) of \(x\) on which \(|f_n(y)-f_n(x)|<\varepsilon/2\) for every \(n\leq N\). Since \(\sum_{n=1}^{N}2^{-n}<1\), for \(y\in V\), \[ d(x,y) <\frac{\varepsilon}{2}\sum_{n=1}^{N}2^{-n}+\frac{\varepsilon}{2} <\varepsilon. \] So \(V\subseteq B_d(x,\varepsilon)\). The two neighborhood comparisons show that the metric topology and original topology agree.
If \(X=\varnothing\), its unique topology is metrizable by the empty distance function; there are no pairs of points for which the metric axioms need checking. If \(X\) is nonempty, the construction above applies, including when \(X\) has just one point.
Worked Examples and the Hypotheses
Worked Example: The Real Line
Take \(\mathbb R\) with its standard topology. The intervals with rational endpoints form a countable basis: for any open \(U\) and \(x\in U\), there is an interval around \(x\) contained in \(U\), and rational numbers can be chosen on either side of \(x\) within that interval. There are only countably many pairs of rational endpoints.
The usual metric is \(d(x,y)=|x-y|\). It makes \(\mathbb R\) \(T_1\): for \(x\ne y\), the open ball of radius \(|x-y|/2\) about \(x\) omits \(y\), so \(\{y\}\) is closed. It is also regular. Given \(x\in U\) open, choose \(r>0\) with \((x-r,x+r)\subseteq U\), and set \(V=(x-r/2,x+r/2)\). Its closure is \([x-r/2,x+r/2]\subseteq(x-r,x+r)\subseteq U\), giving the required shrinking.
Thus the theorem applies. In this example it guarantees a compatible metric, while the usual metric already provides one. The check is that the same open neighborhoods are obtained: each interval around \(x\) contains a sufficiently small ball about \(x\), and each ball is itself an open interval.
Worked Example: A Three-Point Discrete Space
Let \(X=\{a,b,c\}\) have the discrete topology, in which every subset is open. The three singleton sets form a finite, hence countable, basis. The space is \(T_1\), since every singleton is closed.
For a direct metric check, define \(d(x,y)=0\) when \(x=y\), and \(d(x,y)=1\) when \(x\ne y\). If \(x\ne z\), then \(d(x,z)=1\), while \(d(x,y)+d(y,z)\geq1\): at least one of \(x\ne y\) or \(y\ne z\) must hold. This proves the triangle inequality; the other metric conditions follow from the definition. For each \(x\), the ball \(B_d(x,1/2)\) is exactly \(\{x\}\). Every subset is a union of these open balls, so the metric induces the discrete topology. This verifies the theorem’s conclusion in a small explicit case.
Worked Example: Why the \(T_1\) Convention Matters
Let \(X=\{a,b\}\) have the indiscrete topology \(\{\varnothing,X\}\). This topology has a finite basis \(\{X\}\). Under the weaker convention that regularity only requires separating a point from a closed set not containing it, this space is regular: its only closed sets are \(\varnothing\) and \(X\), and the only case with \(x\notin C\) is \(C=\varnothing\). The open sets \(X\) and \(\varnothing\) are disjoint and contain \(x\) and \(C\), respectively.
But the space is not metrizable. In any metric topology on two distinct points, the ball about \(a\) of radius \(d(a,b)/2\) contains \(a\) and omits \(b\). The indiscrete topology has no open set containing \(a\) but not \(b\). Therefore the two-point indiscrete space is not metrizable. The example shows why, if regularity is defined without \(T_1\), the theorem must explicitly include \(T_1\). Under the convention used here, this space is not regular.
What the Hypotheses Do—and Do Not—Say
Second countability gives a countable collection of coordinates, which is what allows the weighted series defining \(d\) to be used. It is a sufficient hypothesis in the theorem, not a necessary condition for every metrizable space. For example, let \(X\) be an uncountable set with the discrete topology. The distance that is \(0\) on equal pairs and \(1\) on distinct pairs induces that topology by the same ball calculation as above. Yet every basis for this topology must contain \(\{x\}\) for each \(x\): a basis element containing \(x\) and contained in the open set \(\{x\}\) can only be \(\{x\}\). Thus no countable basis exists. This example is metrizable but not second countable.
Common pitfall. A countable family of continuous functions does not automatically yield a metric. If all the coordinates on a two-point set \(X=\{a,b\}\) are constant zero, the proposed formula gives \(\ d(a,b)=\sum_{n=1}^{\infty}2^{-n}|0-0|=0\), even though \(a\ne b\). That formula is only a pseudometric in this case. The correction is to ensure the coordinates separate every pair of distinct points; the proof uses the base-pair construction to obtain this separation and also to ensure that the resulting metric topology is not finer or coarser than the original topology.
Check Your Understanding
Answer each question by identifying the relevant hypothesis or checking the definition directly.
- Under the convention used here, what two requirements are included in regularity?
- Explain why, for \(x\in U\) with \(U\) open, regularity gives an open \(V\) with \(x\in V\subseteq\overline V\subseteq U\).
- Why does the weighted sum of coordinate differences fail to be a metric if the coordinates do not separate distinct points?
- Why must an uncountable discrete space have no countable basis, even though it is metrizable?
- What goes wrong if the \(T_1\) condition is omitted from regularity under the weaker separation convention?