Metrizability: A Property of the Open Sets
A metric assigns numerical distances to pairs of points. A topology records which sets are open. These are related, but they are not the same data: a metric produces a topology through its open balls, and different metrics can produce exactly the same open sets. The relevant question is therefore whether a space has at least one metric that generates its topology, rather than whether a particular metric has been specified.
The phrase “some metric” is important. Metrizability is determined by the topology: it asks whether there exists a suitable metric, and it does not select one unique metric. In particular, a statement about distances, such as “every pair of points is at distance at most \(1\),” refers to a chosen metric, not to the topology alone.
Homeomorphisms Preserve Metrizability
A homeomorphism is a bijection that is continuous in both directions. It preserves the open-set structure, so a metric can be transported across it.
Proof. Assume \(X\) is metrizable, and choose a metric \(d\) inducing the topology \(\mathcal{T}_X\) on \(X\). For \(u,v\in Y\), define \(\rho(u,v)=d(h^{-1}(u),h^{-1}(v))\). The function \(\rho\) is nonnegative and symmetric because \(d\) is. If \(\rho(u,v)=0\), then \(d(h^{-1}(u),h^{-1}(v))=0\), so \(h^{-1}(u)=h^{-1}(v)\), and hence \(u=v\); conversely, \(u=v\) gives \(\rho(u,v)=0\). For \(u,v,w\in Y\), the triangle inequality for \(d\) gives \(\rho(u,w)\leq \rho(u,v)+\rho(v,w)\). Thus \(\rho\) is a metric.
We now verify equality of the topology induced by \(\rho\) and the given topology \(\mathcal{T}_Y\); both inclusions are necessary. For every \(u\in Y\) and \(r>0\), the definition gives \(B_\rho(u,r)=h(B_d(h^{-1}(u),r))\). The \(d\)-ball is open in \(X\), and \(h\) sends open sets to open sets because \(h^{-1}\) is continuous. Therefore every \(\rho\)-ball is open in \(Y\), so every set open in the \(\rho\)-topology is open in \(\mathcal{T}_Y\). Conversely, let \(V\in\mathcal{T}_Y\). The set \(h^{-1}(V)\) is open in \(X\), since \(h\) is continuous. For each \(u\in V\), there is an \(r>0\) such that \(B_d(h^{-1}(u),r)\subseteq h^{-1}(V)\). Applying \(h\) and using the ball identity gives \(B_\rho(u,r)\subseteq V\). Thus every point of \(V\) has a \(\rho\)-ball contained in \(V\), so \(V\) is open in the \(\rho\)-topology. The two topologies agree, and \(Y\) is metrizable. Applying the same argument to \(h^{-1}:Y\to X\) proves the converse implication, so metrizability is preserved in both directions. \(\square\)
Worked Example: Transporting the Metric from the Real Line
Let \(X=\mathbb{R}\) and \(Y=(0,\infty)\), both with their standard topologies (the latter as a subspace of \(\mathbb{R}\)). The exponential function \(h(x)=e^x\) is a homeomorphism from \(X\) to \(Y\): it is a continuous bijection, and its inverse \(h^{-1}(u)=\ln u\) is continuous on \(Y\). Starting with the usual metric \(d(x,y)=|x-y|\) on \(X\), the transported metric on \(Y\) is \(\rho(u,v)=|\ln u-\ln v|\).
For example, \(\rho(1,4)=|\ln 1-\ln 4|=|0-\ln 4|=\ln 4\), while the usual distance in \(Y\) between those same points is \(|1-4|=3\). These are different numerical distances. Nevertheless, the transported metric generates the standard topology on \(Y\), because it is obtained by transporting the usual metric through the homeomorphism.
We can check a ball directly. For \(r=\ln 2>0\), membership in \(B_\rho(1,\ln 2)\) means \(|\ln u-\ln 1|<\ln 2\). Since \(\ln 1=0\), this is equivalent to \(-\ln 2<\ln u<\ln 2\). Applying the increasing exponential function gives \(1/2<u<2\), so \(B_\rho(1,\ln 2)=(1/2,2)\). Substitution confirms the endpoints are excluded: \(\rho(1,1/2)=|\ln(1/2)|=\ln 2\) and \(\rho(1,2)=|\ln 2|=\ln 2\), neither of which is strictly less than the radius.
Different Metrics, the Same Topology
Two metrics on the same set may assign different distances while generating identical open sets. The following construction shows that every metric has a bounded alternative with the same topology.
Proof. The function \(d_b\) is nonnegative and symmetric. Also, \(d_b(x,y)=0\) holds exactly when \(d(x,y)=0\), which holds exactly when \(x=y\). To check the triangle inequality, put \(a=d(x,y)\) and \(b=d(y,z)\). Since \(d(x,z)\leq a+b\) and the function \(t\mapsto\min\{t,1\}\) is increasing, \(d_b(x,z)\leq\min\{a+b,1\}\). For all nonnegative \(a,b\), \(\min\{a+b,1\}\leq\min\{a,1\}+\min\{b,1\}\): if either \(a\) or \(b\) is at least \(1\), the right side is at least \(1\); if both are less than \(1\), the right side is \(a+b\). Thus \(d_b(x,z)\leq d_b(x,y)+d_b(y,z)\), proving that \(d_b\) is a metric.
For every \(x\in X\) and \(0<r<1\), \(B_{d_b}(x,r)=B_d(x,r)\), because \(\min\{d(x,y),1\}<r<1\) holds exactly when \(d(x,y)<r\). Now suppose \(U\) is open in the \(d\)-topology and \(x\in U\). Choose \(r>0\) with \(B_d(x,r)\subseteq U\), and set \(s=\min\{r/2,1/2\}\). Then \(0<s<1\), and \(B_{d_b}(x,s)=B_d(x,s)\subseteq B_d(x,r)\subseteq U\). So \(U\) is open in the \(d_b\)-topology. Conversely, if \(U\) is open in the \(d_b\)-topology, choose \(r>0\) with \(B_{d_b}(x,r)\subseteq U\) for each \(x\in U\). Set \(s=\min\{r/2,1/2\}\) again. Then \(B_d(x,s)=B_{d_b}(x,s)\subseteq B_{d_b}(x,r)\subseteq U\), so \(U\) is open in the \(d\)-topology. The topologies are equal. \(\square\)
Worked Example: The Usual and a Bounded Metric on the Real Line
On \(\mathbb{R}\), let \(d(x,y)=|x-y|\) and \(d_b(x,y)=\min\{|x-y|,1\}\). The proposition proves that both induce the standard topology. For a radius below \(1\), their balls agree exactly: \(B_{d_b}(0,1/4)=B_d(0,1/4)=(-1/4,1/4)\). Indeed, \(\min\{|y|,1\}<1/4\) is equivalent to \(|y|<1/4\).
The distances can nevertheless differ. Substitution gives \(d(0,2)=|0-2|=2\), whereas \(d_b(0,2)=\min\{2,1\}=1\). For \(0\) and \(1/4\), the values are \(d(0,1/4)=1/4\) and \(d_b(0,1/4)=\min\{1/4,1\}=1/4\). Every \(d_b\)-distance is at most \(1\) by its definition, while the usual metric is unbounded, since \(d(0,n)=n\) for every \(n\in\mathbb{N}\). These checks illustrate the distinction: the open sets agree although the distances do not.
Check: the open ball of radius \(1/4\) around \(0\) is \((-1/4,1/4)\) for either metric, and the points \(y=\pm1/4\) are not in the ball, since both distances from \(0\) are exactly \(1/4\), not strictly less than \(1/4\).
Worked Example: A Nonmetrizable Two-Point Space
Let \(X=\{p,q\}\) with the indiscrete topology \(\mathcal{T}=\{\varnothing,X\}\). This topology is not induced by any metric. If a metric \(d\) induced it, distinctness of \(p\) and \(q\) would give \(d(p,q)>0\). The ball \(B_d(p,d(p,q))\) would contain \(p\), because \(d(p,p)=0<d(p,q)\), and would not contain \(q\), because \(d(p,q)\) is not strictly less than itself. This is a nonempty proper open set in the metric topology. But the given topology has no nonempty proper open set, a contradiction. Hence \(X\) is nonmetrizable.
The failure can also be seen through the Hausdorff property. A space is Hausdorff if any two distinct points have disjoint open neighborhoods. In this indiscrete space, the only open neighborhood of either point is \(X\), so the two points cannot have disjoint open neighborhoods.
Proof. Let \(d\) induce the topology on \(X\), and take distinct points \(x,y\in X\). Then \(a=d(x,y)>0\). The balls \(B_d(x,a/3)\) and \(B_d(y,a/3)\) are open neighborhoods of \(x\) and \(y\). If they intersected, there would be a point \(z\) with \(d(x,z)<a/3\) and \(d(y,z)<a/3\). The triangle inequality would give \(a=d(x,y)\leq d(x,z)+d(z,y)<2a/3\), contradicting \(a>0\). The balls are disjoint, so \(X\) is Hausdorff. \(\square\)
Common Pitfalls and Why the Distinction Matters
Common pitfall: treating a change of scale as if it left each ball unchanged. If \(\rho(x,y)=2d(x,y)\), it is wrong to assert \(B_\rho(x,r)=B_d(x,r)\) at the same radius \(r\). For example, on \(\mathbb{R}\), with \(x=0\) and \(r=1\), \(B_d(0,1)=(-1,1)\), while \(B_\rho(0,1)=(-1/2,1/2)\), because \(2|y|<1\) is equivalent to \(|y|<1/2\). The correction is \(B_\rho(x,r)=B_d(x,r/2)\). The topologies still agree: scaling the radius gives a ball of either metric inside any prescribed ball of the other.
Another related error is to replace “homeomorphism” by “continuous bijection” in the metrizability theorem. On the two-point set \(X=\{p,q\}\), take the discrete topology on the domain and the indiscrete topology on the codomain. The identity map is a continuous bijection: preimages of the only codomain-open sets, \(\varnothing\) and \(X\), are open in the discrete topology. It is not a homeomorphism, because its inverse would have to take the open singleton \(\{p\}\) in the discrete space to \(\{p\}\), which is not open in the indiscrete space. The target is nonmetrizable by the preceding example, whereas the discrete domain is metrizable by the metric assigning distance \(1\) to distinct points. Continuity and bijectivity alone do not preserve metrizability.
Metrizability matters because it lets us use distance-based tools to study a topology, while the homeomorphism theorem says that this possibility does not depend on the names or coordinates used for the points. At the same time, conclusions about a particular metric—its exact distances or whether it is bounded—must not be mistaken for conclusions about the topology. The topology records the open sets; a metric is one possible way to generate them.
Check Your Understanding
For each question, distinguish carefully between the metric and the topology it induces.
- State the definition of a metric inducing a topology, and explain the existential quantifier in the definition of a metrizable space.
- If \(d\) induces a topology on \(X\), why do \(d\) and \(d_b(x,y)=\min\{d(x,y),1\}\) have the same balls of radius \(r\) whenever \(0<r<1\)?
- In the proof that a homeomorphism preserves metrizability, which continuity direction shows that metric balls transported to \(Y\) are open in \(Y\)? Which direction shows that every given open set in \(Y\) is open in the transported metric?
- Why can the indiscrete topology on a two-point set not be induced by a metric?
- Give an example of two different metrics on \(\mathbb{R}\) with the same topology, and identify one numerical distance on which they disagree.