When Different Distances Give the Same Open Sets
An isometry preserves every distance, while a homeomorphism preserves the open-set structure of two spaces. There is a useful way to connect these ideas: put two metrics on the same set and ask whether they generate the same open sets. If they do, the metrics may assign different values to distances, but they describe the same topology.
Let \(d\) and \(\rho\) be metrics on a nonempty set \(X\). Write \(B_r^d(x)\) for the open ball with center \(x\) and radius \(r\) in the metric \(d\), and use \(B_r^\rho(x)\) for the corresponding ball in \(\rho\). The superscripts help distinguish the two notions of closeness.
The equivalence in the definition follows from the Open-Set Characterization of a Homeomorphism in “Homeomorphisms.” The identity map \(I:X\to X\), \(I(x)=x\), is a bijection whose inverse is also \(I\). Thus it is a homeomorphism precisely when the two metrics give the same open sets. The distances themselves do not have to agree.
A Ball Test for Equivalent Metrics
Open sets in a metric space are detected by balls: around each point of an open set there is a ball contained in that set. This gives a direct test for whether two metrics generate the same open sets. The radii needed in the two metrics may depend on both the center and the desired radius.
- For every \(x\in X\) and every \(\varepsilon>0\), there is a \(\delta>0\) such that \(B_\delta^d(x)\subseteq B_\varepsilon^\rho(x)\).
- For every \(x\in X\) and every \(\varepsilon>0\), there is a \(\delta>0\) such that \(B_\delta^\rho(x)\subseteq B_\varepsilon^d(x)\).
Proof. Suppose first that the metrics generate the same open sets. Fix \(x\in X\) and \(\varepsilon>0\). The ball \(B_\varepsilon^\rho(x)\) is open in the \(\rho\)-metric and therefore open in the \(d\)-metric. Since it contains \(x\), there is a \(\delta>0\) such that \(B_\delta^d(x)\subseteq B_\varepsilon^\rho(x)\). Reversing the roles of the metrics gives the other condition.
Conversely, suppose both ball-containment conditions hold. Let \(U\) be open in the \(\rho\)-metric, and fix \(x\in U\). There is an \(\varepsilon>0\) such that \(B_\varepsilon^\rho(x)\subseteq U\). By the first condition, some \(d\)-ball centered at \(x\) lies inside \(B_\varepsilon^\rho(x)\), and hence inside \(U\). This holds for every \(x\in U\), so \(U\) is open in the \(d\)-metric. The second condition proves that every \(d\)-open set is \(\rho\)-open. The two collections of open sets are therefore equal. \(\square\)
This is a local test: near each fixed point, each metric must be able to control the other. It does not require a single radius that works at every center. That distinction matters when the metrics behave differently in different regions of the space.
Worked Example: Taking the Square Root of a Metric
Let \(d\) be any metric on \(X\), and define \(\rho(x,y)=\sqrt{d(x,y)}\). First, \(\rho\) is a metric. It is nonnegative, symmetric, and zero exactly when \(x=y\), because these properties hold for \(d\). For the triangle inequality, use \(d(x,z)\leq d(x,y)+d(y,z)\) and the inequality \(\sqrt{a+b}\leq\sqrt a+\sqrt b\) for nonnegative \(a,b\): $$ \rho(x,z)=\sqrt{d(x,z)} \leq \sqrt{d(x,y)+d(y,z)} \leq \sqrt{d(x,y)}+\sqrt{d(y,z)} =\rho(x,y)+\rho(y,z). $$
For every \(x\in X\) and \(r>0\), the balls satisfy $$ B_r^\rho(x)=B_{r^2}^d(x), $$ because \(\sqrt{d(x,y)}<r\) if and only if \(d(x,y)<r^2\). Thus the two metrics have exactly the same open sets and are equivalent. Their numerical distances can nevertheless differ: if \(d(x,y)=9\), then \(\rho(x,y)=3\).
Equivalent Metrics and Convergent Sequences
For metric spaces, continuity can be tested using convergent sequences. Applying the Sequential Characterization of a Homeomorphism from “Homeomorphisms” to the identity map gives a useful criterion for equivalence: the two metrics must agree on which sequences converge, and on the limits of those sequences.
Proof. If the metrics are equivalent, the identity map between the two metric spaces is a homeomorphism. The Sequential Characterization of a Homeomorphism then gives the stated equivalence of convergence.
For the reverse direction, suppose that the two metrics have the same convergent sequences with the same limits. Fix \(x\in X\) and \(\varepsilon>0\). If there were no \(\delta>0\) such that \(B_\delta^d(x)\subseteq B_\varepsilon^\rho(x)\), then for every positive integer \(n\) we could choose \(x_n\) satisfying $$ d(x_n,x)<\frac{1}{n} \quad\text{and}\quad \rho(x_n,x)\geq\varepsilon. $$ The first inequality gives \(d(x_n,x)\to0\). By the assumed agreement of convergence, \(\rho(x_n,x)\to0\), contradicting \(\rho(x_n,x)\geq\varepsilon\) for every \(n\). So the first ball-containment condition holds. Applying the same argument with the two metrics interchanged proves the second condition. The Ball Characterization of Equivalent Metrics now shows that \(d\) and \(\rho\) are topologically equivalent. \(\square\)
Worked Example: A Metric from the Cubing Map
On \(\mathbb{R}\), let \(d(x,y)=|x-y|\) and define \(\rho(x,y)=|x^3-y^3|\). The map \(h(x)=x^3\) is injective, so the Metric Induced by an Injective Map theorem shows that \(\rho\) is a metric. In particular, this construction does not require that the formula for \(\rho\) resemble the usual distance.
These metrics are equivalent. The cubing map \(h\) is a homeomorphism from \(\mathbb{R}\) with its usual metric to \(\mathbb{R}\) with its usual metric: it is bijective, continuous, and has the continuous inverse \(h^{-1}(u)=\sqrt[3]{u}\). Also, \(h\) is an isometry from \((\mathbb{R},\rho)\) to \((\mathbb{R},d)\), since $$ d(h(x),h(y))=|x^3-y^3|=\rho(x,y). $$ Consequently, the identity map from \((\mathbb{R},d)\) to \((\mathbb{R},\rho)\) is a homeomorphism. Equivalently, for any sequence \((x_n)\) and any \(x\), continuity of cubing and cube root gives $$ |x_n-x|\longrightarrow0 \quad\text{if and only if}\quad |x_n^3-x^3|\longrightarrow0. $$
Global Bounds: A Convenient Sufficient Condition
Sometimes the ball test can be verified by comparing distances directly. A pair of global bounds is particularly convenient. This condition is stronger than topological equivalence: it forces each metric to control the other at every point with the same constants.
Proof. Fix \(x\in X\) and \(\varepsilon>0\). If \(d(x,y)<\varepsilon/C\), then $$ \rho(x,y)\leq C\,d(x,y)<\varepsilon. $$ Thus \(B_{\varepsilon/C}^d(x)\subseteq B_\varepsilon^\rho(x)\). If \(\rho(x,y)<c\varepsilon\), then $$ d(x,y)\leq \frac{\rho(x,y)}{c}<\varepsilon, $$ so \(B_{c\varepsilon}^\rho(x)\subseteq B_\varepsilon^d(x)\). The Ball Characterization of Equivalent Metrics proves topological equivalence. The choices of radii depend only on \(\varepsilon,c,C\), not on \(x\). They therefore also establish uniform continuity of the identity maps in both directions. \(\square\)
Worked Example: Rescaling All Distances
Let \(d\) be a metric and define \(\rho(x,y)=5d(x,y)\). This is a metric, and the global bounds hold with \(c=C=5\): $$ 5d(x,y)\leq \rho(x,y)\leq5d(x,y). $$ The theorem shows that the metrics are equivalent. More directly, \(B_r^\rho(x)=B_{r/5}^d(x)\). The factor changes every positive distance, but it does not change which sets are open.
Topological Equivalence Is Weaker Than Uniform Equivalence
Equivalent metrics need not satisfy global comparison bounds, and their identity maps need not be uniformly continuous. Topological equivalence requires only local control around each point. Uniform continuity requires a single distance threshold to work throughout the space.
Worked Example: Equivalent Metrics Without Uniform Equivalence
Use the usual metric \(d(x,y)=|x-y|\) on \(\mathbb{R}\) and the equivalent metric \(\rho(x,y)=|x^3-y^3|\) from the earlier example. The identity from \((\mathbb{R},d)\) to \((\mathbb{R},\rho)\) is not uniformly continuous. For each positive integer \(n\), set \(x_n=n\) and \(y_n=n+1/n^2\). Then $$ d(x_n,y_n)=\frac{1}{n^2}\longrightarrow0, $$ but direct expansion gives $$ \rho(x_n,y_n) =\left|\left(n+\frac{1}{n^2}\right)^3-n^3\right| =3+\frac{3}{n^3}+\frac{1}{n^6}\geq3. $$ Thus arbitrarily close pairs in the usual metric can remain separated by at least \(3\) in \(\rho\), contradicting the Sequential Criterion for Uniform Continuity. The metrics generate the same open sets, but the identity in this direction is not uniformly continuous.
By contrast, the identity from \((\mathbb{R},\rho)\) to \((\mathbb{R},d)\) is uniformly continuous: if \(\rho(x,y)<\delta\), then continuity of the cube root, or the estimate for nearby cube roots, can be used locally; uniform continuity of the cube-root function on all of \(\mathbb{R}\) also gives a single suitable \(\delta\) for each desired \(d\)-distance. Uniform continuity in one direction does not make the metrics uniformly equivalent, which requires it in both directions.
A common pitfall is to treat “equivalent” as meaning “bounded above and below by constant multiples.” Such inequalities are sufficient, as the global-bounds theorem shows, but they are not necessary. The ball test allows radii to depend on the center, and the sequential test asks only whether convergence is preserved. Both capture topological equivalence without imposing a global rate of control.
Keep track of which balls and distances are measured by each one.
Use ball containment, convergent sequences, or global distance bounds.
Equivalence requires control from \(d\) to \(\rho\) and from \(\rho\) to \(d\).
Check Your Understanding
Use the definitions and results in this tutorial to answer the following questions.
- What does it mean for two metrics on the same set to be topologically equivalent?
- State both ball-containment conditions in the Ball Characterization of Equivalent Metrics.
- Why does the sequential characterization require the same limits, not merely the same convergent sequences?
- For \(\rho(x,y)=\sqrt{d(x,y)}\), express a \(\rho\)-ball of radius \(r\) as a \(d\)-ball.
- Are global bounds by positive constant multiples necessary for topological equivalence, or only sufficient?
- What additional property is required for two metrics to be uniformly equivalent?