Comparing Metric Topologies Locally
Two metrics on the same set can assign different numerical distances while still producing exactly the same open sets. The useful question is therefore not whether their distances agree, but whether each metric has sufficiently small balls inside the balls of the other. The size of the ball may depend on the point and on the ball being tested.
Let \(d\) and \(\rho\) be metrics on a set \(X\). For either metric \(m\), write \(B_m(x,r)=\{y\in X:m(x,y)<r\}\), where \(r>0\). The topology induced by \(m\), denoted \(\mathcal{T}_m\), consists of the sets \(U\subseteq X\) such that for every \(x\in U\), some \(m\)-ball centered at \(x\) is contained in \(U\). Thus, to show two induced topologies are equal, it is enough to show that their balls can be made to fit inside one another at each point.
The Ball-Inclusion Criterion
Proof. First suppose the criterion holds. Let \(U\in\mathcal{T}_d\), and take any \(x\in U\). By the definition of \(\mathcal{T}_d\), there is \(r>0\) such that \(B_d(x,r)\subseteq U\). The criterion gives an \(s>0\) with \(B_\rho(x,s)\subseteq B_d(x,r)\subseteq U\). This works at every \(x\in U\), so \(U\in\mathcal{T}_\rho\). Therefore \(\mathcal{T}_d\subseteq\mathcal{T}_\rho\). Applying the other inclusion in the criterion to any \(\rho\)-open set gives \(\mathcal{T}_\rho\subseteq\mathcal{T}_d\). Hence \(\mathcal{T}_d=\mathcal{T}_\rho\).
Conversely, suppose \(\mathcal{T}_d=\mathcal{T}_\rho\). Fix \(x\in X\) and \(r>0\). The ball \(B_d(x,r)\) is open in \(\mathcal{T}_d\), hence also open in \(\mathcal{T}_\rho\). It contains \(x\), so the definition of \(\rho\)-openness gives an \(s>0\) such that \(B_\rho(x,s)\subseteq B_d(x,r)\). Interchanging \(d\) and \(\rho\) proves the other required inclusion. Thus the criterion holds. \(\square\)
There is an equivalent version that starts with a ball and allows its center to be any point in that ball: every \(d\)-ball contains a \(\rho\)-ball around each of its points, and every \(\rho\)-ball contains a \(d\)-ball around each of its points. Indeed, if \(x\in B_d(z,r)\), that ball is \(d\)-open, so the equality of topologies gives a \(\rho\)-ball centered at \(x\) inside it. In the other direction, the criterion stated at the same center follows by taking \(x=z\). This is why the center and the radius in the criterion cannot in general be chosen independently of the ball being tested.
Worked Example: A Bounded Transform of the Usual Metric
On \(X=\mathbb{R}\), let \(d(x,y)=|x-y|\) and define \(\rho(x,y)=\frac{|x-y|}{1+|x-y|}\). We verify that \(\rho\) is a metric and then compare its balls with the \(d\)-balls.
Nonnegativity and symmetry follow from the corresponding properties of absolute value. Also \(\rho(x,y)=0\) exactly when \(|x-y|=0\), which is exactly when \(x=y\). For the triangle inequality, put \(a=|x-y|\) and \(b=|y-z|\). Both are nonnegative, and \(|x-z|\leq a+b\). The function \(\phi(t)=t/(1+t)\) is increasing for \(t\geq0\), since for \(u\geq t\) the difference \(\phi(u)-\phi(t)=(u-t)/((1+u)(1+t))\) is nonnegative. Moreover, direct subtraction gives \(\phi(a)+\phi(b)-\phi(a+b)=ab(2+a+b)/((1+a)(1+b)(1+a+b))\geq0\). Consequently \(\rho(x,z)\leq\phi(a+b)\leq\phi(a)+\phi(b)=\rho(x,y)+\rho(y,z)\).
For \(r>0\), set \(s=r/(1+r)>0\). Since all denominators are positive, the inequality \(\frac{|x-y|}{1+|x-y|}<\frac{r}{1+r}\) is equivalent, after multiplying by \((1+|x-y|)(1+r)\) and cancelling the common term \(r|x-y|\), to \(|x-y|<r\). Therefore \(B_\rho(x,s)=B_d(x,r)\).
For the reverse direction, take a \(\rho\)-ball radius \(s>0\). If \(s<1\), multiplying by positive denominators shows that \(\rho(x,y)<s\) is equivalent to \(|x-y|<s/(1-s)\). Thus \(B_\rho(x,s)=B_d(x,s/(1-s))\). If \(s\geq1\), then \(\rho(x,y)<1\leq s\) for every \(y\in\mathbb{R}\), so \(B_\rho(x,s)=\mathbb{R}\). In this case \(B_d(x,1)\subseteq B_\rho(x,s)\). Both directions of the criterion hold, and the metrics induce the same topology.
Check: when \(r=2\), the chosen \(\rho\)-radius is \(2/(1+2)=2/3\), and the ball equality gives \(B_\rho(x,2/3)=B_d(x,2)\). When \(s=2\), the \(\rho\)-ball is all of \(\mathbb{R}\), since every value of \(\rho\) is less than \(1\), and it contains \(B_d(x,1)\).
Working with Explicit Ball Sizes
Worked Example: Two Metrics on a Three-Point Set
Let \(X=\{a,b,c\}\). Define \(d(a,b)=1\), \(d(a,c)=2\), \(d(b,c)=2\), and define \(\rho(a,b)=3\), \(\rho(a,c)=4\), \(\rho(b,c)=4\). Set each distance from a point to itself equal to zero, and make both functions symmetric. These are metrics: the only potentially largest side in a triangle has length \(2\) for \(d\), which is at most \(1+2=3\); for \(\rho\), the largest side has length \(4\), which is at most \(3+4=7\). The other triangle inequalities follow because all distances are nonnegative and the two equal shorter sides are at least the remaining side where needed.
Every nonzero \(d\)-distance is at least \(1\). Therefore, for any center \(x\), \(B_d(x,1)=\{x\}\), using the strict ball condition \(d(x,y)<1\). Every nonzero \(\rho\)-distance is at least \(3\), so \(B_\rho(x,3)=\{x\}\). Now consider any ball in either metric and any point \(x\) in it. The singleton ball centered at \(x\) in the other metric is contained in the original ball, because it is \(\{x\}\) and the original ball contains its center. Thus \(B_d(x,1)\) fits inside every \(\rho\)-ball containing \(x\), and \(B_\rho(x,3)\) fits inside every \(d\)-ball containing \(x\). In particular, the ball-inclusion criterion holds, and the two metrics induce the same topology.
Check: \(B_d(a,1)=\{a\}\), since \(d(a,b)=1\) does not satisfy \(1<1\), and \(d(a,c)=2\) does not satisfy \(2<1\). Also \(B_\rho(a,3)=\{a\}\), since \(\rho(a,b)=3\) does not satisfy \(3<3\), and \(\rho(a,c)=4\) does not satisfy \(4<3\). The strict inequalities are essential in these boundary-radius cases.
The theorem also explains how to detect when the criterion fails: it is enough to find one ball for which no ball of the other metric, centered at a specified point, fits inside it. A single failed direction rules out equality of the topologies.
Worked Example: The Usual Metric and the Discrete Metric
On \(X=\mathbb{R}\), let \(d(x,y)=|x-y|\), and let \(\delta(x,y)=0\) if \(x=y\), while \(\delta(x,y)=1\) if \(x\ne y\). The function \(\delta\) is a metric: for distinct \(x,z\), at least one of \(x\ne y\) or \(y\ne z\) holds, so \(\delta(x,y)+\delta(y,z)\geq1=\delta(x,z)\); when \(x=z\), the triangle inequality holds because its left side is zero.
Fix \(x\in\mathbb{R}\). The \(\delta\)-ball \(B_\delta(x,1)\) is \(\{x\}\), because \(\delta(x,x)=0<1\) but \(\delta(x,y)=1\) is not less than \(1\) for \(y\ne x\). If the topologies were equal, the converse direction of the criterion would require a \(d\)-ball centered at \(x\) contained in \(\{x\}\). But for every \(r>0\), the point \(y=x+r/2\) satisfies \(d(x,y)=r/2<r\) and \(y\ne x\). Thus \(B_d(x,r)\) is not contained in \(\{x\}\), for every \(r>0\). The criterion fails, so the induced topologies differ.
In the other direction, every \(d\)-ball \(B_d(x,r)\) does contain \(B_\delta(x,1)=\{x\}\). This verifies one direction only. It cannot establish equality, which requires both inclusions.
A Common Mistake and a Practical Test
Common pitfall. A common incorrect argument is: “Every usual ball around \(x\) contains the discrete ball \(B_\delta(x,1)=\{x\}\), so the metrics induce the same topology.” The calculation does show \(B_\delta(x,1)\subseteq B_d(x,r)\) for every \(r>0\), because \(x\) belongs to every ball centered at \(x\). But it proves only one direction. The reverse inclusion would require some positive \(r\) with \(B_d(x,r)\subseteq B_\delta(x,1)=\{x\}\), and the point \(x+r/2\) shows that no such \(r\) exists. The correction is to test both directions, not just to find a small ball that fits inside the ball you started with.
When comparing two metrics, keep the target ball fixed first. Choose its center and radius, then find a positive radius for the other metric that gives the required containment. Repeat with the roles of the metrics reversed. There is no requirement that the radii match, or that one radius work for all centers. A proof of a single global relationship between the distances is not needed for this local criterion.
| To test | What must be shown |
|---|---|
| \(\mathcal{T}_d\subseteq\mathcal{T}_\rho\) | For every \(x\) and \(r>0\), find \(s>0\) with \(B_\rho(x,s)\subseteq B_d(x,r)\). |
| \(\mathcal{T}_\rho\subseteq\mathcal{T}_d\) | For every \(x\) and \(s>0\), find \(r>0\) with \(B_d(x,r)\subseteq B_\rho(x,s)\). |
| Equality of topologies | Prove both containments; neither direction alone suffices. |
Check Your Understanding
For each question, identify which ball inclusion is needed and keep the strict ball condition in mind.
- State the ball-inclusion criterion using \(d\)-balls and \(\rho\)-balls with a common center.
- Why does equality of the induced topologies imply that \(B_d(x,r)\) contains a \(\rho\)-ball centered at \(x\)?
- For the metrics \(d\) and \(\rho\) in the bounded-transform example, what \(\rho\)-radius gives the same ball as \(B_d(x,r)\)?
- For the three-point example, why is \(B_\rho(x,3)=\{x\}\), rather than including a point at distance \(3\)?
- For the usual and discrete metrics on \(\mathbb{R}\), which direction of the ball-inclusion criterion fails, and which point witnesses the failure for a proposed usual-ball radius \(r>0\)?