Uniform Comparisons Between Metrics
Two metrics on the same set can assign different numerical distances while still producing exactly the same open sets. A useful sufficient condition is that each distance is bounded above by a fixed constant times the other, with the constants working uniformly for every pair of points. The key is to translate these inequalities into inclusions between open balls.
Let \(d_1\) and \(d_2\) be metrics on a set \(X\). Write \(\mathcal{T}_{d_1}\) and \(\mathcal{T}_{d_2}\) for their induced topologies: a set \(U\subseteq X\) is open in \(\mathcal{T}_{d_i}\) if, for each \(x\in U\), there is \(r>0\) such that \(B_{d_i}(x,r)\subseteq U\), where \(B_{d_i}(x,r)=\{y\in X:d_i(x,y)<r\}\). The strict inequality in this definition means points at distance exactly \(r\) are not in the ball.
Proof. Fix \(x\in X\) and \(r>0\). If \(y\in B_{d_1}(x,r/\alpha)\), then \(d_1(x,y)<r/\alpha\). The first assumed inequality gives \(d_2(x,y)\leq\alpha d_1(x,y)<r\), so \(y\in B_{d_2}(x,r)\). This proves the first inclusion. If \(y\in B_{d_2}(x,r/\beta)\), then \(d_2(x,y)<r/\beta\), and the second assumed inequality gives \(d_1(x,y)\leq\beta d_2(x,y)<r\). Thus \(y\in B_{d_1}(x,r)\), proving the second inclusion.
Now let \(U\in\mathcal{T}_{d_1}\), and fix \(x\in U\). There is \(r>0\) such that \(B_{d_1}(x,r)\subseteq U\). By the second ball inclusion, \(B_{d_2}(x,r/\beta)\subseteq B_{d_1}(x,r)\subseteq U\). Since this holds for every \(x\in U\), \(U\in\mathcal{T}_{d_2}\). Conversely, if \(U\in\mathcal{T}_{d_2}\), then for each \(x\in U\) there is \(r>0\) with \(B_{d_2}(x,r)\subseteq U\). The first inclusion gives \(B_{d_1}(x,r/\alpha)\subseteq U\), so \(U\in\mathcal{T}_{d_1}\). Both inclusions of topologies hold, and hence they are equal. If \(X=\varnothing\), both induced topologies consist only of \(\varnothing\), so the conclusion holds in that case as well. \(\square\)
The constants must be positive, and the bounds must hold for every pair \(x,y\) with the same constants. The theorem gives a sufficient condition for equal topologies; it does not say that every pair of metrics inducing the same topology must satisfy global bounds of this form. Notice also how the radius changes: to fit a \(d_2\)-ball inside a \(d_1\)-ball, the bound \(d_1\leq\beta d_2\) requires dividing the radius by \(\beta\).
Scaled Metrics
Proof. Positivity and symmetry of \(d_c\) follow from those properties of \(d\) and from \(c>0\). Also \(d_c(x,y)=0\) if and only if \(d(x,y)=0\), which holds if and only if \(x=y\). Finally, \(d_c(x,z)=c\,d(x,z)\leq c(d(x,y)+d(y,z))=d_c(x,y)+d_c(y,z)\), so the triangle inequality holds. For the comparison theorem, \(d_c\leq c d\) and \(d\leq (1/c)d_c\) for every pair of points. Thus the two induced topologies agree. \(\square\)
Worked Example: Scaling Distances on the Real Line
Let \(X=\mathbb{R}\), with the standard metric \(d_1(x,y)=|x-y|\), and define \(d_2(x,y)=4|x-y|\). Find the \(d_2\)-ball of radius \(8\) about \(3\), and compare it with a \(d_1\)-ball.
The inequality \(d_2=4d_1\) gives the two global bounds \(d_2\leq4d_1\) and \(d_1\leq\frac14d_2\). The ball-inclusion formulas therefore give \(B_{d_1}(3,2)\subseteq B_{d_2}(3,8)\) and \(B_{d_2}(3,32)\subseteq B_{d_1}(3,8)\). For the requested ball, compute directly: \(d_2(x,3)<8\) if and only if \(4|x-3|<8\), if and only if \(|x-3|<2\). Hence \(B_{d_2}(3,8)=(1,5)=B_{d_1}(3,2)\).
Check: \(x=4\) has \(d_2(4,3)=4<8\), so it is included. The endpoints \(1\) and \(5\) each have \(d_2\)-distance \(4\cdot2=8\) from \(3\), so neither is included. This agrees with the open interval \((1,5)\).
Worked Example: A Scaled Euclidean Metric in the Plane
On \(X=\mathbb{R}^2\), let \(d_1(x,y)=\|x-y\|\) be the Euclidean metric and let \(d_2(x,y)=\frac13\|x-y\|\). Describe \(B_{d_2}((1,-2),\frac23)\).
Here \(d_2=\frac13d_1\), so \(d_2\leq\frac13d_1\) and \(d_1\leq3d_2\). The theorem applies with \(\alpha=\frac13\) and \(\beta=3\). For the ball in question, a point \(y=(y_1,y_2)\) belongs precisely when \(\frac13\sqrt{(y_1-1)^2+(y_2+2)^2}<\frac23\). Multiplying by \(3\), a positive number, gives \(\sqrt{(y_1-1)^2+(y_2+2)^2}<2\). Therefore \(B_{d_2}((1,-2),\frac23)=B_{d_1}((1,-2),2)\), the Euclidean open disk of radius \(2\) centered at \((1,-2)\).
Check: The point \((2,-2)\) is at Euclidean distance \(\sqrt{(2-1)^2+(-2+2)^2}=1\), so its \(d_2\)-distance is \(\frac13\cdot1=\frac13<\frac23\), and it is in the ball. The point \((3,-2)\) is at Euclidean distance \(\sqrt{(3-1)^2+(-2+2)^2}=2\), so its \(d_2\)-distance is \(\frac13\cdot2=\frac23\); it lies on the boundary and is not in the open ball.
Worked Example: Two Different Metrics on a Three-Point Set
Let \(X=\{a,b,c\}\). Define \(d_1\) to be the discrete metric: \(d_1(x,y)=0\) if \(x=y\), and \(d_1(x,y)=1\) otherwise. Define \(d_2(a,b)=2\), \(d_2(a,c)=3\), and \(d_2(b,c)=4\), with symmetry and zero distance from each point to itself. Verify the metric bounds and determine the topology.
For \(d_2\), the three triangle inequalities are \(4\leq2+3=5\), \(3\leq2+4=6\), and \(2\leq3+4=7\). All distances between distinct points are positive, so \(d_2\) is a metric. For every distinct pair, \(d_2\) is one of \(2,3,4\), while \(d_1=1\). Thus \(d_2\leq4d_1\) and \(d_1\leq\frac12d_2\); for equal pairs both sides are zero, so the bounds hold there too. The theorem gives \(\mathcal{T}_{d_1}=\mathcal{T}_{d_2}\).
More explicitly, \(B_{d_1}(a,1)=\{a\}\), since both other points have distance \(1\), which is not strictly less than \(1\). Also \(B_{d_2}(a,2)=\{a\}\), since the distances from \(a\) to \(b,c\) are \(2,3\), neither strictly less than \(2\). The same reasoning gives singleton open balls around \(b\) and \(c\) for both metrics. Every subset of \(X\) is a union of its singleton sets, so every subset is open in both topologies. Both are the discrete topology.
Check: The smallest \(d_2\)-distance between distinct points is \(2\), and \(B_{d_2}(x,2)=\{x\}\) for each \(x\in X\). The smallest \(d_1\)-distance between distinct points is \(1\), and \(B_{d_1}(x,1)=\{x\}\). These strict-radius checks confirm the singleton neighborhoods used above.
Why Both Bounds Matter
A bound in one direction controls only one of the two ball inclusions. It may ensure that balls for one metric contain suitably sized balls for the other, without providing the reverse control needed for equality of the open sets.
Worked Example: One Global Bound Is Not Enough
On \(X=\mathbb{R}\), define \(d_1(x,y)=\min(|x-y|,1)\) and let \(d_2\) be the discrete metric. These are metrics, and \(d_1\leq d_2\) everywhere, but their topologies differ.
To verify the triangle inequality for \(d_1\), put \(s=|x-y|\) and \(t=|y-z|\). The usual triangle inequality gives \(|x-z|\leq s+t\). Since \(u\mapsto\min(u,1)\) is increasing, \(d_1(x,z)\leq\min(s+t,1)\). Also \(\min(s+t,1)\leq\min(s,1)+\min(t,1)\): if either \(s\geq1\) or \(t\geq1\), the right side is at least \(1\); if both are below \(1\), the right side is \(s+t\). Thus \(d_1(x,z)\leq d_1(x,y)+d_1(y,z)\). The other metric properties follow from the absolute-value metric, and \(d_1(x,y)>0\) whenever \(x\ne y\).
For \(x=y\), both distances are zero. For \(x\ne y\), \(d_2(x,y)=1\), while \(d_1(x,y)\leq1\). Hence \(d_1\leq d_2\) on every pair. Nevertheless, \(\{0\}\) is open in the discrete topology: it is \(B_{d_2}(0,\frac12)\). It is not open in \(\mathcal{T}_{d_1}\). For any \(r>0\), set \(y=\min(r/2,1/2)\). Then \(y>0\) and \(d_1(0,y)=y<r\), so every \(d_1\)-ball about \(0\) contains a point other than \(0\). No such ball is contained in \(\{0\}\). The topologies are therefore different.
Check: The missing reverse bound cannot hold for any fixed \(\beta>0\): for \(x=0\) and \(y=2\), \(d_1(0,2)=1\) and \(d_2(0,2)=1\), but for \(y=1/n\), \(d_1(0,1/n)=1/n\) and \(d_2(0,1/n)=1\). The inequality \(d_2\leq\alpha d_1\) would require \(1\leq\alpha/n\) for every \(n\), which fails when \(n>\alpha\).
Common pitfall. If \(d_2=4d_1\), it is incorrect to claim that \(B_{d_2}(x,8)=B_{d_1}(x,8)\). For \(X=\mathbb{R}\), \(d_1(x,y)=|x-y|\), and \(x=0\), direct calculation gives \(B_{d_2}(0,8)=\{y:4|y|<8\}=(-2,2)\), whereas \(B_{d_1}(0,8)=(-8,8)\). The correct radius conversion is \(B_{d_2}(x,8)=B_{d_1}(x,2)\). Equal topologies do not mean equal balls at the same numerical radius; the balls correspond after rescaling their radii.
Check Your Understanding
Use ball inclusions and strict inequalities when checking membership in open balls.
- If \(d_2=7d_1\), express \(B_{d_2}(x,14)\) as a \(d_1\)-ball. Which points at the resulting radius are excluded?
- Suppose \(d_2\leq5d_1\) and \(d_1\leq2d_2\) for all pairs. Which \(d_2\)-ball is guaranteed to lie inside \(B_{d_1}(x,6)\)?
- On a nonempty set, why does multiplying a metric by \(c=0\) fail to produce a metric unless the set has only one point?
- For the three-point example, verify that \(d_2\leq4d_1\) and \(d_1\leq\frac12d_2\) for each of the three distinct unordered pairs.
- In the one-bound example, explain why \(d_1(0,1/n)\) tends to \(0\) while \(d_2(0,1/n)=1\) for every \(n\in\mathbb{N}\).