Comparisons Have Two Different Objects
When comparing metrics, it is easy to slide between statements about distances and statements about open sets. These are related, but they are not the same: a metric assigns numbers to pairs of points, while its induced topology is a collection of subsets. Even when a distance inequality is true, its direction must be translated carefully before drawing a conclusion about which topology is finer.
Let \(d\) be a metric on a set \(X\). Its induced topology, denoted \(\mathcal{T}_d\), consists of the sets \(U\subseteq X\) such that for every \(x\in U\), some open ball \(B_d(x,r)=\{y\in X:d(x,y)<r\}\), with \(r>0\), is contained in \(U\). Thus the topology is a family of subsets of \(X\), not a collection of distances or a single ball.
The notation \(\mathcal{T}_1\subseteq\mathcal{T}_2\) compares collections: every set open in \(\mathcal{T}_1\) must also be open in \(\mathcal{T}_2\). It does not say that every set open in \(\mathcal{T}_2\) is open in \(\mathcal{T}_1\). The following direction rule is useful when the metrics are defined on the same set.
Proof. Let \(U\in\mathcal{T}_{d_1}\), and fix \(x\in U\). By the definition of the induced topology, there is \(r>0\) such that \(B_{d_1}(x,r)\subseteq U\). If \(y\in B_{d_2}(x,r/C)\), then \(d_2(x,y)<r/C\), so \(d_1(x,y)\le C d_2(x,y)<r\). Hence \(y\in B_{d_1}(x,r)\). We have shown \(B_{d_2}(x,r/C)\subseteq U\). This holds for every \(x\in U\), so \(U\in\mathcal{T}_{d_2}\), as required. \(\square\)
In particular, when \(d_1\le d_2\), the topology from \(d_1\) is contained in the topology from \(d_2\). The smaller-distance metric gives balls that can be larger at the same radius, but the topology conclusion comes from the ball inclusion in the proof—not from informally comparing “size.” The inequality alone gives an inclusion of topologies; it does not in general give equality.
Worked Comparisons
Worked Example: A One-Sided Bound Gives a Strict Inclusion
Let \(X=\{0\}\cup\{1/n:n\in\mathbb{N}\}\). Define \(d(x,y)=|x-y|\), the restriction of the usual metric on \(\mathbb{R}\), and define \(\rho(x,y)=0\) if \(x=y\), and \(\rho(x,y)=1\) if \(x\ne y\). The latter is the discrete metric on \(X\).
For distinct \(x,y\in X\), both points lie in \([0,1]\), so \(0<d(x,y)\le1=\rho(x,y)\); for \(x=y\), both distances are \(0\). Thus \(d(x,y)\le\rho(x,y)\) for all pairs. The proposition, with \(d_1=d\), \(d_2=\rho\), and \(C=1\), gives \(\mathcal{T}_d\subseteq\mathcal{T}_\rho\).
This inclusion is strict. The \(\rho\)-ball \(B_\rho(0,1/2)\) is \(\{0\}\), since \(\rho(0,0)=0<1/2\) and \(\rho(0,1/n)=1\not<1/2\) for every \(n\in\mathbb{N}\). So \(\{0\}\in\mathcal{T}_\rho\). But \(\{0\}\notin\mathcal{T}_d\): for every \(r>0\), choose \(n\in\mathbb{N}\) with \(n>1/r\). Then \(0<1/n<r\), so \(1/n\in B_d(0,r)\) and that ball is not contained in \(\{0\}\).
Check: at radius \(1/2\), \(B_d(0,1/2)\) contains \(1/3\), because \(d(0,1/3)=1/3<1/2\), whereas \(B_\rho(0,1/2)=\{0\}\). This agrees with the inclusion of topologies and shows that it can be strict.
Worked Example: Different Distances, the Same Topology
On \(\mathbb{R}\), let \(d(x,y)=|x-y|\) and \(\rho(x,y)=2|x-y|\). These are different metrics: \(d(0,1)=1\), whereas \(\rho(0,1)=2\). Nevertheless, their induced topologies are equal.
For every \(x\in\mathbb{R}\) and \(r>0\), membership in the \(\rho\)-ball satisfies \(\rho(x,y)<r\) if and only if \(2|x-y|<r\), if and only if \(|x-y|<r/2\). Therefore \(B_\rho(x,r)=B_d(x,r/2)\). Conversely, \(B_d(x,r)=B_\rho(x,2r)\), since \(2|x-y|<2r\) is equivalent to \(|x-y|<r\). Each metric’s balls can thus be written as balls for the other metric, so each induced topology is contained in the other.
Check: \(B_\rho(2,1)=\{y:2|y-2|<1\}=(3/2,5/2)\), and \(B_d(2,1/2)=\{y:|y-2|<1/2\}=(3/2,5/2)\). The radii differ, but the sets agree. Equal topologies do not require equal metric values.
Worked Example: Two Metrics on a Two-Point Set
Let \(X=\{a,b\}\). Define \(d(a,b)=d(b,a)=2\) and \(\rho(a,b)=\rho(b,a)=5\); set \(d(a,a)=d(b,b)=\rho(a,a)=\rho(b,b)=0\). Both functions are metrics: they are symmetric, positive on distinct points, and the triangle inequality holds because a route from one point through the other point back to itself has length \(2+2\ge0\) for \(d\), and \(5+5\ge0\) for \(\rho\). For a route from \(a\) to \(b\) through either point, its length is at least the direct distance, since one of the two segments already has length \(2\) for \(d\), or \(5\) for \(\rho\).
For \(0<r\le2\), the strict ball condition \(d(x,y)<r\) gives \(B_d(a,r)=\{a\}\) and \(B_d(b,r)=\{b\}\). For \(r>2\), each of these \(d\)-balls is \(X\). The empty set and \(X\) are open as well, so \(\mathcal{T}_d=\mathcal{P}(X)\), the discrete topology. For \(\rho\), the same calculation with its distance value gives singleton balls when \(0<r\le5\), and \(X\) when \(r>5\). Hence \(\mathcal{T}_\rho=\mathcal{P}(X)=\mathcal{T}_d\), despite \(d(a,b)=2\ne5=\rho(a,b)\).
Check: the endpoint radii matter because the ball condition is strict: at \(r=2\), \(d(a,b)=2\not<2\), so \(b\notin B_d(a,2)\); at \(r=5\), \(\rho(a,b)=5\not<5\), so \(b\notin B_\rho(a,5)\). Both balls are still singletons at those radii.
How to Audit a Comparison
When a problem gives two metrics or two topologies, separate the checks rather than jumping to a conclusion.
A metric comparison concerns values \(d_1(x,y)\) and \(d_2(x,y)\). A topology comparison concerns families of subsets \(\mathcal{T}_{d_1}\) and \(\mathcal{T}_{d_2}\).
For example, \(d_1(x,y)\le C d_2(x,y)\) for every \(x,y\in X\), with \(C>0\), is a global bound. A bound at only one pair of points cannot support a conclusion about the whole topology.
From \(d_1\le C d_2\), the directly justified inclusion is \(B_{d_2}(x,r/C)\subseteq B_{d_1}(x,r)\). Keep both the metric labels and the radii visible.
The proposition gives \(\mathcal{T}_{d_1}\subseteq\mathcal{T}_{d_2}\). To conclude equality, one must also establish the reverse inclusion, for example by a bound in the other direction or by a direct argument with balls.
A common error is to treat \(d_1(x,y)\le d_2(x,y)\) as if it implied \(B_{d_1}(x,r)\subseteq B_{d_2}(x,r)\) at the same radius. That is the wrong direction: a small \(d_2\)-distance forces a small \(d_1\)-distance, but a small \(d_1\)-distance need not force a small \(d_2\)-distance.
Wrong calculation: in the first example, take \(x=0\), \(y=1\), and \(r=3/4\). Then \(d(0,1)=1\), which is not less than \(3/4\), so this pair does not demonstrate the error. Instead take \(x=0\), \(y=1/2\) and \(r=3/4\): \(d(0,1/2)=1/2<3/4\), but \(\rho(0,1/2)=1\not<3/4\). Thus \(1/2\in B_d(0,3/4)\) but \(1/2\notin B_\rho(0,3/4)\), disproving the mistaken same-radius inclusion. The justified inclusion goes the other way: if \(\rho(0,y)<3/4\), then \(d(0,y)\le\rho(0,y)<3/4\), so \(B_\rho(0,3/4)\subseteq B_d(0,3/4)\).
Why the Distinction Matters
Topological questions are about open sets, so metric estimates are useful insofar as they control balls and hence open sets. A metric can change all its numerical distances while leaving its open sets unchanged, as rescaling shows. Conversely, a one-sided estimate can leave a genuine difference in open sets, as the convergent sequence together with its limit shows. The statement “these metrics are close” is not precise enough: specify the inequality, its direction, and whether it holds for all pairs or only locally.
When checking an argument, ask: Are the spaces the same set? Are the metrics defined for every pair being compared? Is the constant positive? Does the distance inequality give the ball inclusion used in the proof? If equality of topologies is claimed, has the reverse inclusion also been established? These checks prevent a correct numerical estimate from being turned into an unsupported topological conclusion.
Check Your Understanding
For each question, keep metric inequalities separate from statements about open sets.
- If \(d_1(x,y)\le 4d_2(x,y)\) for every \(x,y\in X\), which topology inclusion follows?
- On \(\mathbb{R}\), compare \(d(x,y)=|x-y|\) and \(\rho(x,y)=3|x-y|\). Write a ball for each metric as a ball for the other.
- For the discrete metric \(\rho\) on a set with at least two points, what is \(B_\rho(x,1)\)? What is \(B_\rho(x,1/2)\)?
- Explain why a proof of \(\mathcal{T}_{d_1}\subseteq\mathcal{T}_{d_2}\) alone does not prove equality.
- Give a pair of metrics with different distances but the same induced topology, and state how their balls correspond.