Local Bounds Can Compare Topologies
Global bounds give a direct way to compare two metrics: if each distance is bounded above and below by a fixed positive multiple of the other, the metrics induce the same topology. But the constants in a useful comparison may depend on where we are. Local estimates can still prove that the open sets agree, provided they control not only nearby points but also rule out distant points becoming arbitrarily close in the other metric.
Throughout, \(d\) and \(\rho\) are metrics on the same set \(X\). The topology induced by \(d\) has as its open sets those \(U\subseteq X\) such that for every \(x\in U\), some \(d\)-ball \(B_d(x,r)\) is contained in \(U\). The topology induced by \(\rho\) is defined in the same way using \(\rho\)-balls. We prove a pointwise criterion for these topologies to agree.
(i) if \(d(x,y)<r_x\), then \(c_xd(x,y)\leq \rho(x,y)\leq C_xd(x,y)\);
(ii) if \(d(x,y)\geq r_x\), then \(\rho(x,y)\geq s_x\).
The constants may depend on \(x\). Condition (ii) prevents points outside the \(d\)-neighborhood from entering arbitrarily small \(\rho\)-balls about \(x\).
A Pointwise Criterion for Equal Metric Topologies
Proof. We show that a set \(U\subseteq X\) is \(d\)-open if and only if it is \(\rho\)-open. First suppose \(U\) is \(d\)-open, and fix \(x\in U\). Choose \(\varepsilon>0\) such that \(B_d(x,\varepsilon)\subseteq U\). Using the constants at \(x\), set \(\delta=\min\{s_x,C_x\varepsilon\}\), which is positive. If \(\rho(x,y)<\delta\), condition (ii) rules out \(d(x,y)\geq r_x\), since that would give \(\rho(x,y)\geq s_x\geq\delta\). Thus \(d(x,y)<r_x\), and condition (i) gives \(c_xd(x,y)\leq\rho(x,y)<\delta\leq C_x\varepsilon\). To ensure the required \(d\)-bound, instead use \(\delta=\min\{s_x,c_x\varepsilon\}\). Then \(d(x,y)<r_x\) as before, and \(c_xd(x,y)\leq\rho(x,y)<\delta\leq c_x\varepsilon\), so \(d(x,y)<\varepsilon\). Therefore \(B_\rho(x,\delta)\subseteq U\), proving that \(U\) is \(\rho\)-open.
Conversely, suppose \(U\) is \(\rho\)-open and \(x\in U\). Choose \(\eta>0\) such that \(B_\rho(x,\eta)\subseteq U\). Let \(\gamma=\min\{r_x,\eta/C_x\}>0\). If \(d(x,y)<\gamma\), then condition (i) applies and gives \(\rho(x,y)\leq C_xd(x,y)<C_x\gamma\leq\eta\). Hence \(y\in B_\rho(x,\eta)\subseteq U\). Thus \(B_d(x,\gamma)\subseteq U\), so \(U\) is \(d\)-open. The two collections of open sets are equal. \(\square\)
The two parts of the criterion have different jobs. The local upper bound turns sufficiently small \(d\)-distances into small \(\rho\)-distances. The local lower bound works in the opposite direction once a point is known to be in the comparison region. Condition (ii) ensures that a small \(\rho\)-distance really does place the point in that region. Without it, the local lower bound alone need not control every point in a small \(\rho\)-ball.
Metrics Built from Increasing Functions
A convenient source of examples is a strictly increasing function \(f:J\to\mathbb R\), where \(J\) is an interval. Define \(\rho(x,y)=|f(x)-f(y)|\). When \(f\) is injective, \(\rho\) is a metric: positivity and symmetry follow from the absolute value and injectivity, and the triangle inequality follows from \(|f(x)-f(z)|\leq |f(x)-f(y)|+|f(y)-f(z)|\). If \(f\) is differentiable, the Mean Value Theorem often supplies local estimates: for distinct \(x,y\), there is a point \(\xi\) strictly between them with \(\rho(x,y)=|f'(\xi)|\,|x-y|\). The estimates must be checked on the interval between the particular pair of points.
Worked Example: A Cubic Change of Distance
On \(X=\mathbb R\), let \(d(x,y)=|x-y|\), and set \(f(x)=x+x^3\) and \(\rho(x,y)=|f(x)-f(y)|\). The derivative \(f'(t)=1+3t^2\) is positive for every \(t\), so the Mean Value Theorem shows that \(f\) is strictly increasing and hence injective. Thus \(\rho\) is a metric.
Fix \(x\in\mathbb R\) and take \(r_x=1\). If \(|x-y|<1\), every point \(t\) between \(x\) and \(y\) satisfies \(|t|<|x|+1\). Consequently \(1\leq f'(t)\leq1+3(|x|+1)^2\). The Mean Value Theorem gives \[ |x-y|\leq \rho(x,y)\leq \bigl(1+3(|x|+1)^2\bigr)|x-y|. \] Thus we may take \(c_x=1\) and \(C_x=1+3(|x|+1)^2\). If \(|x-y|\geq1\), the same theorem and \(f'(t)\geq1\) give \(\rho(x,y)\geq|x-y|\geq1\). Take \(s_x=1\). The criterion proves that \(d\) and \(\rho\) induce the same topology.
Check: at \(x=2\), the constants above are \(r_2=1\), \(c_2=1\), \(C_2=1+3(3)^2=28\), and \(s_2=1\). For \(y=5/2\), \(d(2,5/2)=1/2<1\), and direct substitution gives \(\rho(2,5/2)=|(5/2+125/8)-(2+8)|=|145/8-10|=65/8\). The bounds read \(1/2\leq65/8\leq28(1/2)=14\), which hold.
There is no global constant \(C\) with \(\rho(x,y)\leq C d(x,y)\) for all \(x,y\). Taking \(x=0\) and \(y=n\in\mathbb N\) gives \(\rho(0,n)/d(0,n)=(n+n^3)/n=1+n^2\), which exceeds any fixed \(C\) for sufficiently large \(n\).
Worked Example: The Arctangent Metric
On \(\mathbb R\), let \(f(x)=\arctan x\) and \(\rho(x,y)=|\arctan x-\arctan y|\). The function is strictly increasing, since \(f'(t)=1/(1+t^2)>0\); therefore \(\rho\) is a metric. Fix \(x\) and set \(r_x=1\). If \(|x-y|<1\), every \(t\) between \(x\) and \(y\) satisfies \(|t|<|x|+1\), so \[ \frac{1}{1+(|x|+1)^2}\leq f'(t)\leq1. \] The Mean Value Theorem gives the required local bounds with \(c_x=1/(1+(|x|+1)^2)\) and \(C_x=1\).
If \(y\geq x+1\), monotonicity gives \(\rho(x,y)\geq f(x+1)-f(x)>0\). If \(y\leq x-1\), it gives \(\rho(x,y)\geq f(x)-f(x-1)>0\). Thus condition (ii) holds with \(s_x=\min\{f(x+1)-f(x),f(x)-f(x-1)\}\), a positive number. The theorem again proves equality of the metric topologies.
Check: at \(x=0\), \(r_0=1\), \(c_0=1/2\), and \(C_0=1\). For \(y=1/2\), the local bounds give \(1/4\leq\arctan(1/2)\leq1/2\). Indeed, the Mean Value Theorem gives \(\arctan(1/2)=\frac12/(1+\xi^2)\) for some \(0<\xi<1/2\), and hence \(2/5<1/(1+\xi^2)<1\), which verifies the stated, weaker bounds.
No global positive lower comparison constant exists: for \(n\in\mathbb N\), \(\rho(0,n)/d(0,n)=\arctan(n)/n\to0\), because \(0<\arctan(n)<\pi/2\) while \(n\to\infty\). Thus local lower constants do not imply a global lower constant.
Worked Example: The Logarithm Metric on the Positive Reals
Take \(X=(0,\infty)\), \(d(x,y)=|x-y|\), and \(\rho(x,y)=|\ln x-\ln y|\). The logarithm is strictly increasing on this interval, with derivative \(1/t>0\), so the pullback construction gives a metric. Fix \(x>0\) and choose \(r_x=x/2\). When \(|x-y|<x/2\), every \(t\) between \(x\) and \(y\) lies between \(x/2\) and \(3x/2\). Therefore \(2/(3x)\leq1/t\leq2/x\), and the Mean Value Theorem gives \[ \frac{2}{3x}|x-y|\leq\rho(x,y)\leq\frac{2}{x}|x-y|. \] Take \(c_x=2/(3x)\) and \(C_x=2/x\).
When \(|x-y|\geq x/2\), either \(y\leq x/2\) or \(y\geq3x/2\). In the first case \(\rho(x,y)=\ln(x/y)\geq\ln2\); in the second case \(\rho(x,y)=\ln(y/x)\geq\ln(3/2)\). Hence condition (ii) holds with \(s_x=\ln(3/2)>0\). The theorem proves that these two metrics induce the same topology.
Check: with \(x=2\) and \(y=5/2\), the local constants are \(r_2=1\), \(c_2=1/3\), and \(C_2=1\). Since \(d(2,5/2)=1/2<1\), the bounds give \(1/6\leq\ln(5/4)\leq1/2\). Directly, \(\ln(5/4)=\int_2^{5/2}(1/t)\,dt\); on this interval \(2/5\leq1/t\leq1/2\), so \(1/5\leq\ln(5/4)\leq1/4\), which confirms the stated bounds.
Neither global comparison direction works here. For the pairs \((x,2x)\), the ratio is \(\rho(x,2x)/d(x,2x)=\ln2/x\). It becomes arbitrarily large as \(x\) decreases to \(0\), ruling out a global upper constant, and tends to \(0\) as \(x\) increases without bound, ruling out a global positive lower constant.
A Common Pitfall: Local Constants Are Not Uniform Constants
A local comparison uses constants chosen separately at each point and only for pairs in the specified neighborhood. It does not say that the same constants work everywhere. In the cubic example, the local upper constant is \(C_x=1+3(|x|+1)^2\), which grows without bound as \(|x|\) grows. The tempting calculation “each \(C_x\) is finite, so choose a finite \(C\) that works for all \(x\)” is invalid: finiteness at each point does not imply a uniform bound over all points.
For a concrete check, the cubic example at \(x=0\), \(y=2\) has \(d(0,2)=2\) and \(\rho(0,2)=10\). This pair is outside the local comparison region \(d(0,y)<r_0=1\), so the local upper estimate with \(C_0=4\) cannot be applied. Indeed, the incorrect substitution would claim \(10\leq4(2)=8\), which is false. The correction is to check the hypothesis before using an estimate: this pair is covered by the remote-point bound, not by the local upper bound. The theorem needs pointwise constants and the remote-point condition; it does not claim global metric inequalities.
The method matters because topology records which sets are open, not the numerical size of distances across the whole space. Point-dependent bounds can be sufficient to preserve all open sets even when distances grow or shrink by arbitrarily large factors in different regions.
Check Your Understanding
For each question, identify which local comparison condition applies before drawing a conclusion.
- In the cubic example, what are \(c_1\), \(C_1\), \(r_1\), and \(s_1\) from the displayed choices?
- For the logarithm metric, why must \(x>0\) before the radius \(r_x=x/2\) is used?
- Which part of the local radial comparison prevents points with \(d(x,y)\geq r_x\) from belonging to every sufficiently small \(\rho\)-ball about \(x\)?
- For the arctangent metric, explain why \(\rho(0,n)/d(0,n)\to0\) rules out a global positive lower comparison constant.
- In the cubic example, may the local upper estimate at \(x=0\) be applied to \(y=2\)? State the distance and radius that decide this.