Same Open Sets, Different Cauchy Sequences
A metric determines which sets are open, but it also determines the sizes of distances. Two metrics can therefore describe the same topology while treating Cauchy sequences differently. Completeness is a property of Cauchy sequences, so it need not be determined by the topology alone.
For a metric \(d\), a sequence \((x_n)\) is Cauchy if for every \(\varepsilon>0\) there is \(N\in\mathbb{N}\) such that \(d(x_m,x_n)<\varepsilon\) whenever \(m,n\ge N\). The metric space is complete if every Cauchy sequence converges to a point of the space. Two metrics on the same set are topologically equivalent if they induce the same open sets. This definition says nothing directly about whether their Cauchy sequences agree.
We use \(\mathbb{R}\) with its usual metric \(d(x,y)=|x-y|\), which is complete by the completeness of the real numbers. Now define a second distance by applying the arctangent function to each point.
Verifying the Metric and Comparing the Topologies
Before using \(d_{\arctan}\), we check that it is a metric. It is nonnegative and symmetric because absolute value has those properties. If \(d_{\arctan}(x,y)=0\), then \(\arctan x=\arctan y\); since arctangent is injective, \(x=y\). Conversely, \(x=y\) gives distance zero. Finally, for any \(x,y,z\in\mathbb{R}\), the triangle inequality for absolute value gives \(d_{\arctan}(x,z)\le d_{\arctan}(x,y)+d_{\arctan}(y,z)\). Thus \(d_{\arctan}\) is a metric.
Proof. We show that the identity map between the two metric spaces is continuous in both directions. Fix \(x\in\mathbb{R}\) and \(\varepsilon>0\). By continuity of \(\arctan\) at \(x\), there is \(\delta>0\) such that \(|y-x|<\delta\) implies \(|\arctan y-\arctan x|<\varepsilon\). This proves continuity of the identity from the usual metric to \(d_{\arctan}\). For the reverse direction, \(\tan\) is continuous at \(\arctan x\), and \(\tan(\arctan x)=x\). Hence there is \(\eta>0\) such that \(u\in(-\pi/2,\pi/2)\) and \(|u-\arctan x|<\eta\) imply \(|\tan u-x|<\varepsilon\). If \(d_{\arctan}(x,y)<\eta\), take \(u=\arctan y\). Then \(|u-\arctan x|<\eta\) and \(\tan u=y\), so \(|y-x|<\varepsilon\). Both identity maps are continuous, so each metric has the same open sets as the other. \(\square\)
Worked Example: A Ball for the Arctangent Metric
Consider the open ball of radius \(1/4\) centered at \(0\) in the arctangent metric. By definition, \(y\) belongs to this ball exactly when \(|\arctan y-\arctan 0|<1/4\). Since \(\arctan 0=0\), this is equivalent to \(|\arctan y|<1/4\). The tangent function is strictly increasing on \((-\pi/2,\pi/2)\), so the inequality is equivalent to \(-\tan(1/4)<y<\tan(1/4)\). Therefore the ball is the usual open interval \((-\tan(1/4),\tan(1/4))\).
Check: If \(y=0\), then \(d_{\arctan}(0,0)=0<1/4\), so the center belongs to the ball. If \(y=\tan(1/4)\), then \(\arctan y=1/4\), so its distance from \(0\) is exactly \(1/4\); it is not in the open ball. This checks the strict endpoint condition.
The preceding theorem concerns open sets. It does not say that distances are comparable by fixed multiplicative constants, or that Cauchy sequences agree. The next example exhibits the difference directly.
Worked Example: A Cauchy Sequence Without a Limit
Take the sequence \(x_n=n\) in \(\mathbb{R}\) with the metric \(d_{\arctan}\). For positive \(t\), the identity \(\pi/2-\arctan t=\arctan(1/t)\) and the inequality \(\arctan s\le s\) for \(s\ge0\) give \(0<\pi/2-\arctan n\le 1/n\). The inequality \(\arctan s\le s\) follows, for example, by integrating its derivative \(1/(1+s^2)\le1\) from \(0\) to \(s\).
Let \(\varepsilon>0\) and choose \(N\in\mathbb{N}\) with \(N>1/\varepsilon\). If \(m,n\ge N\), both \(\arctan m\) and \(\arctan n\) lie in the interval \([\arctan N,\pi/2)\). Consequently, \(|\arctan m-\arctan n|\le\pi/2-\arctan N\le1/N<\varepsilon\). Thus \((n)\) is Cauchy in \(d_{\arctan}\).
It does not converge to any \(x\in\mathbb{R}\) in that metric. Indeed, if \(d_{\arctan}(n,x)\to0\), then \(\arctan n\to\arctan x\). But \(\arctan n\to\pi/2\), while \(\arctan x<\pi/2\) for every real \(x\), a contradiction. Check: the sequence is Cauchy because its transformed values approach one another; the missing limit would have to correspond to the excluded endpoint \(\pi/2\), not to a real number. Therefore \((\mathbb{R},d_{\arctan})\) is incomplete, although it has the usual topology.
A Stronger Comparison That Preserves Completeness
Topological equivalence is not the right strength of comparison for completeness. A useful sufficient condition is that each distance be bounded above and below by fixed positive multiples of the other.
Proof. Suppose \((x_n)\) is Cauchy for \(d\), and let \(\varepsilon>0\). Choose \(N\) so that \(d(x_m,x_n)<\varepsilon/C\) for \(m,n\ge N\). The assumed upper bound gives \(\rho(x_m,x_n)\le C d(x_m,x_n)<\varepsilon\), so the sequence is Cauchy for \(\rho\). Conversely, if it is Cauchy for \(\rho\), choose \(N\) so that \(\rho(x_m,x_n)<c\varepsilon\) for \(m,n\ge N\). The lower bound gives \(d(x_m,x_n)\le\rho(x_m,x_n)/c<\varepsilon\), so it is Cauchy for \(d\).
The same estimates apply with one point fixed as a proposed limit. If \(d(x_n,x)\to0\), then \(\rho(x_n,x)\le C d(x_n,x)\to0\); if \(\rho(x_n,x)\to0\), then \(d(x_n,x)\le\rho(x_n,x)/c\to0\). Thus convergence and limits agree. If \((X,d)\) is complete, every \(\rho\)-Cauchy sequence is \(d\)-Cauchy by the first part, hence converges in \(d\), and therefore in \(\rho\). This proves \(\rho\)-completeness. Reversing the roles proves the converse. \(\square\)
Worked Example: Truncating the Usual Distance
Define \(\rho(x,y)=\min\{|x-y|,1\}\) on \(\mathbb{R}\). This is a metric. Positivity, symmetry, and separation follow from the corresponding properties of absolute value. To check the triangle inequality, put \(a=|x-y|\) and \(b=|y-z|\). Since \(|x-z|\le a+b\), we have \(\min\{|x-z|,1\}\le\min\{a+b,1\}\). If either \(a\ge1\) or \(b\ge1\), then \(\min\{a,1\}+\min\{b,1\}\ge1\), which bounds \(\min\{a+b,1\}\). If both \(a<1\) and \(b<1\), then \(\min\{a,1\}+\min\{b,1\}=a+b\), which also bounds \(\min\{a+b,1\}\). This proves the triangle inequality.
For \(0<r<1\), the \(\rho\)-ball of radius \(r\) about \(x\) is exactly \((x-r,x+r)\): the condition \(\min\{|x-y|,1\}<r\) holds exactly when \(|x-y|<r\). These balls show that \(\rho\) induces the usual topology. Also, \(\rho(x,y)\le |x-y|\), but no constant \(c>0\) satisfies \(c|x-y|\le\rho(x,y)\) for every pair. Given \(c>0\), choose \(x,y\) with \(|x-y|>\max\{1,1/c\}\). Then \(\rho(x,y)=1<c|x-y|\). Thus the two metrics are not comparable by positive lower and upper constants as in the theorem.
Nevertheless, \(\rho\) is complete. If a sequence is Cauchy for \(\rho\), then for every \(\varepsilon>0\) choose \(0<r<\min\{\varepsilon,1\}\). Eventually \(\rho(x_m,x_n)<r\), which implies \(|x_m-x_n|<r<\varepsilon\). The sequence is therefore Cauchy in the usual metric and converges to some \(x\in\mathbb{R}\). Since \(\rho(x_n,x)\le|x_n-x|\), it also converges to \(x\) in \(\rho\). This confirms completeness directly.
What Completeness Depends On
The examples establish two distinctions. First, having the same open sets does not force two metrics to have the same Cauchy sequences: the usual metric and \(d_{\arctan}\) have the same topology, but \((n)\) is Cauchy only in the latter. Second, a comparison by fixed positive multiplicative bounds is strong enough to preserve Cauchy sequences and completeness. Such bounds are sufficient, not necessary: the truncated metric example is complete even though no positive lower comparison constant exists.
Common pitfall. It is incorrect to argue that topologically equivalent metrics must have the same completeness, because completeness is not merely a property of open sets. In the arctangent example, the usual metric is complete, yet \(n\) is \(d_{\arctan}\)-Cauchy and has no \(d_{\arctan}\)-limit. The correction is to check Cauchy behavior in the metric actually being used. A second tempting error is to say that the truncated metric \(\min\{|x-y|,1\}\) is not bounded above by a constant multiple of the usual metric. In fact \(\min\{|x-y|,1\}\le |x-y|\) for every pair; it is the positive lower bound that fails, as the worked calculation shows.
Check Your Understanding
For each question, keep separate the topology, the Cauchy condition, and the existence of a limit in the given metric.
- Verify directly that \(d(x,y)=|\arctan x-\arctan y|\) separates distinct real numbers.
- Explain why the endpoints of the interval \((-\pi/2,\pi/2)\) cannot be values of \(\arctan x\) for real \(x\).
- Show that if \(c\,d\le\rho\le C\,d\) for positive constants \(c,C\), then convergence to a fixed point is equivalent for \(d\) and \(\rho\).
- For \(\rho(x,y)=\min\{|x-y|,1\}\), describe its open ball of radius \(2\) centered at \(0\).
- Give the Cauchy sequence in \(d_{\arctan}\) used above and state why it has no limit in \(\mathbb{R}\) with that metric.