When Cauchy Sequences Have Nowhere to Go
A metric space is complete when every Cauchy sequence converges to a point of that space. The examples of complete spaces in the previous tutorial showed ways this can happen: completeness may follow from finite-dimensional structure, from being closed in a complete ambient space, or from a direct argument about limits. Here we examine what failure looks like. In each example, the key task is to find a Cauchy sequence whose natural limit is missing from the space, or whose limit cannot belong to the space because of its defining properties.
It is important to say “does not converge to any point of the space,” rather than merely “does not have a limit.” The same sequence may converge in a larger ambient space while failing to converge within the given space. The rational numbers with the usual metric provide the standard example: a rational Cauchy sequence can converge in the real numbers to an irrational number.
The Rational Numbers Are Incomplete
We construct rational approximations to \(\sqrt{2}\). For each positive integer \(n\), let \(k_n\) be the greatest integer not exceeding \(10^n\sqrt{2}\), and set \(q_n=k_n/10^n\). Thus \(q_n\) is rational, and the definition of the integer part gives $$ q_n\leq\sqrt{2}<q_n+10^{-n}. $$ The rational numbers \(q_n\) get arbitrarily close to \(\sqrt{2}\), even though \(\sqrt{2}\) is not rational.
Worked Example: A Cauchy Sequence of Rationals with No Rational Limit
For any positive integers \(m,n\), the bounds above imply $$ |q_m-q_n| \leq |q_m-\sqrt{2}|+|\sqrt{2}-q_n| <10^{-m}+10^{-n}. $$ Given \(\varepsilon>0\), choose \(N\) so that \(2\cdot10^{-N}<\varepsilon\). If \(m,n\geq N\), then $$ |q_m-q_n|<10^{-m}+10^{-n}\leq2\cdot10^{-N}<\varepsilon. $$ Therefore \((q_n)\) is Cauchy in \(\mathbb{Q}\) with the usual metric.
The same bounds show that \(q_n\to\sqrt{2}\) in \(\mathbb{R}\), since \(0\leq\sqrt{2}-q_n<10^{-n}\) and \(10^{-n}\to0\). To check that this limit is not rational, suppose \(\sqrt{2}=a/b\) for relatively prime positive integers \(a,b\). Then \(a^2=2b^2\), so \(a^2\) is even and hence \(a\) is even. Write \(a=2c\). Substituting gives \(4c^2=2b^2\), or \(b^2=2c^2\), so \(b\) is even as well. This contradicts the assumption that \(a\) and \(b\) are relatively prime.
If \((q_n)\) had a limit \(q\in\mathbb{Q}\), it would also converge to \(q\) as a real sequence. But it converges in \(\mathbb{R}\) to \(\sqrt{2}\). By uniqueness of metric limits, \(q=\sqrt{2}\), contradicting the irrationality just proved. Thus \(\mathbb{Q}\), with the usual metric, is incomplete.
This argument separates two issues. The Cauchy property concerns distances between terms of the sequence and is unchanged when the rational numbers are viewed inside the real numbers. Convergence, however, requires a candidate limit to belong to the space under consideration. The real numbers provide the missing limit for this particular sequence, but the rational numbers do not contain it.
The worked example proves the theorem by exhibiting a Cauchy sequence in \(\mathbb{Q}\) with no limit in \(\mathbb{Q}\). The example also illustrates a useful general test: if a space sits inside a complete metric space, look for a Cauchy sequence in the smaller space whose ambient limit lies outside it.
A Missing Boundary Point
A simpler version of the same phenomenon occurs when an interval omits one of its endpoints. For the usual metric, the sequence \(x_n=1/(n+1)\) lies in \((0,1)\) for every positive integer \(n\), but its real limit is \(0\), which is not in the interval.
Worked Example: The Open Interval Is Incomplete
For any positive integers \(m,n\), $$ |x_m-x_n|\leq x_m+x_n=\frac{1}{m+1}+\frac{1}{n+1}. $$ Given \(\varepsilon>0\), choose \(N\) so that \(2/(N+1)<\varepsilon\). For \(m,n\geq N\), both terms on the right are at most \(1/(N+1)\), so \(|x_m-x_n|\leq2/(N+1)<\varepsilon\). Hence \((x_n)\) is Cauchy in \((0,1)\).
Also, \(0<x_n=1/(n+1)\to0\) in \(\mathbb{R}\). If the sequence converged to some \(x\in(0,1)\), then it would converge in \(\mathbb{R}\) both to \(x\) and to \(0\). Uniqueness of metric limits would give \(x=0\), which is impossible because \(0\notin(0,1)\). Thus \((0,1)\) is incomplete.
The point is not that open sets are always incomplete; many open subsets, including the whole real line, are complete. Rather, an omitted boundary point can be the destination of a Cauchy sequence. The closed-subset result from the previous tutorial gives a useful contrast: a closed subset of a complete metric space is complete. The open interval does not meet that condition in \(\mathbb{R}\), since it is not closed there.
Density and Incompleteness
The rational numbers are dense in the real numbers: every real number can be approximated arbitrarily closely by rational numbers. Density alone does not mean that a subset is incomplete. For example, a dense subset might equal the whole space if the space itself is being considered. The following consequence of the Complete Subsets of a Complete Space Theorem gives a precise criterion when the ambient space is complete.
Proof. Suppose, to the contrary, that \(A\) is complete in its restricted metric. By the Complete Subsets of a Complete Space Theorem from Tutorial 663, a complete subset of a complete metric space is closed in that space. Thus \(A\) is closed in \(X\). But \(A\) is also dense in \(X\), so its closure is \(X\). Because \(A\) is closed, its closure is \(A\), giving \(A=X\). This contradicts the assumption that \(A\) is proper. Therefore \(A\) is incomplete. \(\square\)
This theorem gives a short route to incompleteness when density is already known. For instance, \(\mathbb{Q}\) is a proper dense subset of the complete space \(\mathbb{R}\). The explicit sequence in the earlier example reveals exactly how incompleteness occurs, while the theorem explains why some Cauchy sequence must fail to converge inside \(\mathbb{Q}\).
A Cauchy Sequence Whose Limit Cannot Be Continuous
Incompleteness is not restricted to subsets that omit points from a familiar interval. It can also occur when the metric measures functions differently from the supremum norm. Consider \(C[0,1]\), the space of continuous real-valued functions on \([0,1]\), with the metric $$ d_1(f,g)=\int_0^1 |f(x)-g(x)|\,dx. $$ This is a metric on continuous functions: if the integral of \(|f-g|\) is zero, continuity forces \(f=g\) everywhere. By contrast with the supremum norm, this metric allows a discrepancy on a short interval to have small distance, even if the discrepancy there is not small in height.
Worked Example: A Cauchy Sequence in \(C[0,1]\) with the Integral Metric
Let \(c=1/2\). For each integer \(n\geq3\), define \(f_n:[0,1]\to\mathbb{R}\) by $$ f_n(x)= \begin{cases} 0,&x\leq c-1/n,\\ \frac{n}{2}\bigl(x-(c-1/n)\bigr),&c-1/n\leq x\leq c+1/n,\\ 1,&x\geq c+1/n. \end{cases} $$ At \(x=c-1/n\), the middle formula gives \((n/2)(0)=0\). At \(x=c+1/n\), it gives \((n/2)(2/n)=1\). Thus the formulas agree at both joining points, and \(f_n\) is continuous. Its values lie between \(0\) and \(1\).
If \(m,n\geq N\), both functions equal \(0\) to the left of \(c-1/N\) and equal \(1\) to the right of \(c+1/N\). Their difference is therefore zero outside \([c-1/N,c+1/N]\). Everywhere, \(|f_m-f_n|\leq1\), and this interval has length \(2/N\). Hence $$ d_1(f_m,f_n)=\int_0^1|f_m(x)-f_n(x)|\,dx\leq\frac{2}{N}. $$ Given \(\varepsilon>0\), choose \(N\geq3\) with \(2/N<\varepsilon\). This proves that \((f_n)\) is Cauchy in the integral metric.
Suppose it converged in that metric to some \(g\in C[0,1]\). Fix \(\delta\) with \(0<\delta<1/2\). Whenever \(n\geq1/\delta\), the definition gives \(f_n(x)=0\) throughout \([0,c-\delta]\). Consequently, $$ \int_0^{c-\delta}|g(x)|\,dx \leq\int_0^1|g(x)-f_n(x)|\,dx =d_1(g,f_n). $$ As \(n\to\infty\), the right-hand side tends to zero, so the integral on the left is zero. A continuous nonnegative function with integral zero on an interval must vanish there: if it were positive at a point, continuity would make it bounded below by a positive number on a subinterval, giving a positive integral. Thus \(g=0\) on \([0,c-\delta]\).
For the other side, whenever \(n\geq1/\delta\), \(f_n(x)=1\) on \([c+\delta,1]\). The same argument applied to \(|g-1|\) shows that \(g=1\) on \([c+\delta,1]\). Since this holds for every \(0<\delta<1/2\), it follows that \(g(x)=0\) for every \(x<c\) and \(g(x)=1\) for every \(x>c\). Such a function cannot be continuous at \(c\): values to the left approach \(0\), while values to the right approach \(1\). This contradiction shows that \((f_n)\) has no limit in \(C[0,1]\) under \(d_1\). Therefore this metric space is incomplete.
The sequence increasingly concentrates its change from \(0\) to \(1\) in a narrower interval around \(c\). Its integral distances become small because that interval becomes short. The limiting step-shaped behavior cannot be represented by a continuous function, which is why there is no limit in the specified space.
Completeness depends on both the underlying set and the chosen metric. In the previous tutorial, \(C[0,1]\) with the supremum norm was shown to be complete, since it is \(C_b([0,1])\). Here the same set of functions is incomplete under the integral metric. A Cauchy condition in one metric need not control distances in another metric strongly enough to preserve the space’s defining properties.
How to Recognize Incompleteness
These examples use two complementary strategies. In \(\mathbb{Q}\) and \((0,1)\), a Cauchy sequence has a limit in a larger complete space, but that limit is missing from the subset. In \(C[0,1]\) with the integral metric, the sequence becomes Cauchy because the region where it changes shrinks; a possible limit would have to behave like a discontinuous step function and so cannot lie in the space.
A common pitfall is to identify a plausible ambient limit and stop there. To establish incompleteness, one must show both that the sequence is Cauchy in the stated metric and that no point of the space is its limit. In the function example, identifying a step-shaped pointwise limit is not enough: the proof rules out every continuous metric limit by examining intervals on either side of the jump.
Check Your Understanding
Use the examples and results in this tutorial to answer the following questions.
- Why does the rational approximation sequence \((q_n)\) satisfy the Cauchy condition?
- Why does convergence of \(x_n=1/(n+1)\) to \(0\) in \(\mathbb{R}\) rule out convergence to a point of \((0,1)\)?
- What earlier theorem is used to prove that a proper dense subset of a complete space is incomplete?
- For the functions \(f_n\), why is the integral distance between two terms bounded by \(2/N\) when both indices are at least \(N\)?
- Why can no continuous function be the \(d_1\)-limit of the sequence \((f_n)\)?