When Does a Cauchy Sequence Have Its Limit in the Set?
The Cauchy Criterion for Real Sequences, proved in the previous tutorial, says that a real sequence converges to a finite real limit if and only if it is Cauchy. The phrase “real limit” matters: the sequence is considered as a sequence in \(\mathbb{R}\). If its terms are required to lie in a smaller set, its real limit may not belong to that set.
Completeness describes exactly whether this can happen. A complete set has no Cauchy sequences whose limits are missing from the set. We will define completeness for subsets of \(\mathbb{R}\), prove that every nonempty closed subset is complete, and then show that the converse holds as well. Throughout, “closed” refers to closedness in \(\mathbb{R}\), not merely relative to some larger set.
The requirement that \(E\) be nonempty is part of this definition. It also keeps the characterization below precisely stated: we will show that a nonempty subset \(E\) is complete if and only if it is closed in \(\mathbb{R}\). The empty set is closed, but it is not included in the definition being used here.
Closed Sets Capture Limits of Their Sequences
We will use the sequential description of closedness in \(\mathbb{R}\). It says that a set is closed exactly when it contains the limit of every convergent sequence whose terms lie in the set. Here is a proof, so the connection between the topological definition and the sequence arguments is explicit.
Proof. Suppose first that \(E\) is closed. If a convergent sequence \((x_n)\) has all its terms in \(E\), assume for contradiction that its limit \(x\) is not in \(E\). Since \(\mathbb{R}\setminus E\) is open, there is an \(r>0\) such that \((x-r,x+r)\subseteq\mathbb{R}\setminus E\). But \(x_n\to x\), so for all sufficiently large \(n\), \(x_n\in(x-r,x+r)\). This contradicts \(x_n\in E\).
Conversely, suppose every convergent sequence in \(E\) has its limit in \(E\). If \(E\) were not closed, then \(\mathbb{R}\setminus E\) would not be open. Thus there would be some \(x\notin E\) such that every open interval centered at \(x\) contains a point of \(E\). For each \(n\in\mathbb{N}_0\), choose \(x_n\in E\) with $$ |x_n-x|<\frac{1}{n+1}. $$ Given \(\varepsilon>0\), choose \(N\in\mathbb{N}_0\) so that \(1/(N+1)<\varepsilon\). For every \(n\geq N\), we then have \(|x_n-x|<1/(n+1)\leq1/(N+1)<\varepsilon\), so \(x_n\to x\). The assumed property forces \(x\in E\), contradicting \(x\notin E\). Therefore \(E\) is closed. \(\square\)
Closed Subsets Are Complete
The proof now combines two facts already established in the course. A Cauchy sequence of real numbers converges in \(\mathbb{R}\), by the Cauchy Criterion for Real Sequences. And a closed set contains the limit of each convergent sequence whose terms lie in that set, by the theorem just proved. Together, these facts give the first main result.
Proof. Let \((x_n)\) be any Cauchy sequence with \(x_n\in E\) for every \(n\in\mathbb{N}_0\). Regard it as a real sequence. By the Cauchy Criterion for Real Sequences, there is an \(x\in\mathbb{R}\) such that \(x_n\to x\). Since \(E\) is closed and every \(x_n\) belongs to \(E\), the Sequential Characterization of Closed Sets gives \(x\in E\). Thus the sequence converges to a point of \(E\), as required. \(\square\)
Worked Example: A Closed Interval Is Complete
Let \(E=[-2,3]\). This set is nonempty and closed in \(\mathbb{R}\), so the theorem shows that it is complete. In detail, if \((x_n)\) is any Cauchy sequence with \(-2\leq x_n\leq3\) for every \(n\), then it converges in \(\mathbb{R}\) to some \(x\). The closed-set theorem ensures that \(-2\leq x\leq3\), so the limit still belongs to the interval. No monotonicity or explicit formula for the sequence is needed.
Closedness is the condition that prevents a limit from escaping the set. For example, the interval \([-2,3]\) includes both of its endpoints. By contrast, an open interval can omit a point that is the limit of a Cauchy sequence of its own elements.
Worked Example: The Open Interval \((0,1)\) Is Not Complete
For \(n\in\mathbb{N}_0\), let \(x_n=1/(n+2)\). Each term belongs to \((0,1)\). If \(m,n\geq N\), then $$ |x_m-x_n| \leq \frac{1}{m+2}+\frac{1}{n+2} \leq \frac{2}{N+2}. $$ Given \(\varepsilon>0\), choose \(N\) large enough that \(2/(N+2)<\varepsilon\). This proves that \((x_n)\) is Cauchy. In \(\mathbb{R}\), it converges to \(0\), since \(1/(n+2)\to0\). But \(0\notin(0,1)\). Thus it is a Cauchy sequence in \((0,1)\) that does not converge to a point of \((0,1)\), so this set is not complete.
Completeness Also Forces Closedness
The other direction uses the same ideas in reverse. A convergent sequence is Cauchy, as established by the theorem Every Convergent Sequence Has the Cauchy Property. If its terms lie in a complete set, completeness supplies a limit inside that set. Uniqueness of limits then identifies that limit with the original real limit.
Proof. If \(E\) is closed, it is complete by the theorem A Nonempty Closed Subset of \(\mathbb{R}\) Is Complete.
For the converse, suppose \(E\) is nonempty and complete. Let \((x_n)\) be a convergent sequence with \(x_n\in E\) for every \(n\), and write \(x_n\to x\in\mathbb{R}\). The theorem Every Convergent Sequence Has the Cauchy Property shows that \((x_n)\) is Cauchy. Completeness of \(E\) therefore gives a \(y\in E\) such that \(x_n\to y\). By the Theorem of Uniqueness of Limits, \(x=y\). Hence \(x\in E\). Every convergent sequence in \(E\) has its limit in \(E\), so the Sequential Characterization of Closed Sets implies that \(E\) is closed. \(\square\)
Worked Example: The Rational Numbers Are Not Complete
For each \(n\in\mathbb{N}_0\), define the rational number $$ q_n=\frac{\lfloor 10^n\sqrt{2}\rfloor}{10^n}. $$ The floor is an integer, and \(10^n\) is a positive integer, so \(q_n\in\mathbb{Q}\). The defining property of the floor gives $$ 0\leq\sqrt{2}-q_n<\frac{1}{10^n}. $$ For \(m,n\geq N\), it follows that $$ |q_m-q_n| \leq |q_m-\sqrt{2}|+|q_n-\sqrt{2}| <\frac{1}{10^m}+\frac{1}{10^n} \leq\frac{2}{10^N}. $$ For any \(\varepsilon>0\), choose \(N\) with \(2/10^N<\varepsilon\). Then \((q_n)\) is Cauchy, and the same error estimate shows that \(q_n\to\sqrt{2}\) in \(\mathbb{R}\).
This real limit is not rational. Indeed, if \(\sqrt{2}=p/q\) for relatively prime positive integers \(p,q\), then \(p^2=2q^2\). Thus \(p^2\) is even, which forces \(p\) to be even; writing \(p=2k\) gives \(q^2=2k^2\), so \(q\) is even as well. This contradicts the assumption that \(p\) and \(q\) are relatively prime. If \((q_n)\) converged to a rational number, that number would also be its real limit, contradicting uniqueness of limits. Consequently, \(\mathbb{Q}\) is not complete.
What Completeness Does—and Does Not—Mean
Completeness is about where Cauchy sequences converge, not about the size or boundedness of a set. The real line is complete even though it is unbounded. The interval \((0,1)\) is bounded but not complete. The rational numbers are dense in \(\mathbb{R}\), yet are not complete because some rational Cauchy sequences have irrational real limits. A set can therefore be large, dense, or bounded without being complete.
A useful way to apply the characterization is to check closedness in the ambient real line. For example, \([0,\infty)\) is nonempty and closed, so it is complete. The set \((-\infty,0)\) is not closed in \(\mathbb{R}\): the sequence \(-1/(n+1)\) belongs to it and converges to \(0\), which does not. The same sequence is Cauchy, so it also directly witnesses the failure of completeness.
There is an important distinction between convergence in \(\mathbb{R}\) and convergence within \(E\). Every Cauchy sequence in a subset \(E\) is still a real Cauchy sequence, and so has a real limit. Completeness asks for the additional fact that this limit lies in \(E\). When proving completeness, it is not enough to establish convergence in \(\mathbb{R}\); one must also verify membership of the limit in the set. Closedness is precisely what supplies that step.
Check Your Understanding
Use the definition and the closed-set characterization to answer the following questions.
- In the definition of completeness, where must the limit of a Cauchy sequence in \(E\) lie?
- Which two results combine to prove that a nonempty closed subset of \(\mathbb{R}\) is complete?
- Why does the sequence \(1/(n+2)\) show that \((0,1)\) is not complete?
- In the proof that completeness implies closedness, why does the limit supplied by completeness equal the original real limit?
- Why does the characterization specify nonempty subsets, even though the empty set is closed?