When Do the Terms of a Sequence Get Close Together?
A convergent sequence approaches a point in its space. A Cauchy sequence has a related property that does not name a proposed limit: its terms eventually get close to one another. The distinction matters when a space has a “missing” limit. A sequence can satisfy the Cauchy condition inside the space even though its natural limit lies outside it.
Throughout, \((X,d)\) is a metric space. The Cauchy condition is about distances between pairs of terms, with both indices sufficiently large. Completeness asks whether every sequence meeting that condition actually converges to a point of \(X\).
The order of the quantifiers is essential: after choosing any positive tolerance \(\varepsilon\), one index \(N\) must work for every pair \(m,n\ge N\). The point of convergence is not part of the Cauchy definition. In particular, completeness is a property of a metric space, not of a sequence alone. The empty metric space is complete because it has no sequences to test; a one-point space is complete because every sequence in it is constant.
Basic Facts About Cauchy Sequences
Proof. Suppose first that \(x_n\to x\). Given \(\varepsilon>0\), convergence gives \(N\) such that \(d(x_n,x)<\varepsilon/2\) for every \(n\ge N\). For \(m,n\ge N\), the triangle inequality gives \(d(x_m,x_n)\le d(x_m,x)+d(x,x_n)<\varepsilon/2+\varepsilon/2=\varepsilon\). Thus \((x_n)\) is Cauchy.
Now suppose \((x_n)\) is Cauchy. Use the definition with \(\varepsilon=1\) to choose \(N\) such that \(d(x_n,x_N)<1\) for every \(n\ge N\). The terms from index \(N\) onward lie in the ball \(B(x_N,1)\), so their distances from \(x_N\) are bounded by \(1\). There are only finitely many earlier terms \(x_1,\ldots,x_{N-1}\), and each has a finite distance from \(x_N\). Therefore all the numbers \(d(x_n,x_N)\) are bounded. By the triangle inequality, \(d(x_m,x_n)\le d(x_m,x_N)+d(x_N,x_n)\), so the sequence is bounded. \(\square\)
A sequence can be Cauchy without having a limit in its space; that is exactly what incompleteness permits. When a limit does exist, it is unique: if \(x_n\to x\) and \(x_n\to y\), then for every \(n\), \(d(x,y)\le d(x,x_n)+d(x_n,y)\). The right side tends to \(0\), so \(d(x,y)=0\), which implies \(x=y\).
Worked Example: A Cauchy Sequence of Rationals with No Rational Limit
Give a sequence in \(\mathbb{Q}\), with the usual metric \(d(x,y)=|x-y|\), that is Cauchy but does not converge in \(\mathbb{Q}\).
Let \(q_n\) be the decimal truncation of \(\sqrt{2}\) after \(n\) digits following the decimal point: \(q_n=\lfloor 10^n\sqrt{2}\rfloor/10^n\). The first terms are \(q_1=1.4\), \(q_2=1.41\), \(q_3=1.414\), and \(q_4=1.4142\). Each \(q_n\) is rational. The definition of the floor function gives \(0\le \sqrt{2}-q_n<10^{-n}\). Thus, for \(m,n\ge N\), the triangle inequality gives \(|q_m-q_n|\le |q_m-\sqrt{2}|+|\sqrt{2}-q_n|<10^{-m}+10^{-n}\le 2\cdot 10^{-N}\). For any \(\varepsilon>0\), choose \(N\) large enough that \(2\cdot 10^{-N}<\varepsilon\). This proves that \((q_n)\) is Cauchy.
The same error bound shows that \(q_n\to\sqrt{2}\) in \(\mathbb{R}\). To see why the sequence cannot converge to a rational \(r\), suppose it did. Convergence in \(\mathbb{Q}\) uses the same distance as convergence in \(\mathbb{R}\), so it would also converge to \(r\) in \(\mathbb{R}\). Uniqueness of limits would force \(r=\sqrt{2}\). But \(\sqrt{2}\notin\mathbb{Q}\): if \(\sqrt{2}=a/b\) for relatively prime positive integers \(a,b\), then \(a^2=2b^2\), so \(a\) is even. Writing \(a=2k\) gives \(b^2=2k^2\), so \(b\) is even too, contradicting that \(a,b\) are relatively prime. Therefore \((q_n)\) has no limit in \(\mathbb{Q}\), and \(\mathbb{Q}\) with its usual metric is not complete.
Check: the first four truncations satisfy \(1.4\le\sqrt{2}<1.5\), \(1.41\le\sqrt{2}<1.42\), \(1.414\le\sqrt{2}<1.415\), and \(1.4142\le\sqrt{2}<1.4143\), as required by the truncation bounds.
The Real Line Is Complete
Here the completeness of \(\mathbb{R}\) follows from the least-upper-bound property: every nonempty set of real numbers that is bounded above has a least upper bound. This property is part of the standard construction of the real numbers. We prove the Cauchy-sequence conclusion from it.
Proof. Let \((x_n)\) be Cauchy in \(\mathbb{R}\). The boundedness proposition shows that its terms form a bounded set. For each \(n\), let \(s_n=\sup\{x_k:k\ge n\}\), the supremum of the \(n\)th tail. The tails are nonempty and bounded above, so each \(s_n\) exists. Since each tail contains the next tail, \(s_{n+1}\le s_n\). The set \(\{s_n:n\in\mathbb{N}\}\) is bounded below by any lower bound for all the \(x_k\), so let \(L=\inf\{s_n:n\in\mathbb{N}\}\).
Given \(\varepsilon>0\), choose \(N\) such that \(|x_j-x_k|<\varepsilon/2\) whenever \(j,k\ge N\). For each \(n\ge N\), every term \(x_k\) with \(k\ge n\) satisfies \(x_k<x_n+\varepsilon/2\). Therefore \(s_n\le x_n+\varepsilon/2\), while \(x_n\le s_n\) because \(x_n\) belongs to the \(n\)th tail. Hence \(0\le s_n-x_n\le\varepsilon/2\).
By the defining property of an infimum, some \(n_0\) has \(s_{n_0}<L+\varepsilon/2\); otherwise \(L+\varepsilon/2\) would be a lower bound larger than \(L\). For \(n\ge n_0\), monotonicity and the lower-bound property of \(L\) give \(L\le s_n\le s_{n_0}<L+\varepsilon/2\). For \(n\ge\max(N,n_0)\), we now have \(|x_n-L|\le |x_n-s_n|+|s_n-L|\le\varepsilon/2+|s_n-L|<\varepsilon\). Thus \(x_n\to L\in\mathbb{R}\). \(\square\)
Worked Example: A Cauchy Sequence in the Real Line
Consider \(x_n=3+1/n\) in \(\mathbb{R}\). For \(m,n\ge N\), \(|x_m-x_n|=|1/m-1/n|\le 1/m+1/n\le 2/N\). Given \(\varepsilon>0\), choose \(N\in\mathbb{N}\) with \(N>2/\varepsilon\). Then \(2/N<\varepsilon\), so the sequence is Cauchy. Also \(|x_n-3|=1/n\to0\), hence \(x_n\to3\).
Check: substituting the proposed limit gives \(3+1/n-3=1/n\), which tends to \(0\). This sequence illustrates convergence in \(\mathbb{R}\); the theorem proves that every Cauchy sequence in \(\mathbb{R}\), not just this one, has a real limit.
Completeness Is Not Boundedness
Boundedness concerns the distances between all points of a space: a metric space is bounded if there is a finite \(M\) such that \(d(x,y)\le M\) for every \(x,y\in X\). Completeness concerns the limits of Cauchy sequences. Neither condition implies the other.
Worked Example: A Bounded Space That Is Not Complete
Take \(X=(0,1)\) with the metric inherited from \(\mathbb{R}\). It is bounded: for any \(x,y\in X\), \(0<x<1\) and \(0<y<1\), so \(|x-y|<1\). Consider \(x_n=1/(n+1)\). Every term lies strictly between \(0\) and \(1\). For \(m,n\ge N\), both terms lie in \((0,1/(N+1)]\), so \(|x_m-x_n|<1/(N+1)\). Given \(\varepsilon>0\), choose \(N\) with \(1/(N+1)<\varepsilon\); this proves the sequence is Cauchy.
In \(\mathbb{R}\), the sequence converges to \(0\), since \(|x_n-0|=1/(n+1)\to0\). It cannot converge to any point of \((0,1)\): if it converged there, its limit in \(\mathbb{R}\) would have to be \(0\) by uniqueness, but \(0\notin(0,1)\). Thus this bounded metric space is not complete.
Check: the strict inequality in the membership test is satisfied: \(0<1/(n+1)<1\) for every \(n\in\mathbb{N}\). The missing limit is the endpoint \(0\), not a point of the space.
Worked Example: An Infinite Bounded Complete Discrete Metric Space
Let \(X=\mathbb{N}\) and define the discrete metric by \(d(x,y)=0\) if \(x=y\), and \(d(x,y)=1\) if \(x\ne y\). This metric induces the discrete topology: for any \(x\in X\), the open ball \(B(x,1/2)\) is exactly \(\{x\}\). The space is bounded, since all distances are at most \(1\), and it is infinite.
To test completeness, let \((x_n)\) be Cauchy and use \(\varepsilon=1\). There is \(N\) such that \(d(x_m,x_n)<1\) for all \(m,n\ge N\). The only possible value of this distance below \(1\) is \(0\), so \(x_m=x_n\) for all \(m,n\ge N\). Thus the sequence is eventually constant and converges to that constant value in \(X\). Every Cauchy sequence in \(X\) therefore converges, proving completeness.
Check: an eventually constant sequence has distance \(0\) between any two terms after its constant tail begins; \(0<\varepsilon\) for every \(\varepsilon>0\). The argument works for any set with the discrete metric, including an infinite one.
Closed Subspaces and Completeness
There is a useful link between completeness and closedness. For a subset \(A\) of a metric space, use the restricted metric \(d|_{A\times A}\). Closedness here refers to the topology of the ambient space \(X\).
Proof. Suppose first that \(A\) is closed in \(X\), and let \((a_n)\) be Cauchy in \(A\). The same distances are used in \(A\) and \(X\), so it is Cauchy in \(X\). Completeness of \(X\) gives a limit \(x\in X\). If \(x\notin A\), then \(X\setminus A\) is open and contains \(x\), so some open ball \(B(x,r)\) lies in \(X\setminus A\). Convergence gives \(a_n\in B(x,r)\) for all sufficiently large \(n\), contradicting \(a_n\in A\). Hence \(x\in A\), and \(A\) is complete.
Conversely, suppose \(A\) is complete. If \(A=\varnothing\), it is closed, so assume it is nonempty. Take \(x\in\overline{A}\). For each \(n\in\mathbb{N}\), the open ball \(B(x,1/n)\) meets \(A\), so choose \(a_n\in A\) with \(d(a_n,x)<1/n\). For \(m,n\ge N\), the triangle inequality gives \(d(a_m,a_n)<1/m+1/n\le2/N\); hence \((a_n)\) is Cauchy in \(A\). By completeness of \(A\), it converges to some \(a\in A\). It also converges to \(x\) in \(X\), since \(d(a_n,x)<1/n\). Uniqueness of limits in \(X\) gives \(a=x\), so \(x\in A\). Thus \(\overline{A}\subseteq A\), and since \(A\subseteq\overline{A}\), the set \(A\) is closed. \(\square\)
A Common Mistake
Pitfall: Boundedness does not make a sequence Cauchy. For example, \(x_n=(-1)^n\) is bounded in \(\mathbb{R}\), since every term has absolute value \(1\). But for every \(N\), choose \(n\ge N\) and set \(m=n+1\). Then \(|x_m-x_n|=2\), which is not less than \(\varepsilon=1\). No index \(N\) works for this tolerance, so the sequence is not Cauchy.
The reverse confusion is just as misleading: \(\mathbb{R}\) with its usual metric is complete but unbounded, since the distance between \(0\) and \(k\) is \(k\) for every \(k\in\mathbb{N}\). Completeness tests what happens to Cauchy sequences, not how far apart arbitrary points can be.
Check Your Understanding
In each question, use the definitions and keep track of the space in which a limit is required.
- Show directly that \(x_n=2-1/n\) is Cauchy in \(\mathbb{R}\), and identify its limit.
- In a metric space, prove that a Cauchy sequence has at most one limit.
- Is the closed interval \([0,1]\) with the usual metric complete? Explain using the closed-subspace theorem.
- Show that the sequence \(x_n=n\) is not Cauchy in \(\mathbb{R}\).
- For the discrete metric on any set \(X\), prove that a sequence is Cauchy if and only if it is eventually constant.