Completeness Is About Limits Inside the Space
A sequence can draw closer and closer to itself without appearing to have a limit in the space where it lives. Completeness makes this distinction precise: it requires every Cauchy sequence to converge to a point of the given metric space. We will use that test for Euclidean spaces, discrete metric spaces, and intervals with their usual distance.
A metric \(d\) on \(X\) induces a topology: a set \(U\subseteq X\) is open if every \(x\in U\) lies in some open ball \(B_d(x,r)=\{y\in X:d(x,y)<r\}\) contained in \(U\). Unless specified otherwise, \(\mathbb{R}\) has its standard topology and metric \(d(x,y)=|x-y|\), while \(\mathbb{R}^n\) has the Euclidean metric \(d(x,y)=\|x-y\|\). Intervals carry the restricted metric and subspace topology from \(\mathbb{R}\).
The Cauchy condition compares pairs of terms far out in the sequence; it does not name a candidate limit. Convergence, by contrast, requires a specific point belonging to \(X\). For the empty metric space there are no sequences indexed by \(\mathbb{N}\), so the condition “every Cauchy sequence converges” holds vacuously. For nonempty spaces, the examples below give concrete tests.
The Real Line and Euclidean Spaces
We use the completeness theorem for \(\mathbb{R}\): every Cauchy sequence of real numbers converges to a real number. This standard real-analysis result follows from the least-upper-bound property. For completeness of the argument here, its proof is included.
Proof. Let \((x_n)\) be Cauchy. It is bounded: choose \(N_0\) so that \(|x_n-x_{N_0}|<1\) for \(n\ge N_0\). The terms from \(N_0\) onward lie between \(x_{N_0}-1\) and \(x_{N_0}+1\), and the finitely many earlier terms are bounded too. For each \(N\), let \(i_N=\inf\{x_n:n\ge N\}\) and \(s_N=\sup\{x_n:n\ge N\}\); these are finite by boundedness and the least-upper-bound property. Given \(\varepsilon>0\), choose \(N\) so that \(|x_m-x_n|<\varepsilon/2\) for all \(m,n\ge N\). Every two terms in this tail differ by less than \(\varepsilon/2\), so \(s_N-i_N\le\varepsilon/2\). The intervals \([i_N,s_N]\) are nested: removing initial terms from a tail cannot decrease its infimum or increase its supremum. Define \(L=\sup\{i_N:N\in\mathbb{N}\}\). For fixed \(N\), every \(i_k\le s_N\): if \(k\ge N\), then \(i_k\le s_k\le s_N\); if \(k<N\), then \(i_k\le i_N\le s_N\). Thus \(i_N\le L\le s_N\). For every \(n\ge N\), \(x_n\in[i_N,s_N]\), so \(|x_n-L|\le s_N-i_N\le\varepsilon/2<\varepsilon\). Hence \(x_n\to L\) in \(\mathbb{R}\). \(\square\)
Proof. Let \((x_k)\) be Cauchy in \(\mathbb{R}^n\), and write \(x_k=(x_{k,1},\ldots,x_{k,n})\). For each coordinate \(j\), the Euclidean distance satisfies \(|x_{m,j}-x_{k,j}|\le\|x_m-x_k\|\). Therefore the \(j\)-th coordinate sequence is Cauchy in \(\mathbb{R}\). By the lemma, it converges to some \(x_j\in\mathbb{R}\). Put \(x=(x_1,\ldots,x_n)\in\mathbb{R}^n\). Given \(\varepsilon>0\), for each \(j\) choose \(N_j\) so that \(|x_{k,j}-x_j|<\varepsilon/\sqrt n\) whenever \(k\ge N_j\). Take \(N\) to be the largest of the finitely many \(N_j\). If \(k\ge N\), then \[ \|x_k-x\|^2=\sum_{j=1}^n|x_{k,j}-x_j|^2 <n\left(\frac{\varepsilon}{\sqrt n}\right)^2=\varepsilon^2. \] Since distances are nonnegative, \(\|x_k-x\|<\varepsilon\). Thus \(x_k\to x\) in \(\mathbb{R}^n\), as required. \(\square\)
Worked Example: A Cauchy Sequence in \(\mathbb{R}^2\)
Use the Euclidean metric on \(\mathbb{R}^2\), and set \(x_k=(1/k,\,2-1/k)\) for \(k\in\mathbb{N}\). For \(m,k\in\mathbb{N}\), compute the distance between terms: \[ \|x_m-x_k\|=\sqrt{(1/m-1/k)^2+(-1/m+1/k)^2} =\sqrt{2}\,|1/m-1/k|. \] If \(m,k\ge N\), then both \(1/m\) and \(1/k\) lie in \((0,1/N]\), so \(|1/m-1/k|\le 1/N\). Hence \(\|x_m-x_k\|\le\sqrt{2}/N\). Given \(\varepsilon>0\), choose \(N\in\mathbb{N}\) with \(N>\sqrt{2}/\varepsilon\); then \(\|x_m-x_k\|<\varepsilon\) for all \(m,k\ge N\). The sequence is Cauchy.
Its coordinate limits are \(0\) and \(2\), so the proposed limit is \(x=(0,2)\). Directly, \[ \|x_k-x\|=\sqrt{(1/k)^2+(-1/k)^2}=\frac{\sqrt{2}}{k}\longrightarrow 0. \] This verifies convergence to a point of \(\mathbb{R}^2\). The coordinate argument in the theorem proves the same conclusion for every Cauchy sequence in every finite-dimensional Euclidean space.
Discrete Metrics
For any set \(X\), the discrete metric is \(d(x,y)=0\) if \(x=y\) and \(d(x,y)=1\) if \(x\ne y\). Its induced topology is the discrete topology: for each \(x\), the ball \(B_d(x,1/2)\) is the singleton \(\{x\}\), so every subset is open.
Proof. Let \((x_n)\) be Cauchy. Apply the definition with \(\varepsilon=1\). There is \(N\) such that \(d(x_m,x_n)<1\) whenever \(m,n\ge N\). The only possible distance strictly less than \(1\) in the discrete metric is \(0\), so \(x_m=x_n\) for all \(m,n\ge N\). In particular, \(x_n=x_N\) for every \(n\ge N\), and the sequence converges to \(x_N\in X\): its distance from \(x_N\) is eventually \(0\). If \(X\) is empty, the result holds vacuously because there is no sequence in \(X\). \(\square\)
Worked Example: Eventual Constancy in a Finite Discrete Space
Let \(X=\{p,q,r\}\) with the discrete metric, and define \(x_1=q\), \(x_2=r\), and \(x_k=p\) for \(k\ge3\). For \(m,k\ge3\), both terms equal \(p\), so \(d(x_m,x_k)=0\). Thus for any \(\varepsilon>0\), \(N=3\) makes all these distances \(0<\varepsilon\), and the sequence is Cauchy. It converges to \(p\), since \(d(x_k,p)=0<\varepsilon\) for every \(k\ge3\) and every \(\varepsilon>0\). This calculation illustrates the general proof: in a discrete metric, the Cauchy condition with threshold \(1\) forces all sufficiently late terms to be identical.
Closed and Open Intervals
Proof. Let \((x_n)\) be Cauchy in \([a,b]\). Its distances are the same as in \(\mathbb{R}\), so it is Cauchy as a real sequence. By the real-line lemma it converges to some \(L\in\mathbb{R}\). We show that \(L\in[a,b]\). If \(L<a\), set \(\delta=(a-L)/2>0\). Eventually \(|x_n-L|<\delta\), which implies \(x_n<L+\delta=(L+a)/2<a\), contradicting \(x_n\in[a,b]\). If \(L>b\), set \(\delta=(L-b)/2>0\). Eventually \(|x_n-L|<\delta\), which implies \(x_n>L-\delta=(L+b)/2>b\), again a contradiction. Therefore \(a\le L\le b\), so \(L\in[a,b]\). The convergence in \(\mathbb{R}\) is consequently convergence in the restricted metric on \([a,b]\). This argument also covers \(a=b\), when every term equals \(a\). \(\square\)
Worked Example: A Sequence Approaching an Endpoint of a Closed Interval
Use the restricted standard metric on \([2,5]\), and let \(x_n=2+1/n\). For every \(n\ge1\), \(0<1/n\le1\), so \(2<x_n\le3<5\); thus every term belongs to \([2,5]\). For \(m,n\ge N\), \[ |x_m-x_n|=|1/m-1/n|\le 1/N. \] Given \(\varepsilon>0\), choose \(N\in\mathbb{N}\) with \(1/N<\varepsilon\). Then the sequence is Cauchy. Also \(|x_n-2|=1/n\to0\), and \(2\in[2,5]\), so it converges in the interval. The endpoint is allowed: completeness requires the limit to belong to the space, not to be an interior point.
An open interval can fail the same test because a sequence can approach an excluded endpoint. The following example verifies the failure without relying on a picture.
Worked Example: An Incomplete Open Interval
Give \((0,1)\) the restricted standard metric, and set \(x_n=1/(n+1)\). For every \(n\ge1\), \(0<1/(n+1)\le1/2<1\), so \(x_n\in(0,1)\). If \(m,n\ge N\), then \(0<x_m,x_n\le1/(N+1)\), and therefore \[ |x_m-x_n|\le 1/(N+1). \] Given \(\varepsilon>0\), choose \(N\in\mathbb{N}\) with \(1/(N+1)<\varepsilon\). This proves that \((x_n)\) is Cauchy.
In \(\mathbb{R}\), \(x_n\to0\), since \(|x_n-0|=1/(n+1)\to0\); but \(0\notin(0,1)\). To rule out a different limit inside the interval, suppose \(x_n\to c\) for some \(c\in(0,1)\). For every \(n\), the triangle inequality gives \(|c|\le|c-x_n|+|x_n|\). Both terms on the right tend to \(0\), so \(|c|=0\), hence \(c=0\), a contradiction. The sequence has no limit in \((0,1)\), and that metric space is not complete.
A Common Mistake and a Practical Test
Common pitfall. A sequence may be Cauchy and converge in an ambient space, yet fail to converge in the space being tested. For the example above, a mistaken membership check might say “the limit \(0\) satisfies \(0\le0<1\), so \(0\in(0,1)\).” This uses the wrong left-hand inequality: membership in \((0,1)\) requires \(0<0<1\), and the first strict inequality is false. The corrected conclusion is that the limit lies in \(\mathbb{R}\) but not in \((0,1)\). Testing completeness always asks whether the limit belongs to the specified metric space.
The examples can be summarized by the question each one answers. For a Cauchy sequence in \(\mathbb{R}^n\), coordinate limits produce a point in \(\mathbb{R}^n\). In a discrete metric, the sequence is eventually constant. In a closed interval, the real limit cannot lie beyond either endpoint. In an open interval, a Cauchy sequence can approach an endpoint that the interval excludes.
| Metric space | Test for a Cauchy sequence | Conclusion |
|---|---|---|
| \(\mathbb{R}^n\), Euclidean metric | Each coordinate is a Cauchy real sequence | Complete |
| Any set with the discrete metric | Use the Cauchy condition with \(\varepsilon=1\) | Eventually constant, hence complete |
| \([a,b]\), restricted standard metric | The real limit stays between the endpoints | Complete |
| \((0,1)\), restricted standard metric | \(1/(n+1)\) is Cauchy but has no limit in the interval | Not complete |
Check Your Understanding
For each question, state which metric is being used and check whether the proposed limit belongs to the space.
- Why does the Cauchy condition with \(\varepsilon=1\) force a sequence in a discrete metric space to be eventually constant?
- Is \([0,3]\) complete with its restricted standard metric? Which result above applies?
- Show that \(y_n=1-1/(n+1)\) is a Cauchy sequence in \((0,1)\), and identify why its limit does not give convergence in that interval.
- In \(\mathbb{R}^2\), explain why a Cauchy sequence must be Cauchy in each coordinate.
- Can a sequence be Cauchy in a metric space while failing to converge in that space? Give the sequence and space from this tutorial that demonstrate this.