
# 3.13: Cauchy Sequences. Completeness


A convergent sequence is characterized by the fact that its terms $$x_{m}$$ become (and stay) arbitrarily close to its limit, as $$m \rightarrow+\infty .$$ Due to this, however, they also get close to each other; in fact, $$\rho\left(x_{m}, x_{n}\right)$$ can be made arbitrarily small for sufficiently large $$m$$ and $$n .$$ It is natural to ask whether the latter property, in turn, implies the existence of a limit. This problem was first studied by Augustin-Louis Cauchy $$(1789-1857) .$$ Thus we shall call sequences Cauchy sequences. More precisely, we formulate the following.

Definition

A sequence $$\left\{x_{m}\right\} \subseteq(S, \rho)$$ is called a Cauchy sequence (we briefly say that $$"\left\{x_{m}\right\}$$ is Cauchy") iff, given any $$\varepsilon>0$$ (no matter how small), we have $$\rho\left(x_{m}, x_{n}\right)<\varepsilon$$ for all but finitely many $$m$$ and $$n .$$ In symbols,

$(\forall \varepsilon>0)(\exists k)(\forall m, n>k) \quad \rho\left(x_{m}, x_{n}\right)<\varepsilon.$

Observe that here we only deal with terms $$x_{m}, x_{n},$$ not with any other point. The limit (if any) is not involved, and we do not have to know it in advance. We shall now study the relationship between property $$(1)$$ and convergence.

Theorem $$\PageIndex{1}$$

Every convergent sequence $$\left\{x_{m}\right\} \subseteq(S, \rho)$$ is Cauchy.

Proof

Let $$x_{m} \rightarrow p .$$ Then given $$\varepsilon>0,$$ there is a $$k$$ such that

$(\forall m>k) \quad \rho\left(x_{m}, p\right)<\frac{\varepsilon}{2}.$

As this holds for any $$m>k,$$ it also holds for any other term $$x_{n}$$ with $$n>k$$.

Thus

$(\forall m, n>k) \quad \rho\left(x_{m}, p\right)<\frac{\varepsilon}{2} \text{ and } \rho\left(p, x_{n}\right)<\frac{\varepsilon}{2}.$

Adding and using the triangle inequality, we get

$\rho\left(x_{m}, x_{n}\right) \leq \rho\left(x_{m}, p\right)+\rho\left(p, x_{n}\right)<\varepsilon,$

and $$(1)$$ is proved. $$\square$$

Theorem $$\PageIndex{2}$$

Every Cauchy sequence $$\left\{x_{m}\right\} \subseteq(S, \rho)$$ is bounded.

Proof

We must show that all $$x_{m}$$ are in some globe. First we try an arbitrary radius $$\varepsilon$$ . Then by $$(1),$$ there is $$k$$ such that $$\rho\left(x_{m}, x_{n}\right)<\varepsilon$$ for $$m, n>k .$$ Fix some $$n>k .$$ Then

$(\forall m>k) \rho\left(x_{m}, x_{n}\right)<\varepsilon, \text{ i.e., } x_{m} \in G_{x_{n}}(\varepsilon).$

Thus the globe $$G_{x_{n}}(\varepsilon)$$ contains all $$x_{m}$$ except possibly the $$k$$ terms $$x_{1}, \ldots, x_{k}$$. To include them as well, we only have to take a larger radius $$r,$$ greater than $$\rho\left(x_{m}, x_{n}\right), m=1, \ldots, k .$$ Then all $$x_{m}$$ are in the enlarged globe $$G_{x_{n}}(r) .$$ $$\square$$

Note 1. In $$E^{1},$$ under the standard metric, only sequences with finite limits are regarded as convergent. If $$x_{n} \rightarrow \pm \infty,$$ then $$\left\{x_{n}\right\}$$ is not even a Cauchy sequence in $$E^{1}(\text { in view of Theorem } 2) ;$$ but in $$E^{*},$$ under a suitable metric (cf. Problem 5 in §11, it is convergent (hence also Cauchy and bounded).

Theorem $$\PageIndex{3}$$

If a Cauchy sequence $$\left\{x_{m}\right\}$$ clusters at a point $$p,$$ then $$x_{m} \rightarrow p$$.

Proof

We want to show that $$x_{m} \rightarrow p,$$ i.e., that

$(\forall \varepsilon>0)(\exists k)(\forall m>k) \quad \rho\left(x_{m}, p\right)<\varepsilon.$

Thus we fix $$\varepsilon>0$$ and look for a suitable $$k .$$ Now as $$\left\{x_{m}\right\}$$ is Cauchy, there is a $$k$$ such that

$(\forall m, n>k) \quad \rho\left(x_{m}, x_{n}\right)<\frac{\varepsilon}{2}.$

Also, as $$p$$ is a cluster point, the globe $$G_{p}\left(\frac{\varepsilon}{2}\right)$$ contains infinitely many $$x_{n},$$ so we can fix one with $$n>k\left(k \text { as above). Then } \rho\left(x_{n}, p\right)<\frac{\varepsilon}{2} \text { and, as noted above, }\right.$$ also $$\rho\left(x_{m}, x_{n}\right)<\frac{\varepsilon}{2}$$ for $$m>k .$$ Hence

$(\forall m>k) \quad \rho\left(x_{m}, x_{n}\right)+\rho\left(x_{n}, p\right)<\varepsilon,$

implying $$\rho\left(x_{m}, p\right) \leq \rho\left(x_{m}, x_{n}\right)+\rho\left(x_{n}, p\right)<\varepsilon,$$ as required. $$\square$$

Note 2. It follows that a Cauchy sequence can have at most one cluster point $$p,$$ for $$p$$ is also its limit and hence unique; see §14, Corollary 1.

These theorems show that Cauchy sequences behave very much like convergent ones. Indeed, our next theorem (a famous result by Cauchy) shows that, in $$E^{n}\left(^{*} \text { and } C^{n}\right)$$ the two kinds of sequences coincide.

Theorem $$\PageIndex{4}$$

(Cauchy's convergence criterion). A sequence $$\left\{\overline{x}_{m}\right\}$$ in $$E^{n}$$ (*or $$C^{n}$$ ) converges if and only if it is a Cauchy sequence.

Proof

Conversely, let $$\left\{x_{m}\right\}$$ be a Cauchy sequence. Then by Theorem $$2,$$ it is bounded. Hence by the Bolzano-Weierstrass theorem (Theorem 2 of §16, it has a cluster point $$\overline{p} .$$ Thus by Theorem 3 above, it converges to $$\overline{p},$$ and all is proved. $$\square$$

Unfortunately, this theorem (along with the Bolzano-Weierstrass theorem used in its proof) does not hold in all metric spaces. It even fails in some subspaces of $$E^{1} .$$ For example, we have

$x_{m}=\frac{1}{m} \rightarrow 0 \text{ in } E^{1}.$

By Theorem 1 , this sequence, being convergent, is also a Cauchy sequence. Moreover, it still preserves $$(1)$$ even if we remove the point 0 from $$E^{1}$$ since the distances $$\rho\left(x_{m}, x_{n}\right)$$ remain the same. However, in the resulting subspace $$S=E^{1}-\{0\},$$ the sequence no longer converges because its limit (and unique cluster point) 0 has disappeared, leaving a "gap" in its place. Thus we have a Cauchy sequence in $$S,$$ without a limit or cluster points, so Theorem 4 fails in $$S$$ (along with the Bolzano-Weierstrass theorem).

Quite similarly, both theorems fail in $$(0,1)$$ (but not in $$[0,1] )$$ as a subspace of $$E^{1}$$ . By analogy to incomplete ordered fields, it is natural to say that $$S$$ is "incomplete" because of the missing cluster point $$0,$$ and call a space (or subspace) "complete" if it has no such "gaps," i.e., if Theorem 4 holds in it.

Thus we define as follows.

Definition

A metric space (or subspace) $$(S, \rho)$$ is said to be complete iff every Cauchy sequence in $$S$$ converges to some point $$p$$ in $$S .$$

Similarly, a set $$A \subseteq(S, \rho)$$ is called complete iff each Cauchy sequence $$\left\{x_{m}\right\} \subseteq A$$ converges to some point $$p$$ in $$A,$$ i.e., iff $$(A, \rho)$$ is complete as a metric subspace of $$(S, \rho) .$$

In particular, $$E^{n}\left(^{*} \text { and } C^{n}\right)$$ are complete by Theorem $$4 .$$ The sets $$(0,1)$$ and $$E^{1}-\{0\}$$ are incomplete in $$E^{1},$$ but $$[0,1]$$ is complete. Indeed, we have the following theorem.

Theorem $$\PageIndex{5}$$

(i) Every closed set in a complete space is complete itself.

(ii) Every complete set $$A \subseteq(S, \rho)$$ is necessarily closed.

Proof

(i) Let $$A$$ be a closed set in a complete space $$(S, \rho) .$$ We have to show that Theorem 4 holds in $$A(\text { as it does in } S) .$$ Thus we fix any Cauchy sequence $$\left\{x_{m}\right\} \subseteq A$$ and prove that it converges to some $$p$$ in $$A .$$

Now, since $$S$$ is complete, the Cauchy sequence $$\left\{x_{m}\right\}$$ has a limit $$p$$ in $$S .$$ As $$A$$ is closed, however, that limit must be in $$A$$ by Theorem 4 in §16. Thus (i) is proved.

(ii) Now let $$A$$ be complete in a metric space $$(S, \rho) .$$ To prove that $$A$$ is closed, we again use Theorem 4 of §16. Thus we fix any convergent sequence $$\left\{x_{m}\right\} \subseteq A, x_{m} \rightarrow p \in S,$$ and show that $$p$$ must be $$i n A .$$

Now, since $$\left\{x_{m}\right\}$$ converges in $$S,$$ it is a Cauchy sequence, in $$S$$ as well as in A. Thus by the assumed completeness of $$A,$$ it has a limit $$q$$ in $$A .$$ Then, however, the uniqueness of $$\lim _{x_{m}} x_{m}(\text { in } S)$$ implies that $$p=q \in A,$$ so that $$p$$ is in $$A,$$ indeed. $$\square$$