# 3.13: Cauchy Sequences. Completeness

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- 20146

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. \(\square\)