Cauchy Sequences in Metric Spaces

When Sequence Terms Draw Close to One Another

Convergence in a metric space compares the terms of a sequence with a specified point of the space. The uniqueness arguments in the previous tutorial used this comparison to show that a sequence cannot have two different limits. But in many situations, a candidate limit is not yet known, or may not even belong to the space. We can still ask whether the sequence’s terms become close to one another as the indices grow.

That question leads to the definition of a Cauchy sequence. It expresses an internal form of closeness: sufficiently late terms must be near each other, regardless of which two late indices are chosen. We will show that convergence implies this condition, and that a Cauchy sequence with a convergent subsequence must itself converge. The converse to the first statement does not hold in every metric space, as an example will illustrate.

Definition and First Interpretation

Definition (Cauchy Sequence): Let $(X,d)$ be a metric space. A sequence $(x_n)$ in $X$ is a Cauchy sequence if, for every $\varepsilon>0$, there is an index $N$ such that $$ d(x_m,x_n)<\varepsilon $$ whenever $m,n\geq N$.

The indices $m$ and $n$ are chosen independently, and both must be at least $N$. The condition is therefore stronger than checking that consecutive terms become close: it controls every pair in the tail. The index $N$ may depend on $\varepsilon$, but it may not depend on the particular pair $m,n$ once both are in that tail.

Compare this definition with convergence. To say $x_n\to x$ requires that the terms eventually lie within $\varepsilon$ of one fixed point $x$. To say $(x_n)$ is Cauchy requires that any two sufficiently late terms lie within $\varepsilon$ of each other. The definition of a Cauchy sequence does not name a point in $X$, so it does not by itself promise that a limit exists in $X$.

Worked Example: A Cauchy Sequence in the Real Line

Consider the sequence $a_n=4+\frac{(-1)^n}{n+2}$ in $\mathbb{R}$ with its usual metric. For any positive integers $m,n$, the triangle inequality gives $$ |a_m-a_n| =\left|\frac{(-1)^m}{m+2}-\frac{(-1)^n}{n+2}\right| \leq \frac{1}{m+2}+\frac{1}{n+2}. $$ If $m,n\geq N$, then $$ |a_m-a_n|\leq\frac{2}{N+2}. $$ Given $\varepsilon>0$, choose a positive integer $N$ such that $2/(N+2)<\varepsilon$. Then every $m,n\geq N$ satisfies $|a_m-a_n|<\varepsilon$. Thus $(a_n)$ is Cauchy. This estimate works directly with pairs of terms; it does not need a proposed limit.

Convergence Implies the Cauchy Condition

The first connection between the two ideas follows from the triangle inequality. If late terms are each close to a limit, then they are close to one another by traveling from one term to the limit and then from the limit to the other term.

Theorem (Every Convergent Sequence Is Cauchy): Let $(X,d)$ be a metric space. If $x_n\to x$ in $X$, then $(x_n)$ is a Cauchy sequence.

Proof. Let $\varepsilon>0$. Since $x_n\to x$, there is an index $N$ such that $d(x_n,x)<\varepsilon/2$ whenever $n\geq N$. Take any $m,n\geq N$. By the triangle inequality, $$ d(x_m,x_n)\leq d(x_m,x)+d(x,x_n) <\frac{\varepsilon}{2}+\frac{\varepsilon}{2} =\varepsilon. $$ The same $N$ works for every such pair $m,n$. This is exactly the Cauchy condition, so $(x_n)$ is Cauchy. $\square$

The use of $\varepsilon/2$ allocates half of the allowed distance to each leg of the triangle. More generally, when a triangle inequality bounds a quantity by a sum of two errors, each error can be made smaller than half the desired total. The conclusion requires no special property of $X$ beyond the metric axioms.

Worked Example: A Cauchy Sequence Without a Limit in Its Space

Let $X=(0,1)$ with the usual metric, and define $x_n=1/(n+1)$. For $m,n\geq N$, both terms lie between $0$ and $1/(N+1)$. Therefore $$ |x_m-x_n|\leq\max\{x_m,x_n\}\leq\frac{1}{N+1}. $$ Given $\varepsilon>0$, choose $N$ so that $1/(N+1)<\varepsilon$. Then $|x_m-x_n|<\varepsilon$ for all $m,n\geq N$, so the sequence is Cauchy in $X$.

However, it has no limit in $X$. Viewed as a sequence in $\mathbb{R}$, it converges to $0$. If it also converged to some $x\in(0,1)$, it would have two limits in $\mathbb{R}$. By the Uniqueness of Metric Limits Theorem, those limits would be equal, so $x=0$, contrary to $x\in(0,1)$. Thus the sequence is Cauchy in $X$ but has no limit there. The issue is that its natural limiting point is not an element of the space.

Subsequences Preserve the Cauchy Property

A subsequence keeps terms from the original sequence in their original order. If the original sequence is Cauchy, then any sufficiently late terms selected from it must also be close to one another. This fact lets us use information about a subsequence without losing the Cauchy condition.

Theorem (Every Subsequence of a Cauchy Sequence Is Cauchy): Let $(x_n)$ be Cauchy in a metric space, and let $(x_{n_k})$ be a subsequence. Then $(x_{n_k})$ is Cauchy.

Proof. Let $\varepsilon>0$. Since $(x_n)$ is Cauchy, choose $N$ such that $d(x_m,x_n)<\varepsilon$ whenever $m,n\geq N$. The indices of a subsequence are strictly increasing positive integers, so $n_k\geq k$. In particular, if $k,\ell\geq N$, then $n_k,n_\ell\geq N$. Applying the Cauchy estimate for the original sequence gives $$ d(x_{n_k},x_{n_\ell})<\varepsilon. $$ This holds for all $k,\ell\geq N$, which proves that the subsequence is Cauchy. $\square$

The argument depends on the subsequence indices eventually passing every fixed index of the original sequence. It does not require that the subsequence contain every term, or that it have any special pattern beyond being a subsequence.

Worked Example: An Alternating Sequence in the Discrete Metric

Let $X=\{p,q\}$ have the discrete metric $\delta$, where $\delta(u,v)=0$ if $u=v$ and $\delta(u,v)=1$ otherwise. Define $x_n=p$ when $n$ is even and $x_n=q$ when $n$ is odd. This sequence is not Cauchy. For $\varepsilon=1/2$, every tail contains an even-indexed term $p$ and an odd-indexed term $q$. The distance between these two terms is $$ \delta(p,q)=1\geq\frac12. $$ Thus no index $N$ can make all pairs of terms in the tail have distance less than $1/2$.

The even-indexed subsequence is constantly equal to $p$, and the odd-indexed subsequence is constantly equal to $q$; each is Cauchy, since any two terms within either subsequence have distance $0$. This does not conflict with the theorem: it says a subsequence of a Cauchy sequence is Cauchy, not that a sequence whose subsequences are Cauchy must itself be Cauchy. The two different kinds of terms remain a fixed positive distance apart.

A Convergent Subsequence Determines the Whole Sequence

A particularly useful strengthening combines the Cauchy condition with convergence of just one subsequence. The subsequence provides a candidate limit, and the Cauchy condition ensures that all sufficiently late terms of the original sequence are close to a sufficiently late term of that subsequence.

Theorem (A Cauchy Sequence with a Convergent Subsequence Converges): Let $(x_n)$ be Cauchy in a metric space $(X,d)$. If a subsequence $(x_{n_k})$ converges to $x\in X$, then $x_n\to x$.

Proof. Let $\varepsilon>0$. Since $(x_n)$ is Cauchy, choose $N_1$ such that $$ d(x_m,x_n)<\frac{\varepsilon}{2} $$ whenever $m,n\geq N_1$. Since $x_{n_k}\to x$, choose $K$ such that $d(x_{n_k},x)<\varepsilon/2$ whenever $k\geq K$. The subsequence indices increase without bound, so choose $k\geq K$ with $n_k\geq N_1$. Fix this $k$. For any $n\geq N_1$, both $n$ and $n_k$ are at least $N_1$, and hence $$ d(x_n,x)\leq d(x_n,x_{n_k})+d(x_{n_k},x) <\frac{\varepsilon}{2}+\frac{\varepsilon}{2} =\varepsilon. $$ This holds for every $n\geq N_1$, so $x_n\to x$. $\square$

The order of choices matters. First obtain a tail on which every pair of original terms is close. Then choose one subsequence term that is both in that tail and close to $x$. That fixed term serves as a bridge from any sufficiently late $x_n$ to $x$. This theorem is a useful way to turn subsequential convergence into convergence of a whole sequence, provided the Cauchy condition is available.

Cauchy Sequences Are Bounded

In a metric space, a sequence is bounded if there are a point $z\in X$ and a finite number $R$ such that $d(x_n,z)\leq R$ for every $n$. A Cauchy sequence always has this property. The proof uses the Cauchy condition for the tail and handles the finitely many earlier terms separately.

Theorem (Every Cauchy Sequence Is Bounded): Every Cauchy sequence in a metric space is bounded.

Proof. Apply the Cauchy condition with $\varepsilon=1$. There is an index $N$ such that $d(x_n,x_N)<1$ whenever $n\geq N$. For the finitely many terms with $n<N$, each distance $d(x_n,x_N)$ is a finite real number. Choose $R$ to be the maximum of $1$ and those finitely many distances; if there are no indices $n<N$, take $R=1$. Then $d(x_n,x_N)\leq R$ for every $n$: for $n\geq N$ the distance is less than $1\leq R$, and for $n<N$ it is included in the maximum. Thus the sequence is bounded, with center $x_N$. $\square$

This result is one-way: boundedness alone does not guarantee that terms draw closer together. The alternating sequence in the discrete metric is bounded, since every point is at distance at most $1$ from $p$, but it is not Cauchy. Boundedness controls distance from a center; the Cauchy condition controls distances between every pair of late terms.

Why the Cauchy Condition Matters

The Cauchy condition is useful precisely because it can be checked without first knowing a limit. It also identifies a requirement that a metric space must satisfy if every Cauchy sequence is to converge within that space. The next tutorial develops this idea through completeness. For now, keep the distinction clear: convergence implies the Cauchy condition, but the example in $(0,1)$ shows that a Cauchy sequence need not converge to a point of the space.

A common proof error is to show only that $d(x_n,x_{n+1})\to0$ and conclude that the sequence is Cauchy. The definition asks for control of $d(x_m,x_n)$ for arbitrary late $m,n$, not just consecutive indices. Another error is to assume that being Cauchy automatically supplies a limit in $X$. The definition contains no such point; a limit may be missing from the space, as it is for $1/(n+1)$ in $(0,1)$.

Takeaway: A sequence is Cauchy when all pairs of sufficiently late terms are close. Every convergent sequence is Cauchy, every subsequence of a Cauchy sequence is Cauchy, and a Cauchy sequence with a convergent subsequence converges to the same limit. Whether every Cauchy sequence has a limit in its space is a further property of that space.

Check Your Understanding

Use the definition and proved results to answer the following questions.

In the Cauchy definition, why must the same index $N$ work for every pair $m,n\geq N$?
How does the triangle inequality show that a sequence converging to $x$ is Cauchy?
Why does the sequence $1/(n+1)$ have no limit in $(0,1)$, even though it is Cauchy there?
Which property of subsequence indices is used to prove that a subsequence of a Cauchy sequence is Cauchy?
In the theorem about a Cauchy sequence with a convergent subsequence, what role does the selected subsequence term play?
Why does boundedness not imply the Cauchy condition, as illustrated by the alternating sequence in the discrete metric?