---
title: Convergence, Cauchy Sequences, and Completeness
module: Metric Spaces and Topology
moduleNumber: 3
lessonNumber: 3
order: 303
summary: >
  The $\varepsilon$-$N$ definition of a limit transfers verbatim to any metric
  space once $|x-y|$ is replaced by $d(x,y)$. Convergent sequences characterize
  closed sets and closures; Cauchy sequences and completeness capture spaces
  with no missing limits, with $\mathbb{R}^n$ and $C[a,b]$ complete and
  $\mathbb{Q}$ and $(0,1]$ not.
topics: [Metric Spaces and Topology]
draft: false
sources:
  - book: Lebl
    ref: "Ch. 7 — Metric Spaces; §7.3 Sequences and convergence, §7.4 Completeness"
  - book: Rosenlicht
    ref: "Ch. 3 — Metric spaces"
---

The definition of a limit for real
[sequences](/real-analysis/sequences-series/sequences-limits) used only two
things: the terms $x_n$, and the distance $|x_n - L|$ from the candidate limit.
Replace $|x_n - L|$ by $d(x_n, p)$ and the definition, together with almost
every proof about it, carries over to an arbitrary metric space unchanged. What
is new is that some spaces have "gaps" a Cauchy sequence can fall into.
**Completeness** is the property of having none, and it is the metric-space form
of the [least-upper-bound axiom](/real-analysis/foundations/ordered-fields-completeness)
that made $\mathbb{R}$ usable.

## Sequences and their limits

> **Definition (Convergence).** A sequence $\{x_n\}$ in a metric space
> $(X, d)$ **converges** to $p \in X$ if for every $\varepsilon > 0$ there is an
> $M \in \mathbb{N}$ such that $d(x_n, p) < \varepsilon$ for all $n \ge M$. Then
> $p$ is the **limit**, written $\lim_{n \to \infty} x_n = p$, and the sequence
> is **convergent**.

The condition $d(x_n, p) < \varepsilon$ says $x_n \in B(p, \varepsilon)$, so
convergence means: every ball about $p$, however small, contains all but
finitely many terms. The number $M$ marks the tail from which the sequence
stays inside the $\varepsilon$-ball.

$$
% caption: A sequence converging to $p$: for the drawn $\varepsilon$, all
% terms from $x_7$ on lie inside the ball $B(p,\varepsilon)$, so $M = 7$.
\begin{tikzpicture}[scale=1.0, font=\small]
\definecolor{acc}{HTML}{4A6FA5}
\fill[acc] (0,0) circle (1.6pt);
\node[right] at (0.08,0) {$p$};
\draw[acc, thick] (0,0) circle (1.3);
\draw[acc] (0,0) -- ++(115:1.3) node[midway, left, font=\scriptsize]{$r$};
% far scattered early terms
\foreach \pt/\lab in {(-3.0,1.4)/{x_1}, (2.7,-1.5)/{x_2}, (-2.6,-1.7)/{x_3},
  (3.1,1.6)/{x_4}, (-1.9,1.9)/{x_5}, (2.0,1.8)/{x_6}} {
  \fill[black] \pt circle (1.4pt);
  \node[black, font=\scriptsize] at \pt {};
}
\node[black, font=\scriptsize] at (-3.0,1.65) {$x_1$};
\node[black, font=\scriptsize] at (2.7,-1.75) {$x_2$};
% close terms inside ball
\foreach \pt in {(0.7,0.5),(-0.5,0.6),(0.4,-0.6),(-0.6,-0.3),(0.2,0.3)} {
  \fill[black] \pt circle (1.4pt);
}
\node[acc, font=\scriptsize] at (1.1,0.85) {tail terms};
\end{tikzpicture}
$$

The limit is unique, and the proof is the real-line proof with $d$ in place of
$|\cdot|$: if $x$ and $y$ are both limits, then for any $\varepsilon$ some tail
satisfies $d(x, x_n) < \varepsilon/2$ and $d(x_n, y) < \varepsilon/2$, so
$d(x, y) \le d(x, x_n) + d(x_n, y) < \varepsilon$; since $\varepsilon$ was
arbitrary, $d(x, y) = 0$ and $x = y$. Two more facts transfer intact:

- **Convergent sequences are bounded** (the tail sits in one ball, the finite
  head in another).
- **Squeeze by a null real sequence.** $x_n \to p$ iff there are reals
  $a_n \to 0$ with $d(x_n, p) \le a_n$; this reduces metric convergence to
  ordinary convergence of the distances.

**Componentwise convergence in $\mathbb{R}^n$.** A sequence $x_m =
(x_{m,1}, \dots, x_{m,n})$ in Euclidean space converges iff each coordinate
sequence converges, in which case the limit is taken coordinate by coordinate:

$$
\lim_{m \to \infty} x_m
= \Bigl( \lim_m x_{m,1},\ \dots,\ \lim_m x_{m,n} \Bigr).
$$

One direction follows from $|y_k - x_{m,k}| \le d(y, x_m)$, the other from
$d(y, x_m) < \varepsilon$ once every coordinate is within
$\varepsilon/\sqrt{n}$. This reduces limits in $\mathbb{R}^n$ to $n$ limits in
$\mathbb{R}$, and identifies convergence in $\mathbb{C} \cong \mathbb{R}^2$ with
convergence of real and imaginary parts.

> **Worked example ($\varepsilon$-$M$ convergence in $\mathbb{R}^2$).** Take
> $x_m = \bigl( \tfrac1m,\ \tfrac{m+1}{m} \bigr)$ and $p = (0, 1)$. Coordinatewise,
> $\tfrac1m \to 0$ and $\tfrac{m+1}{m} = 1 + \tfrac1m \to 1$. Directly,
> $$
> d(x_m, p) = \sqrt{\Bigl(\tfrac1m\Bigr)^2 + \Bigl(\tfrac1m\Bigr)^2}
> = \frac{\sqrt{2}}{m}.
> $$
> Given $\varepsilon > 0$, choose
> $M = \bigl\lceil \sqrt{2}/\varepsilon \bigr\rceil + 1$. For $m \ge M$,
> $d(x_m, p) = \sqrt{2}/m \le \sqrt{2}/M < \varepsilon$, so $x_m \to (0, 1)$.

$$
% caption: Componentwise convergence in $\mathbb{R}^2$: as $x_m \to (0,1)$, the
% two coordinate sequences converge on the axes, and conversely.
\begin{tikzpicture}[scale=1.0, font=\small]
\definecolor{acc}{HTML}{4A6FA5}
\draw[->, black] (-0.4,0) -- (3.6,0) node[right]{$x_1$};
\draw[->, black] (0,-0.3) -- (0,3.0) node[above]{$x_2$};
% limit point p=(0,1) scaled: put axis at 0, y=1 -> y=1.4 in drawing
\fill[acc] (0,1.4) circle (1.8pt);
\node[left] at (-0.05,1.4) {$p$};
% terms x_m: x=1/m scaled by 3, y=1+1/m scaled
\foreach \mx/\xd/\yd in {1/3.0/2.8, 2/1.5/2.1, 3/1.0/1.87, 4/0.75/1.75, 6/0.5/1.63} {
  \fill[black] (\xd,\yd) circle (1.5pt);
}
\node[black, right, font=\scriptsize] at (3.05,2.8) {$m = 1$};
% dashed projections for the first term
\draw[black, dashed] (3.0,2.8) -- (3.0,0);
\draw[black, dashed] (3.0,2.8) -- (0,2.8);
\node[black, below, font=\scriptsize] at (3.0,0) {$\frac{1}{m}$};
\node[black, left, font=\scriptsize] at (0,2.8) {$1 + \frac{1}{m}$};
\end{tikzpicture}
$$

## Convergence and the topology

Convergence can be stated with no reference to $\varepsilon$ at all, using only
open neighborhoods, which is why it is a topological notion.

> **Proposition (Neighborhood form of convergence).** $\{x_n\}$ converges to
> $p$ iff for every open neighborhood $U$ of $p$ there is an $M$ with $x_n \in
> U$ for all $n \ge M$.

Any open $U \ni p$ contains a ball $B(p, \varepsilon)$, and the tail inside that
ball is inside $U$; conversely each ball is itself an open neighborhood. The two
formulations are interchangeable, and they connect convergence to
[closed sets and closures](/real-analysis/metric-spaces/open-closed-sets).

> **Proposition (Sequential description of closed sets).** Let $E \subseteq X$.
>
> - If $E$ is closed and $\{x_n\} \subseteq E$ converges to $p \in X$, then
>   $p \in E$.
> - For any $A \subseteq X$, a point $p \in \overline{A}$ iff some sequence in
>   $A$ converges to $p$.

A closed set contains the limits of all its convergent sequences: if the limit
$p$ lay in the open set $E^c$, some tail would be trapped in $E^c$, contradicting
$\{x_n\} \subseteq E$. And the closure consists precisely of the sequential limits
from $A$: given $p \in \overline{A}$, each ball $B(p, 1/n)$ meets $A$ in some
$x_n$, and those $x_n \to p$. This makes "closed" and "contains its sequential
limits" the same statement, and recovers the closure as the reachable points.

## Cauchy sequences

Convergence names a limit. Often one wants to certify that a sequence _settles_
without naming where. The Cauchy condition does this using only the terms.

> **Definition (Cauchy sequence).** A sequence $\{x_n\}$ in $(X, d)$ is
> **Cauchy** if for every $\varepsilon > 0$ there is an $M$ with
> $d(x_n, x_k) < \varepsilon$ for all $n, k \ge M$.

The terms bunch together: past $M$, any two of them are within $\varepsilon$.
Every convergent sequence is Cauchy, since if $x_n \to p$ then for $n, k \ge M$,

$$
d(x_n, x_k) \le d(x_n, p) + d(p, x_k) < \tfrac{\varepsilon}{2} +
\tfrac{\varepsilon}{2} = \varepsilon.
$$

The converse is the question. In $\mathbb{R}$ it holds, and that was the
substance of [Cauchy completeness of $\mathbb{R}$](/real-analysis/sequences-series/cauchy-completeness).
In a general metric space it can fail: the terms bunch, but the point they
bunch around may not exist in the space.

$$
% caption: A Cauchy sequence in $(0,1]$: the terms $1/n$ bunch together and
% aim at $0$, but $0$ lies outside the space, so there is no limit in $(0,1]$.
\begin{tikzpicture}[scale=1.0, font=\small]
\definecolor{acc}{HTML}{4A6FA5}
\draw[black, thick] (0,0) -- (6,0);
% open endpoint at 0
\draw[acc, thick, fill=white] (0,0) circle (2.4pt);
\node[below] at (0,-0.1) {$0$};
\node[below, acc, font=\scriptsize] at (0.0,-0.5) {excluded};
% closed endpoint at 1
\fill[black] (6,0) circle (2.4pt);
\node[below] at (6,-0.1) {$1$};
% terms 1/n
\foreach \x/\lab in {6/1, 3/{\frac12}, 2/{\frac13}, 1.5/{\frac14}, 1.2/{}, 1.0/{}, 0.85/{}} {
  \fill[black] (\x,0) circle (1.8pt);
}
\node[black, above, font=\scriptsize] at (6,0.1) {$1$};
\node[black, above, font=\scriptsize] at (3,0.1) {$\tfrac12$};
\node[black, above, font=\scriptsize] at (2,0.1) {$\tfrac13$};
\draw[acc, ->] (1.4,0.55) .. controls (0.8,0.5) .. (0.25,0.2);
\node[acc, font=\scriptsize] at (1.9,0.7) {bunching toward $0$};
\end{tikzpicture}
$$

## Complete metric spaces

> **Definition (Completeness).** A metric space $(X, d)$ is **complete** if
> every Cauchy sequence in $X$ converges to a point of $X$.

Completeness says the space has no gaps for a Cauchy sequence to fall through.
The central examples are complete; the instructive failures are subspaces that
omit a limit point.

> **Theorem (Standard complete spaces).**
> $\mathbb{R}^n$ with the Euclidean metric is complete, and $C[a,b]$ with the
> sup metric is complete.

For $\mathbb{R}^n$ the proof reduces to coordinates: a Cauchy sequence in
$\mathbb{R}^n$ is Cauchy in each coordinate by $|x_{i,k} - x_{j,k}| \le
d(x_i, x_j)$, each coordinate converges by completeness of $\mathbb{R}$, and
componentwise convergence assembles the limit. For $C[a,b]$, a Cauchy sequence
in the sup metric is a uniformly Cauchy sequence of functions; it converges
uniformly to a limit, and the uniform limit of continuous functions is
continuous, so the limit lies back in $C[a,b]$.[^lebl-74]

> **Worked example ($(0,1]$ is not complete).** The sequence $x_n = 1/n$ lies in
> $(0, 1]$. It is Cauchy: for $n, k \ge M$,
> $$
> \Bigl| \tfrac1n - \tfrac1k \Bigr| \le \tfrac1n + \tfrac1k \le \tfrac2M,
> $$
> which is below any $\varepsilon$ once $M > 2/\varepsilon$. In $\mathbb{R}$ the
> sequence converges to $0$, and a limit is unique, so $0$ is the only candidate
> limit in $(0, 1]$ as well. But $0 \notin (0, 1]$, so $\{1/n\}$ has no limit in
> the space. Thus $(0, 1]$ is not complete.

The rationals $\mathbb{Q}$ fail for the same reason: a sequence of rationals
converging to $\sqrt{2}$ in $\mathbb{R}$ is Cauchy in $\mathbb{Q}$ with no
rational limit, since $\sqrt{2}$ is irrational. Completeness is precisely the
property that separates $\mathbb{R}$ from $\mathbb{Q}$, and the
[least-upper-bound property](/real-analysis/foundations/ordered-fields-completeness)
is what supplies it.

$$
% caption: The rationals are incomplete: rational truncations of $\sqrt{2}$
% bunch together yet aim at a point the line of rationals does not contain.
\begin{tikzpicture}[scale=1.0, font=\small]
\definecolor{acc}{HTML}{4A6FA5}
\draw[black, thick] (0,0) -- (6.4,0);
\node[black, right] at (6.45,0) {$\mathbb{Q}$};
% target gap at sqrt2 ~ 1.414, place at x=4.4
\draw[acc, thick, fill=white] (4.4,0) circle (2.4pt);
\node[below] at (4.4,-0.12) {$L$};
\node[acc, above, font=\scriptsize] at (5.3,0.55) {no rational here};
% rational terms approaching from left: 1, 1.4, 1.41, 1.414
\foreach \x/\lab in {1.6/{p_1}, 3.2/{p_2}, 3.9/{p_3}, 4.2/{}} {
  \fill[black] (\x,0) circle (1.8pt);
}
\node[black, below, font=\scriptsize] at (1.6,-0.1) {$p_1$};
\node[black, below, font=\scriptsize] at (3.2,-0.1) {$p_2$};
\node[black, below, font=\scriptsize] at (3.9,-0.1) {$p_3$};
\draw[acc, ->] (2.2,0.5) .. controls (3.4,0.45) .. (4.15,0.2);
\node[acc, font=\scriptsize] at (1.7,0.85) {Cauchy, no limit in $\mathbb{Q}$};
\end{tikzpicture}
$$

| Space | Complete? | Witness |
| --- | --- | --- |
| $\mathbb{R}^n$, Euclidean | yes | reduces to $\mathbb{R}$ coordinatewise |
| $C[a,b]$, sup metric | yes | uniform limit of continuous is continuous |
| discrete metric, any set | yes | Cauchy sequences are eventually constant |
| $\mathbb{Q}$, standard metric | no | rationals approaching $\sqrt{2}$ |
| $(0,1]$, subspace of $\mathbb{R}$ | no | the sequence $\{1/n\}$ |

The pattern in the failures is that both omit a limit point of themselves.
Adding back exactly the missing limits repairs completeness.

> **Proposition (Closed subspaces inherit completeness).** A closed subset $E$
> of a complete metric space $(X, d)$ is complete in the subspace metric.

A Cauchy sequence in $E$ is Cauchy in $X$, so it converges in $X$; because $E$
is closed, the limit lies in $E$. This is why $(0, 1]$ fails but $[0, 1]$
succeeds: the latter is closed in the complete space $\mathbb{R}$.

> **Worked example (A Cauchy sequence in $C[0,1]$).** Let $f_n(t) = \sum_{k=0}^n
> t^k / k!$, the partial sums of the exponential series. For $m > n$ and every
> $t \in [0, 1]$,
> $$
> \Bigl| f_m(t) - f_n(t) \Bigr| = \Bigl| \sum_{k=n+1}^m \frac{t^k}{k!} \Bigr|
> \le \sum_{k=n+1}^m \frac{1}{k!} \le \sum_{k=n+1}^\infty \frac{1}{k!},
> $$
> using $0 \le t \le 1$. Taking the supremum over $t$,
> $$
> d_\infty(f_m, f_n) \le \sum_{k=n+1}^\infty \frac{1}{k!} \xrightarrow{\ n \to \infty\ } 0,
> $$
> because the numeric series $\sum 1/k!$ converges. So $\{f_n\}$ is Cauchy in the
> sup metric, and completeness of $C[0,1]$ guarantees a continuous limit, namely
> $e^t$.

The abstract theorem replaces an $\varepsilon$-$\delta$ argument about the limit
function with a single appeal to completeness.

## Completion

A space with gaps can be enlarged to fill them, exactly as $\mathbb{Q}$ is
enlarged to $\mathbb{R}$.

> **Theorem (Completion).** Every metric space $(X, d)$ embeds isometrically as
> a dense subset of a complete metric space $(\widehat{X}, \widehat{d})$, unique
> up to isometry. The larger space $\widehat{X}$ is the **completion** of $X$.

The construction mirrors the [Cauchy-sequence build of
$\mathbb{R}$](/real-analysis/sequences-series/cauchy-completeness): points of
$\widehat{X}$ are equivalence classes of Cauchy sequences in $X$, two sequences
identified when the distance between corresponding terms tends to $0$, and
$\widehat{d}$ is the limit of the term-by-term distances. The original space
sits inside as the classes of constant sequences, and it is dense because every
Cauchy sequence is, by definition, approached by its own terms. The completion
of $\mathbb{Q}$ is $\mathbb{R}$; the completion of $(0, 1]$ is $[0, 1]$; the
completion of the polynomials on $[a, b]$ under the sup metric is all of
$C[a, b]$, since by the Weierstrass approximation theorem every continuous
function is a uniform limit of polynomials. Completion lets analysis assume completeness without
loss of generality.

## Nested sets and completeness

Completeness has a geometric consequence: nested closed balls whose radii tend
to zero always contain a common point. This is the metric analogue of the
[nested interval property](/real-analysis/foundations/intervals-uncountability).

$$
% caption: In a complete space, closed balls $C_1 \supseteq C_2 \supseteq
% \cdots$ with radii shrinking to $0$ have exactly one common point $p$.
\begin{tikzpicture}[scale=1.0, font=\small]
\definecolor{acc}{HTML}{4A6FA5}
\draw[black, thick] (0,0) circle (2.0);
\node[black] at (-1.5,1.5) {$C_1$};
\draw[black, thick] (0.5,0.2) circle (1.35);
\node[black] at (1.5,1.1) {$C_2$};
\draw[black, thick] (0.9,0.35) circle (0.8);
\node[black] at (0.35,0.3) {$C_3$};
\draw[black, thick] (1.15,0.45) circle (0.4);
\fill[acc] (1.35,0.5) circle (1.8pt);
\node[right, acc] at (1.42,0.5) {$p$};
\end{tikzpicture}
$$

> **Theorem (Nested closed balls).** In a complete metric space, if $C(x_1,
> r_1) \supseteq C(x_2, r_2) \supseteq \cdots$ are closed balls with $r_n \to
> 0$, then $\bigcap_n C(x_n, r_n)$ is a single point.

The centers $\{x_n\}$ form a Cauchy sequence, because for $n, k \ge M$ both lie
in $C(x_M, r_M)$ and so are within $2 r_M \to 0$; completeness gives a limit
$p$, and $p$ lies in every $C(x_n, r_n)$ since each is closed and contains the
tail. Two common points would be within every $2 r_n$, forcing them equal.

This nested-ball property, completeness, and the convergence of Cauchy
sequences are three views of the same fact about $\mathbb{R}^n$ and $C[a,b]$.
Completeness together with a finiteness condition yields
[compactness](/real-analysis/metric-spaces/compactness), where every sequence
has a convergent subsequence.

[^lebl-74]: **Lebl**, _Basic Analysis I_, §7.3 (sequences and convergence) and §7.4.1 (Cauchy sequences and completeness): uniqueness of limits, sequential characterization of closed sets, completeness of $\mathbb{R}^n$ and $C[a,b]$, and completeness of closed subspaces.