Completeness in Metric Spaces

A metric space is complete if every Cauchy sequence converges to a limit in the space. Completeness generalizes the completeness of $\mathbb{R}$ and is essential for fixed-point theorems, existence of solutions to differential equations, and functional analysis. Complete spaces are "without gaps" — there are no missing limit points.

Definition

Definition8.1Cauchy sequence

A sequence $(x_n)$ in a metric space $(X, d)$ is Cauchy if for every $\epsilon > 0$ , there exists $N$ such that $d(x_m, x_n) < \epsilon$ for all $m, n \geq N$ .

Definition8.2Complete metric space

A metric space $(X, d)$ is complete if every Cauchy sequence in $X$ converges to a limit in $X$ .

ExampleR is complete

By the Cauchy criterion (Theorem 2.2), $\mathbb{R}$ with the usual metric is complete.

ExampleQ is not complete

$\mathbb{Q}$ with the usual metric is not complete. For example, the sequence $1, 1.4, 1.41, 1.414, \ldots$ (decimal approximations to $\sqrt{2}$ ) is Cauchy in $\mathbb{Q}$ but does not converge to a rational.

ExampleC([a, b]) is complete

The space $C([a, b])$ of continuous functions with the sup norm $d(f, g) = \sup |f - g|$ is complete. If $(f_n)$ is Cauchy in sup norm, it converges uniformly to a continuous function $f \in C([a, b])$ .

Example(0, 1) is not complete

$(0, 1)$ with the Euclidean metric is not complete. The sequence $x_n = 1/n$ is Cauchy, but its limit $0 \notin (0, 1)$ .

Banach Fixed-Point Theorem

Theorem8.1Banach Fixed-Point Theorem (Contraction Mapping)

Let $(X, d)$ be a complete metric space and $f : X \to X$ a contraction: there exists $0 < \lambda < 1$ such that $d(f(x), f(y)) \leq \lambda d(x, y)$ for all $x, y \in X$ . Then $f$ has a unique fixed point: there exists unique $x^* \in X$ with $f(x^*) = x^*$ .

Moreover, for any $x_0 \in X$ , the sequence $x_{n+1} = f(x_n)$ converges to $x^*$ .

RemarkApplications

The Banach Fixed-Point Theorem is used to prove existence and uniqueness of solutions to ODEs (Picard-Lindelöf), solve integral equations, and analyze iterative algorithms.

Examplef(x) = (x + 2)/2 on [0, 1]

$f(x) = (x + 2)/2$ maps $[0, 1]$ to itself. For $x, y \in [0, 1]$ ,

$|f(x) - f(y)| = \frac{|x - y|}{2} \leq \frac{1}{2} |x - y|.$

By Banach's theorem, $f$ has a unique fixed point: $x^* = 2$ (but $2 \notin [0, 1]$ , so this example fails). Corrected: $f : [1, 2] \to [1, 2]$ has fixed point $x^* = 2$ .

Summary

Completeness ensures Cauchy sequences converge:

$\mathbb{R}^n$ , $C([a, b])$ are complete.
$\mathbb{Q}$ , $(0, 1)$ are not complete.
Banach Fixed-Point Theorem: contractions on complete spaces have unique fixed points.

See Metric Spaces and Banach Spaces.