Complete Metric Space¶

Definition¶

A metric space $(X, d)$ is termed complete if each Cauchy sequence contained within converges in $(X, d)$. When a complete normed vector space is encountered, it's also referred to as a Banach space.

Interpretation¶

A complete metric space is one with no "missing points": every Cauchy sequence, a sequence that looks as if it should converge, actually has a limit inside the space.

The displayed Cauchy sequence converges to $\sqrt{2}$ in $\mathbb{R}$. It has no limit in $\mathbb{Q}$ because $\sqrt{2}\notin\mathbb{Q}$$\mathbb{Q}$, illustrating the incompleteness of the rational numbers.

Completeness lets one prove that a limit exists without knowing it in advance: it is enough to show that the sequence is Cauchy. For $\mathbb{R}$ it comes from the supremum property, which $\mathbb{Q}$ lacks: $\{q\in\mathbb{Q}\colon q^2<2\}$ is bounded above but has no least upper bound in $\mathbb{Q}$.

Completeness depends on the metric, not only on which sequences converge. On $(-\tfrac{\pi}{2},\tfrac{\pi}{2})$, the metric $d_{\tan}(x,y)=|\tan x-\tan y|$ has the same convergent sequences as $|x-y|$$y|$, because $\tan$ and its inverse are continuous. Yet $d_{\tan}$ is complete, since $\tan$ matches it with $\mathbb{R}$ and its usual distance, while $|x-y|$ is not: $x_k=\tfrac{\pi}{2}-\tfrac{1}{k}$ is Cauchy for $|x-y|$ without a limit in the interval.

Every finite-dimensional real normed space, in particular $\mathbb{R}^n$ with any norm, is complete, so the term Banach space matters mainly for spaces of functions: the bounded functions on a set, with the uniform norm, form a Banach space.

A common slip is to link completeness with boundedness. $\mathbb{R}$ is unbounded and complete, while the bounded interval $(0,1)$ is not complete.

Key Results¶

• In any metric space every convergent sequence is a Cauchy sequence; in a complete metric space the converse also holds, so a sequence converges if and only if it is Cauchy.

Examples¶

• The Euclidean space $\mathbb{R}^n$ is a complete normed vector space.
• The open interval $(0,1)$ with the usual metric $d(x,y)=|x-y|$ is not complete: $(1/n)_{n\ge2}$ is a Cauchy sequence whose limit $0$ does not belong to $(0,1)\text{.}$
• The $n$-dimensional real coordinate space equipped with the discrete metric is complete: a Cauchy sequence is eventually constant, because $d(\boldsymbol{x}_j,\boldsymbol{x}_k)<\tfrac{1}{2}$ forces $\boldsymbol{x}_j=\boldsymbol{x}_k\text{.}$

Prerequisites¶

• Cauchy Sequence in a Metric Space
• Convergence in a Metric Space
• Norm and Normed Vector Space

Related Concepts¶

• Euclidean space

Sources¶

• Math III: Analysis · Lecture 4

Aliases: complete space Symbols: element of, pi, rational numbers, real numbers, square root

Last updated: 20 October 2026 03:02