Completeness of a Normed Vector Space Definition

Scientific claim

Completeness of a Normed Vector Space Definition

Claim

A normed vector space $(V,\|\cdot\|)$ is complete if every Cauchy sequence $(v_n)_{n\in\mathbb{N}}$ in $V$ , with distance defined by $d(u,v)=\|u-v\|$ , converges to a limit $v\in V$ .

Theoretical: Not applicableExperimental: Not applicable

Assumptions

No assumptions recorded.

Proof

This is a definition, so it is not proved from more fundamental propositions within this node. Its purpose is to state exactly when a normed vector space has no missing limits.

Let $V$ be a vector space over $\mathbb{R}$ or $\mathbb{C}$ . Its elements are called vectors.

Let $\|\cdot\|:V\to\mathbb{R}$ be a norm. For each vector $v\in V$ , the value $\|v\|$ measures its size.

For two vectors $u,v\in V$ , define their distance by:

d(u,v)=\|u-v\|.

The vector $u-v$ is the displacement from $v$ to $u$ . Its norm therefore measures how far apart $u$ and $v$ are.

Let $(v_n)_{n\in\mathbb{N}}$ be a sequence of vectors in $V$ , where $\mathbb{N}=\{1,2,3,\ldots\}$ and $v_n\in V$ is the vector at index $n$ .

The sequence $(v_n)_{n\in\mathbb{N}}$ is Cauchy when its later vectors become arbitrarily close to one another. Formally:

\forall\varepsilon>0,\ \exists N\in\mathbb{N}\ \text{such that}\ \forall n,m\geq N,\quad \|v_n-v_m\|<\varepsilon.

Here, $\varepsilon$ is any chosen positive tolerance. The index $N$ is the point after which every pair of sequence terms is closer than that tolerance.

A sequence $(v_n)_{n\in\mathbb{N}}$ converges to a vector $v\in V$ when its later vectors become arbitrarily close to one particular vector $v$ . Formally:

\forall\varepsilon>0,\ \exists N\in\mathbb{N}\ \text{such that}\ \forall n\geq N,\quad \|v_n-v\|<\varepsilon.

A Cauchy sequence guarantees that its later terms increasingly agree with one another, but it does not by itself guarantee that the vector they approach belongs to $V$ .

A normed vector space is complete when every Cauchy sequence in that space converges to a vector that is also an element of the space:

\forall (v_n)_{n\in\mathbb{N}}\subseteq V,\quad \left[ \forall\varepsilon>0,\ \exists N\in\mathbb{N}\ \text{such that}\ \forall n,m\geq N,\quad \|v_n-v_m\|<\varepsilon \right] \Rightarrow \left[ \exists v\in V,\quad \lim_{n\to\infty}\|v_n-v\|=0 \right].

This condition ensures that when a sequence of vectors stabilises to arbitrary precision, its limiting vector is not missing from the vector space.

Examples

Finite-Dimensional Real Vector Space — Complete
Let $V=\mathbb{R}^n$ with the Euclidean norm:
$\|\mathbf{v}\|_2 = \sqrt{\sum_{i=1}^{n}v_i^2}.$
The normed vector space $(\mathbb{R}^n,\|\cdot\|_2)$ is complete.
Every Cauchy sequence of vectors has Cauchy coordinate sequences in $\mathbb{R}$ . Since every Cauchy sequence of real numbers converges to a real number, the coordinate limits form a vector in $\mathbb{R}^n$ .
Finite-Support Sequences with the $\ell^2$ Norm — Not Complete
Let $c_{00}$ be the vector space of real sequences with only finitely many non-zero terms.
For $a=(a_1,a_2,\ldots)\in c_{00}$ , define:
$\|a\|_2 = \sqrt{\sum_{k=1}^{\infty}|a_k|^2}.$
Let $e_k$ denote the sequence whose $k$ -th term is $1$ and whose other terms are $0$ .
Define the sequence of vectors:
$s_n = \sum_{k=1}^{n}2^{-k}e_k.$
The sequence $(s_n)_{n\in\mathbb{N}}$ is Cauchy in the $\ell^2$ norm because the norm of its remaining tail can be made arbitrarily small.
Its limit is:
$s = \left( \frac{1}{2}, \frac{1}{4}, \frac{1}{8}, \ldots \right).$
The vector $s$ has infinitely many non-zero terms, so:
$s\notin c_{00}.$
Therefore, $(c_{00},\|\cdot\|_2)$ is not complete.

Evidence

01

What is a Hilbert Space?Abide by Reason · 2025-08-22 · supporting

Related Concepts

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