CorollaryStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-09-04

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Every finite-dimensional normed space is Banach

Statement

Let $X$ be a normed space over $K \in {R, C}$ and assume $X$ admits an ordered basis of finite length. Then $X$ is a Banach space in the sense of Banach space.

Facts & Assumptions

Given: A normed space $X$ over $K \in {R, C}$ with an ordered basis $e : n \to X$ .

[L1]

The basis map $T : K^{n} \to X$ is a topological isomorphism (A chosen algebraic basis identifies a finite-dimensional normed space with a coordinate space).

[L2]

For $m \geq 1$ , $R^{m}$ is complete in every norm (For $n \geq 1$ a sequence in $R^{n}$ converges iff each coordinate sequence converges, is Cauchy iff each coordinate sequence is Cauchy, and $R^{n}$ is complete in every norm).

[L3]

$C$ is the real coordinate plane ( $C$ is the real coordinate plane, with coordinate arithmetic).

[L4]

A Banach space is a normed space complete for its norm metric (Banach space).

Proof

technique · direct

1.1

By [L1], it is enough to prove that $K^{n}$ is complete for the coordinate $ℓ^{1}$ norm, because a bounded bijection with bounded inverse preserves Cauchy sequences and their limits.

L1L4

1.2

In the real case $K = R$ , if $n = 0$ then $R^{0} = {0}$ is complete trivially. If $n \geq 1$ , [L2] applies directly to the $ℓ^{1}$ norm on $R^{n}$ , so $R^{n}$ is complete.

1.3

In the complex case $K = C$ , if $n = 0$ the same trivial argument applies. If $n \geq 1$ , [L3] identifies $C^{n}$ with $R^{2 n}$ by real and imaginary parts, and the proof of A chosen algebraic basis identifies a finite-dimensional normed space with a coordinate space already shows that the complex coordinate $ℓ^{1}$ norm is equivalent to a real norm on $R^{2 n}$ . By [L2], that real norm is complete, hence so is $C^{n}$ with the complex coordinate $ℓ^{1}$ norm.

L2L3

2.1

Let $(x_{m})$ be a Cauchy sequence in $X$ , and write $a_{m} : = T^{- 1} x_{m}$ . Since $T^{- 1}$ is bounded, $(a_{m})$ is Cauchy in $K^{n}$ ; by steps 1.2 and 1.3 it converges to some $a \in K^{n}$ . Since $T$ is bounded, $x_{m} = T (a_{m}) \to T (a)$ in $X$ . Thus every Cauchy sequence in $X$ converges in $X$ .

step 1.1step 1.2step 1.3L1

3.1

By [L4], step 2.1 says exactly that $X$ is Banach.

L4step 2.1∎

Remarks

The only substantive input is completeness of finite-dimensional real coordinate space. Everything else is transport of structure.

Depends on

Used by

A finite-dimensional normed subspace is closed Corollary

Dependency tree · two levels

33 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Daniel Daners, Introduction to Functional Analysis (standard reference, not scraped)
Tomasz Kochanek, Functional analysis, Lecture 1 (standard reference, not scraped)