CorollaryStatement: Literature-sourcedProof: AI-generatedPipeline-generatedverified 2026-09-24 (gpt-6-sol)

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Finite-dimensional subspaces are complemented

Statement

Assume the Axiom of Choice. Every finite-dimensional linear subspace $M$ of a normed space $X$ is complemented in $X$ .

Facts & Assumptions

Given: The Axiom of Choice, A finite-dimensional subspace $M \subseteq X$ .

[A1]

AC is used through functional extension [F2] in step 1.1.

[F1]

Relative to a fixed finite basis, every coordinate functional on a finite-dimensional normed space is bounded (A chosen algebraic basis identifies a finite-dimensional normed space with a coordinate space).

[F2]

A bounded functional on a subspace extends norm-preservingly to the ambient normed space (A bounded linear functional on an arbitrary subspace extends with the same norm, without assuming the subspace is closed).

[F3]

A subspace is complemented exactly when it is the range of a bounded projection (A closed subspace is complemented exactly when it is the range of a bounded projection).

Proof

technique · direct

1.1givenF1F2choose

Choose a basis $e_{1}, \dots, e_{n}$ of $M$ , and let $ϕ_{j} : M \to K$ be its coordinate maps. By [F1]--[F2], extend each $ϕ_{j}$ to $f_{j} \in X^{*}$ .

2.1step 1.1algebra

Define $P : X \to X$ by $P (x) = \sum_{j = 1}^{n} f_{j} (x) e_{j}$ . It is bounded, has range in $M$ , and for $m = \sum a_{j} e_{j} \in M$ satisfies $P (m) = m$ .

3.1step 2.1F3∎

Thus $P^{2} = P$ and $ran P = M$ ; [F3] makes $M$ complemented.

Depends on

Dependency tree · 2 levels

Sources

Piotr Hajlasz, Functional Analysis, Theorem 10.17(a)