CorollaryStatement: AI-adaptedProof: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-28

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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.

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Every spanning subset of a vector space contains a basis

Statement

Assume the Axiom of Choice, through Zorn's lemma gives a basis between any linearly independent set and any spanning set containing it: if $L \subseteq S \subseteq V$ with $L$ independent and $span (S) = V$ , there is a basis $B$ of $V$ with $L \subseteq B \subseteq S$ . Let $V$ be a vector space over a field $F$ (Vector space over a field) and let $S \subseteq V$ span $V$ (Linear combination of a finite list, and the span $span (S)$ as the smallest linear subspace containing $S$ ). Then there is a basis $B$ of $V$ (Basis of a vector space: a linearly independent spanning subset; and ordered basis: an injective finite list whose image is a basis) with $B \subseteq S$ .

Facts & Assumptions

Given: A field $F$ , a vector space $V$ over $F$ , and a subset $S \subseteq V$ with $span (S) = V$ .

[L1]

$\emptyset$ is linearly independent (Linear independence: a finite list $v : n \to V$ is independent when $\sum_{i < n} λ_{i} v_{i} = 0_{V}$ forces every $λ_{i} = 0_{F}$ , and a subset $S \subseteq V$ is independent when every injective finite list into $S$ is independent).

[L2]

If $L \subseteq S \subseteq V$ with $L$ linearly independent and $span (S) = V$ , there is a basis $B$ of $V$ with $L \subseteq B \subseteq S$ (Zorn's lemma gives a basis between any linearly independent set and any spanning set containing it: if $L \subseteq S \subseteq V$ with $L$ independent and $span (S) = V$ , there is a basis $B$ of $V$ with $L \subseteq B \subseteq S$ ).

Proof

technique · direct

1.1L1L2

The empty set is linearly independent and satisfies $\emptyset \subseteq S \subseteq V$ , and $S$ spans $V$ by hypothesis, so the hypotheses of the extension theorem hold with $L : = \emptyset$ .

1.2L2

The extension theorem therefore supplies a basis $B$ of $V$ with $\emptyset \subseteq B \subseteq S$ .

2.1step 1.1step 1.2∎

That $B$ is a basis of $V$ contained in $S$ , which is the claim.

Remarks

What this costs. The proof spends the Axiom of Choice, once, inside Zorn's lemma. For a finite spanning set the conclusion is also reachable without any choice principle: among the subsets of $S$ that span $V$ there is one of least size, by The well-ordering principle applied to the set of sizes of spanning subsets of $S$ , and a spanning subset of least size is a minimal spanning set, hence a basis by For $B \subseteq V$ the following are equivalent: $B$ is a basis; $B$ is a maximal linearly independent subset of $V$ ; $B$ is a minimal spanning subset of $V$ — maximality and minimality being in the inclusion order. That route is sketched here rather than carried out, because the page needs the general statement in any case and the general statement subsumes it.
The basis obtained is not unique. A spanning set typically contains many bases: the companion page exhibits a three-element spanning set of $F^{2}$ each of whose three two-element subsets is a basis. Nothing above singles one out; Zorn's lemma produces a maximal element, not a canonical one.
Spanning alone is not enough to be a basis, which is exactly what this corollary repairs: it says a spanning set can always be cut down, not that it need not be.

Every spanning subset of a vector space contains a basis

Statement

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 2 levels

Sources