LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-13

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Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
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Assuming choice, if $v \notin U \leq V$ , some $f \in V^{*}$ vanishes on $U$ and satisfies $f (v) = 1$

Statement

Assume the axiom of choice. If $U$ is a linear subspace of $V$ and $v \in V ∖ U$ , then there exists $f \in V^{*}$ such that $f ∣_{U} = 0$ and $f (v) = 1$ .

Facts & Assumptions

Given: The axiom of choice, a subspace $U \leq V$ , and $v \notin U$ .

[L1]

Assuming choice, every linearly independent subset of a vector space extends to a basis (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 $L \subseteq B \subseteq S$ ).

[L2]

A subspace is closed under finite linear combinations (Linear subspace of a vector space).

[L3]

Elements of $V^{*}$ are linear maps $V \to F$ (Linear functionals and the algebraic dual $V^{*} = L (V, F)$ ).

Proof

technique · basis extension

1.1L1L2givenchoose

By [L1], choose a basis $C$ of $U$ . The set $C \cup {v}$ is linearly independent: a relation with nonzero coefficient of $v$ would put $v$ in the span of $C$ , which is $U$ by [L2].

2.1step 1.1L1L3choose

Extend $C \cup {v}$ by [L1] to a basis $B$ of $V$ . Prescribe $f (v) = 1$ and $f (b) = 0$ for every $b \in B ∖ {v}$ , then extend by the unique finite basis expansion. This defines a linear functional $f \in V^{*}$ .

3.1step 2.1L2∎

Every element of $U$ is a linear combination of elements of $C$ , so $f ∣_{U} = 0$ , while the prescription gives $f (v) = 1$ .

Depends on

Dependency tree · 2 levels

Sources

K. Conrad, Infinite-Dimensional Dual Spaces

Assuming choice, if v∉U≤Vv∈/U≤V, some f∈V∗f∈V∗ vanishes on UU and satisfies f(v)=1f(v)=1