Alphabeta MathHow 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.
Assuming choice, if , some vanishes on and satisfies
Statement
Assume the axiom of choice. If is a linear subspace of and , then there exists such that and .
Facts & Assumptions
Given: The axiom of choice, a subspace , and .
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 with independent and , there is a basis of with ).
A subspace is closed under finite linear combinations (Linear subspace of a vector space).
Elements of are linear maps (Linear functionals and the algebraic dual ).
Proof
By [L1], choose a basis of . The set is linearly independent: a relation with nonzero coefficient of would put in the span of , which is by [L2].
Extend by [L1] to a basis of . Prescribe and for every , then extend by the unique finite basis expansion. This defines a linear functional .
Every element of is a linear combination of elements of , so , while the prescription gives .
Depends on
- Linear functionals and the algebraic dual $V^*=\mathcal L(V,F)$
- 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 $\operatorname{span}(S) = V$, there is a basis $B$ of $V$ with $L \subseteq B \subseteq S$
- Linear subspace of a vector space
Used by
- In infinite dimension, distinct subspaces of V^* can have the same preannihilator Counterexample
- For the polynomial space F[x], the canonical map to the algebraic double dual is injective but not surjective Example
- Assuming choice, ^∘(U^∘)=U; in finite dimension, dim U^∘=dim V-dim U Theorem
- Assuming choice, J_V:V→ V^** is surjective if and only if V is finite-dimensional Theorem
- Assuming choice, the canonical map J_V:V→ V^** is linear and injective Theorem
Dependency tree · two levels
22 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
- K. Conrad, Infinite-Dimensional Dual Spaces (standard reference, not scraped)