Alphabeta Math
TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-13
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.

Assuming choice, ∘(U∘)=U; in finite dimension, dim⁡U∘=dim⁡V−dim⁡U

Statement

Assume the axiom of choice. For every subspace U≤V,

∘(U∘)=U.

If V is finite-dimensional, then

dim⁡U∘=dim⁡V−dim⁡U.

Facts & Assumptions

Given: The axiom of choice, an F-vector space V, and a subspace U≤V.

[L1]

The annihilator U∘ consists of the functionals vanishing on U, and ∘S consists of vectors annihilated by every member of S (The annihilator U∘≤V∗ of U≤V and the preannihilator ∘S≤V of S≤V∗).

[L2]

If v∉U, some f∈V∗ vanishes on U and has f(v)=1 (Assuming choice, if v∉U≤V, some f∈V∗ vanishes on U and satisfies f(v)=1).

[L3]

The dual of a finite-dimensional space has the same dimension (The dual family of a finite basis is a basis of the dual space, with the same dimension).

[L4]

Rank-nullity gives dim⁡X=dim⁡ker⁡R+dim⁡im⁡R for a linear map with finite-dimensional domain X (Rank-nullity: dim⁡FV=nullity⁡T+rank⁡T).

[L5]

In finite dimension, a basis of a subspace extends without Choice to a basis of the ambient space (If dim⁡FV=n and U is a linear subspace of V, then U is finite-dimensional, dim⁡FU≤n, and dim⁡FU=n if and only if U=V, clause 3).

Proof

technique · direct
1.1L1L2

Every u∈U is killed by every f∈U∘, so [L1] gives U⊆∘(U∘). If v∉U, [L2] gives f∈U∘ with f(v)=1, so v∉∘(U∘). Hence ∘(U∘)=U.

1.2L1L5choose

Now suppose V is finite-dimensional and consider restriction R:V∗→U∗, R(f)=f∣U. Its kernel is U∘ by [L1]. It is surjective: extend a basis of U to one of V by [L5], prescribe any given functional's values on the former basis, and prescribe zero on the added vectors.

2.1step 1.2L3L4algebra

Rank-nullity and surjectivity give dim⁡V∗=dim⁡U∘+dim⁡U∗. Applying [L3] to V and U, then rearranging, gives dim⁡U∘=dim⁡V−dim⁡U.

3.1step 1.1step 2.1∎

Step 1.1 proves the double-annihilator identity in arbitrary dimension under Choice, and step 2.1 proves the finite-dimensional formula, including U=0 and U=V.

Depends on

Dependency tree · 2 levels

Sources