CounterexampleConstruction: Literature-sourcedVerification: AI-adaptedprecheck passaudited 2026-08-13

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In infinite dimension, distinct subspaces of $V^{*}$ can have the same preannihilator

Statement refuted

Refuted claim: Distinct subspaces of an algebraic dual always have distinct preannihilators.

Facts & Assumptions

Given: The axiom of choice, an infinite-dimensional vector space $V$ , an infinite Hamel basis $B$ , and $Φ = span {b^{*} : b \in B} \leq V^{*}$ .

[L1]

The preannihilator $^{\circ} S$ consists of vectors killed by every functional in $S$ (The annihilator $U^{\circ} \leq V^{*}$ of $U \leq V$ and the preannihilator $^{\circ} S \leq V$ of $S \leq V^{*}$ ).

[L2]

For an infinite Hamel basis, the span $Φ$ of its coordinate functionals is a proper subspace of $V^{*}$ (For an infinite Hamel basis, its dual family is linearly independent but does not span the algebraic dual).

[L3]

Assuming choice, every nonzero vector is detected by some linear functional (Assuming choice, if $v \notin U \leq V$ , some $f \in V^{*}$ vanishes on $U$ and satisfies $f (v) = 1$ , with the zero subspace).

Counterexample

technique · explicit subspaces

1.1

If $v \neq 0$ , some coordinate of its finite basis expansion is nonzero, so the corresponding $b^{*} \in Φ$ does not kill $v$ . Hence [L1] gives $^{\circ} Φ = {0}$ .

L1givenalgebra

1.2

By [L3], every nonzero $v$ is detected by some member of $V^{*}$ , so $^{\circ} (V^{*}) = {0}$ .

L1L3

2.1

Yet [L2] gives $Φ \neq V^{*}$ , while steps 1.1 and 1.2 give equal preannihilators. These are the required distinct subspaces.

step 1.1step 1.2L2∎

Depends on

Used by

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Dependency tree · two levels

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Sources

K. Conrad, Infinite-Dimensional Dual Spaces (standard reference, not scraped)

In infinite dimension, distinct subspaces of V∗V∗ can have the same preannihilator