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TheoremStatement: AI-adaptedProof: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-07
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Double annihilators give norm and weak-star closures

Statement

Let K=R or C. For a normed X and a linear subspace NX, (N)=Nσ(X,X). Consequently N is weak-star closed if and only if N=(N), and weak-star dense in X if and only if N={0}. For a linear subspace MX, the primal formula is (M)=M.

Facts & Assumptions

Given: The spaces, maps, scalar field, and hypotheses in the statement above. All duals consist of linear functionals over the ambient field; evaluation has no conjugation.

[F1]

From The weak-star topology from finite evaluations, with its stated hypotheses: Let K=R or C. For a normed X with continuous dual X from def-dual-space-of-a-normed-space, the weak-star topology σ(X,X) is the initial topology of all evaluations ff(x) into K with its usual topology. At f0, a neighbourhood basis consists of U(f0;x1,,xn;ε)={fX:(ff0)(xj)<ε (1jn)}, where n is finite and ε>0. For n=0 the set is all of X. Finite intersections of inverse images of scalar open sets form the initial-topology basis; at the given point, finitely many disks can be refined using their smallest positive radius. Weak-star closure means closure in this topology, not merely sequential closure.

[F2]

From Annihilator notation and the preannihilator, with its stated hypotheses: Let K=R or C. For a normed X and arbitrary subsets MX, NX, define M={fX:f(m)=0 for all mM},N={xX:f(x)=0 for all fN}. Here X is def-dual-space-of-a-normed-space. The first notation agrees with def-continuous-annihilator-of-a-subspace on spanM, since linearity makes vanishing on M equivalent to vanishing on its span. The preannihilator lies in X, not in X. Empty sets impose no conditions: =X and =X.

[F3]

From Finite evaluations separate a functional from a dual subspace, with its stated hypotheses: Let K=R or C. Let X be normed, NX a linear subspace, and x1,,xnX with n1. Define E(f)=(f(x1),,f(xn)). If f0X satisfies E(f0)E(N), there is xspan{x1,,xn} such that g(x)=0 for all gN and f0(x)=1.

[F4]

From The annihilator detects the closure of a subspace, with its stated hypotheses: For every linear subspace MX, M=fMkerf.

Proof

1.1

If xN, evaluation at x vanishes on N. Its kernel is weak-star closed by the definition of that topology. Intersecting these kernels shows Nσ(X,X)(N).

F1F2
1.2

Let f0 lie outside the weak-star closure. Choose a basic neighbourhood U(f0;x1,,xn;ε) disjoint from N. Since 0N, this neighbourhood cannot have n=0. Its finite-coordinate map satisfies E(f0)E(N), since equality with E(g) would put that gN in U.

F1
2.1

Finite-evaluation separation supplies xN with f0(x)=1. Hence f0(N). Together with step 1.1, this proves equality.

F2F3step 1.1step 1.2
3.1

A set is closed exactly when it equals its closure, so the first equivalence follows in both directions. For density, if N=0, the equality gives closure X. Conversely if the closure is X, any xN is annihilated by all of X. The published primal formula with M=0 says X=0, so x=0.

F2F4step 2.1
4.1

The primal formula in the statement is exactly the published annihilator-closure identity, with the preannihilator notation unpacked. It also checks the extremes N=0 and N=X: their closures and double annihilators are respectively 0 and X. For X=0 these coincide.

F2F4step 2.1

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