Alphabeta Math
LemmaStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedjudge pass (gpt-6-luna)
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A separable Banach space has a countable norming family

Statement

Assume HB and countable choice. Let B be a real separable Banach space. There is a sequence (ℓj)j≥1⊆B∗ with ∥ℓj∥=1 such that ∥x∥=sup⁡j≥1∣ℓj(x)∣(x∈B). If B={0}, the conclusion is interpreted using the zero functional and the supremum is 0.

Facts & Assumptions

[F1]

Separability supplies a countable dense subset of B (Separability: the existence of an at most countable dense subset).

[F2]

For each nonzero x∈B, Hahn–Banach supplies ℓ∈B∗ with ∥ℓ∥=1 and ℓ(x)=∥x∥ (The real dominated-extension principle as an additional hypothesis over ZF, Relative dual norming, point separation, and recovery of the norm).

[A1]

Countable choice selects one such functional for each element of a countable dense subset (The Axiom of Countable Choice (ACω)).

Proof

Given: HB, countable choice, and a real separable Banach space B.

1.1givenF1F2A1choose

If B={0}, take the constant zero sequence. Otherwise let D be a countable dense subset of B. The normalized set {d/∥d∥:d∈D, d≠0} is countable and dense in the unit sphere: for ∥x∥=1 and d→x, eventually d≠0 and ∥d/∥d∥−x∥≤2∥d−x∥. Enumerate it as (xj). By [F2] and [A1], choose ℓj∈B∗ with ∥ℓj∥=1 and ℓj(xj)=1 for each j.

2.1givenstep 1.1algebra∎

For every x∈B and every j, ∣ℓj(x)∣≤∥x∥, so the supremum is at most ∥x∥. If x≠0, choose unit vectors xjk→x/∥x∥. Then ℓjk(x/∥x∥)≥1−∥xjk−x/∥x∥∥→1, so sup⁡j∣ℓj(x)∣≥∥x∥. The zero vector gives equality directly.

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Sources