CorollaryStatement: Literature-sourcedProof: AI-generatedprecheck passverified 2026-09-24 (gpt-6-sol)

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The dual space separates points of a normed space

Statement

Assume the Axiom of Choice. Let $X$ be a normed space over $R$ or $C$ . If $x, y \in X$ with $x \neq y$ , then there exists $f \in X^{*}$ such that $f (x) \neq f (y)$ .

Facts & Assumptions

Given: The Axiom of Choice, A normed space $X$ over $R$ or $C$ and two vectors $x, y \in X$ with $x \neq y$ .

[A1]

AC is used through the norming functional [L1] in step 2.1.

[L1]

Every nonzero vector admits a norming functional (Every nonzero vector has a norming functional).

Proof

technique · direct

1.1given

Since $x \neq y$ , the vector $x - y$ is nonzero.

2.1L1step 1.1

By [L1], there exists $f \in X^{*}$ with $f (x - y) = ∥ x - y ∥ .$ In particular $f (x - y) \neq 0$ .

3.1step 2.1algebra∎

If $f (x) = f (y)$ , then linearity would give $f (x - y) = 0$ , contradicting step 2.1. Therefore $f (x) \neq f (y)$ , so the dual space separates $x$ and $y$ .

Depends on

Used by

Banach-valued Cauchy integral vanishes Lemma
Atkinson Theorem
Compact operator sends weakly convergent sequences to norm convergent sequences Theorem
RNP and almost-everywhere differentiability of Lipschitz curves Theorem
Spectrum is nonempty compact and norm bounded Theorem

Dependency tree · two levels

6 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

Gerald Teschl, Topics in Real and Functional Analysis, Corollary 4.16 (standard reference, not scraped)
Daniel Daners, Introduction to Functional Analysis, Remark 26.6 (standard reference, not scraped)