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 $y$ .

The dual space separates points of a normed space

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources