TheoremStatement: Literature-sourcedProof: AI-generatedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-06

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c_0 is not complemented in ell-infinity

Statement

Assuming ${A C}_{ω}$ , the closed subspace $c_{0}$ is not complemented in $ℓ^{\infty}$ .

Facts & Assumptions

Given: The quotient map $π : ℓ^{\infty} \to ℓ^{\infty} / c_{0}$ .

[F1]

Under ${A C}_{ω}$ , $(ℓ^{\infty} / c_{0})^{*}$ has no countable separating family (The quotient ell-infinity/c_0 has no countable separating family).

Proof

technique · contradiction

1.1assume-contragivenconstruct

Suppose a bounded projection $P : ℓ^{\infty} \to c_{0}$ exists. For each coordinate $n$ , define $φ_{n} (π x) = x_{n} - (P x)_{n}$ . This is well defined because $P z = z$ for $z \in c_{0}$ , and it is a bounded functional on the quotient.

2.1step 1.1given

If $φ_{n} (π x) = 0$ for every $n$ , then $(I - P) x = 0$ coordinatewise, so $x = P x \in c_{0}$ and $π x = 0$ . Thus $(φ_{n})$ is a countable separating family.

3.1step 2.1F1discharge-contradiction∎

This contradicts [F1]. Hence no such projection exists, and $c_{0}$ is not complemented.

Depends on

Used by

A closed uncomplemented subspace Example

Dependency tree · two levels

18 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

Piotr Hajlasz, Functional Analysis, Theorem 10.19 (standard reference, not scraped)