CounterexampleConstruction: Literature-sourcedVerification: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-04‡ rests on unproved material (inherited)

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

‡ Rests on 1 statement not proved in this library, by way of the results it cites. This item cites no such statement directly; it depends on results that do. The unproved premises it inherits are ‡ The Baire category theorem is four inequivalent statements over ZF. Each is recorded with a citation to the literature and is not established here, because the track that would prove it has not yet been developed in this library. Everything else in this proof is proved here.

The polynomial space admits no complete norm

Statement refuted

Refuted claim: the polynomial space over a scalar field can be made into a Banach space by some norm.

Let $K \in {R, C}$ and let $K [x]$ be the polynomial ring of The polynomial ring over a commutative ring as finitely supported coefficient sequences with convolution, regarded as a vector space over $K$ . Then no norm on $K [x]$ is complete.

Facts & Assumptions

Given: A scalar field $K \in {R, C}$ and the vector space $K [x]$ of polynomials in one indeterminate.

[L1]

A Banach space has no countably infinite Hamel basis (A Banach space has no countably infinite Hamel basis).

[L2]

A basis is a linearly independent spanning subset (Basis of a vector space: a linearly independent spanning subset; and ordered basis: an injective finite list whose image is a basis).

[L3]

The polynomial ring $K [x]$ is the set of finite sums $a_{0} + a_{1} x + \dots + a_{m} x^{m}$ (The polynomial ring over a commutative ring as finitely supported coefficient sequences with convolution).

Counterexample

technique · direct

1.1

The set ${1, x, x^{2}, \dots}$ spans $K [x]$ by [L3], because every polynomial is a finite linear combination of monomials. It is linearly independent: if $a_{0} + a_{1} x + \dots + a_{m} x^{m} = 0$ as a polynomial, then every coefficient is $0$ . Hence [L2] makes ${1, x, x^{2}, \dots}$ a countably infinite Hamel basis of $K [x]$ .

L2L3algebra

2.1

If some norm on $K [x]$ were complete, then [L1] would forbid the countably infinite Hamel basis from step 1.1. This contradiction shows that no norm on $K [x]$ is complete.

L1step 1.1assume-contradischarge-contradiction∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

24 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

Christopher Heil, A Basis Theory Primer (standard reference, not scraped)