Alphabeta Math
CounterexampleConstruction: Literature-sourcedVerification: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-04
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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∈{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∈{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).

[L3]

The polynomial ring K[x] is the set of finite sums a0+a1x+⋯+amxm (The polynomial ring over a commutative ring as finitely supported coefficient sequences with convolution).

Counterexample

technique · direct
1.1L2L3algebra

The set {1,x,x2,… } spans K[x] by [L3], because every polynomial is a finite linear combination of monomials. It is linearly independent: if a0+a1x+⋯+amxm=0 as a polynomial, then every coefficient is 0. Hence [L2] makes {1,x,x2,… } a countably infinite Hamel basis of K[x].

2.1L1step 1.1assume-contradischarge-contradiction∎

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.

Depends on

Used by

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Dependency tree · two levels

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Sources