Alphabeta Math
CounterexampleConstruction: Literature-sourcedVerification: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-02
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Polynomials vanishing at zero separate points of [0,1] but are not dense in C([0,1])

Statement refuted

Refuted: point separation alone makes a real function algebra dense in C([0,1]).

Facts & Assumptions

Given: A={p∈R[x]:p(0)=0}, restricted to [0,1].

[L1]

Point separation and the unital condition are distinct requirements in A unital point-separating real subalgebra of C(K,R).

Proof

technique · direct
1.1

The set A is an algebra, and the function x↦x in A separates every distinct pair of points of [0,1].

givenL1algebra
1.2

Every p∈A vanishes at 0, so A contains no constant-one function and is not unital.

givenL1algebra
2.1

For every p∈A, ∥p−1∥∞≥∣p(0)−1∣=1. Thus 1 is not in the uniform closure of A.

step 1.2algebra
3.1

Therefore A separates points but is not dense.

step 1.1step 2.1algebra∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 17 results over 9 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources