Alphabeta Math
CounterexampleConstruction: Literature-sourcedVerification: AI-adaptedprecheck passaudited 2026-08-02
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  • 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.
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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
1.1givenL1algebra

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

1.2givenL1algebra

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

2.1step 1.2algebra

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

3.1step 1.1step 2.1algebra∎

Therefore A separates points but is not dense.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

6 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