CounterexampleConstruction: Literature-sourcedVerification: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-02

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Even polynomials on $[-1,1]$ form a unital algebra but are not dense because they do not separate $-1$ and $1$

Statement refuted

Refuted: a unital real function algebra on a compact space is dense without the point-separation hypothesis.

Facts & Assumptions

Given: $A$ is the algebra of even polynomials restricted to $[-1,1]$ .

[L1]

A unital separating real function algebra has the properties in A unital point-separating real subalgebra of $C(K,\mathbb R)$ .

Proof

technique · direct

1.1

The constants belong to $A$ , and sums and products of even polynomials are even, so $A$ is a unital algebra.

givenL1algebra

1.2

Every $p\in A$ has $p(-1)=p(1)$ ; therefore $A$ does not separate these two points.

givenL1algebra

2.1

If $p\in A$ , then $\max\{|p(-1)+1|,|p(1)-1|\}\ge1$ , so $\lVert p-\operatorname{id}\rVert_\infty\ge1$ .

step 1.2algebra

3.1

Hence the identity function is not in the closure of $A$ , and $A$ is not dense.

step 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

Stone-Weierstrass Theorem (University of Chicago) (standard reference, not scraped)

Even polynomials on [−1,1][-1,1][−1,1] form a unital algebra but are not dense because they do not separate −1-1−1 and 111