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

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.

Polynomials are uniformly dense in $C([0,1],\mathbb R)$

Statement

For every $f\in C([0,1],\mathbb R)$ and $\varepsilon>0$ , there is a polynomial $p$ with $\sup_{x\in[0,1]}|p(x)-f(x)|<\varepsilon$ .

Facts & Assumptions

Given: $f\in C([0,1],\mathbb R)$ and $\varepsilon>0$ .

[L1]

The Bernstein polynomials of $f$ converge uniformly to $f$ (Bernstein polynomials converge uniformly to every continuous function on $[0,1]$ ).

Proof

technique · direct

1.1

Choose $n$ with $\sup_x|B_n(f)(x)-f(x)|<\varepsilon$ .

L1choose

2.1

The finite defining sum for $B_n(f)$ is a polynomial in $x$ , so $p:=B_n(f)$ has the required property.

step 1.1algebra∎

Depends on

Bernstein polynomials converge uniformly to every continuous function on $[0,1]$

Used by

Polynomials are uniformly dense in C([a,b],ℝ) for every closed interval Corollary

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 22 results over 11 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

Bernstein polynomial (Encyclopedia of Mathematics) (standard reference, not scraped)

Polynomials are uniformly dense in C([0,1],R)C([0,1],\mathbb R)C([0,1],R)