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LemmaStatement: AI-adaptedProof: AI-generatedSession-authored (Fable 5 assisted)precheck passverified 2026-08-10 (gpt-5.6-terra-codex-subscription)
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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.

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Uniform convergence of real-valued functions implies pointwise convergence

Statement

Let XX be a set. If a sequence of functions fk:XRf_k:X\to\mathbb{R} converges uniformly to f:XRf:X\to\mathbb{R}, then it converges pointwise to ff (Pointwise convergence, uniform convergence, and the uniformly Cauchy condition for sequences of real-valued functions).

Facts & Assumptions

Given: A set XX, functions fk,f:XRf_k,f:X\to\mathbb{R}, and uniform convergence fkff_k\to f on XX.

[A1]

Uniform convergence means that for every real ε>0\varepsilon>0 there is NNN\in\mathbb{N} such that fk(x)f(x)<ε|f_k(x)-f(x)|<\varepsilon for every kNk\ge N and every xXx\in X (Pointwise convergence, uniform convergence, and the uniformly Cauchy condition for sequences of real-valued functions).

Proof

technique · direct
1.1

Fix xXx\in X and a real ε>0\varepsilon>0. By [A1] choose NNN\in\mathbb{N} such that fk(y)f(y)<ε|f_k(y)-f(y)|<\varepsilon for every kNk\ge N and every yXy\in X.

A1choose
2.1

In particular, fk(x)f(x)<ε|f_k(x)-f(x)|<\varepsilon for every kNk\ge N.

step 1.1
3.1

Since xx and ε\varepsilon were arbitrary, fk(x)f(x)f_k(x)\to f(x) for every xXx\in X, which is pointwise convergence.

step 2.1A1

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · next 3 levels

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Sources