LemmaStatement: AI-adaptedProof: AI-adaptedPipeline-generatedjudge pass (gpt-6-luna)

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.

Sequence-limit criterion with explicit choice use

Statement

Let $A \subseteq R$ , let $c \in R$ be a limit point of $A$ , let $f : A \to R$ and $L \in R$ . Assume the indexed axiom of countable choice ${A C}_{ω}$ of The Axiom of Countable Choice ( ${A C}_{ω}$ ). Then the epsilon-delta condition $\lim_{x \to c} f (x) = L$ of The $ε$ - $δ$ limit $\lim_{x \to c} f (x) = L$ of $f : A \to R$ at a limit point $c$ of $A$ is equivalent to the following: for every sequence $(x_{n})_{n \in N}$ with $x_{n} \in A ∖ {c}$ and $x_{n} \to c$ , one has $f (x_{n}) \to L$ . Sequence and sequence convergence mean Sequences of reals: bounded, eventually, frequently, tails, subsequences and Limits and Cauchy sequences of reals. The forward direction uses no choice; the reverse direction uses only the stated indexed ${A C}_{ω}$ , not arbitrary-index choice.

Facts & Assumptions

Given: $A, c, f, L$ and ${A C}_{ω}$ as stated; all inequalities use the real ordered field.

[F1]

The $ε$ - $δ$ limit $\lim_{x \to c} f (x) = L$ of $f : A \to R$ at a limit point $c$ of $A$ , Limit point, isolated point, adherent point, derived set, and dense subset of $R$ and The $ε$ -neighbourhood and the punctured $ε$ -neighbourhood of a point of $R$ specify the punctured real epsilon-delta neighborhoods, with $c$ a limit point, without requiring $c \in A$ .

[F2]

Sequences of reals: bounded, eventually, frequently, tails, subsequences and Limits and Cauchy sequences of reals define a real sequence and convergence using positive rational tolerances. By The rationals embed densely in the reals, for every real $δ > 0$ there is a rational $q$ with $0 < q < δ$ ; hence convergence also gives eventual error below every positive real tolerance.

[F3]

For every $ε > 0$ in a complete ordered field there is a natural $n \geq 1$ with $1 / n < ε$ , Every complete ordered field is Archimedean, Inverses of positives are positive, and reciprocation reverses order and Canonical naturals are positive and strictly increasing give $1 / (n + 1) \to 0$ and eventual domination by any positive real bound. The real numbers, Ordered field, Order on the reals and Basic properties of the absolute value supply the ordered-field and absolute-value arithmetic.

[A1]

The Axiom of Countable Choice ( ${A C}_{ω}$ ): for every specified sequence of nonempty sets $(X_{n})$ , there is a function $x$ on $N$ with $x (n) \in X_{n}$ . Only this Definition is assumed; its Remarks implication from full choice is not a premise.

Proof

1.1givenF1F2

Assume the epsilon-delta condition and fix any stated sequence $x_{n} \to c$ . For positive rational $ε$ , [F1] gives a real $δ > 0$ such that $0 < ∣ x - c ∣ < δ$ implies $∣ f (x) - L ∣ < ε$ . By [F2], eventually $∣ x_{n} - c ∣ < δ$ , and $x_{n} \neq c$ gives strict positivity. Thus eventually $∣ f (x_{n}) - L ∣ < ε$ , proving convergence in [F2]. No family of witnesses is selected: the sequence and one tolerance are fixed.

1.2givenF1F3construct

Conversely suppose the sequential condition holds but the epsilon-delta condition fails. Its negation yields one real $ε_{0} > 0$ such that for every $δ > 0$ there exists $x \in A$ with $0 < ∣ x - c ∣ < δ$ and $∣ f (x) - L ∣ \geq ε_{0}$ . For each $n \in N$ , define $X_{n} = {x \in A : 0 < ∣ x - c ∣ < 1 / (n + 1), ∣ f (x) - L ∣ \geq ε_{0}}$ . Positivity of $n + 1$ and the negated condition make every $X_{n}$ nonempty. These sets are a specified indexed family of subsets of the given set $A$ .

2.1A1F2F3step 1.2

Apply [A1] exactly once to obtain $x_{n} \in X_{n}$ for every $n$ . Then $x_{n} \in A ∖ {c}$ and $∣ x_{n} - c ∣ < 1 / (n + 1)$ . For every positive rational $q$ , [F3] gives a threshold beyond which $1 / (n + 1) < q$ , so $x_{n} \to c$ by [F2]. The sequential hypothesis therefore implies $f (x_{n}) \to L$ .

3.1F1F2step 1.2step 2.1∎

Choose one rational $q$ with $0 < q < ε_{0}$ by [F2]. Convergence in step 2.1 makes $∣ f (x_{n}) - L ∣ < q < ε_{0}$ eventually, contradicting membership in $X_{n}$ . Hence the epsilon-delta condition holds. The only countable selection was step 2.1; the forward implication and the two single rational choices require no choice axiom. The limit-point hypothesis and exclusion of $c$ are retained throughout.

Sequence-limit criterion with explicit choice use

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources