Alphabeta Math
CorollaryStatement: AI-adaptedProof: AI-generatedSession-authored (Fable 5 assisted)precheck passjudge pass (z-ai/glm-5.2)audited 2026-07-27
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A function has no limit at cc as soon as two sequences in A{c}A \setminus \{c\} tending to cc give different limits of the values

Statement

Let ARA \subseteq \mathbb{R}, let f:ARf : A \to \mathbb{R} and let cc be a limit point of AA (Limit point, isolated point, adherent point, derived set, and dense subset of R\mathbb{R}). Then ff has no limit at cc — that is, no LRL \in \mathbb{R} satisfies limxcf(x)=L\lim_{x \to c} f(x) = L (The ε\varepsilon-δ\delta limit limxcf(x)=L\lim_{x \to c} f(x) = L of f:ARf : A \to \mathbb{R} at a limit point cc of AA) — as soon as either of the following occurs.

  1. There are sequences (xk)(x_k) and (yk)(y_k) with all terms in A{c}A \setminus \{c\}, both converging to cc, and reals PQP \ne Q with f(xk)Pf(x_k) \to P and f(yk)Qf(y_k) \to Q (Sequences of reals: bounded, eventually, frequently, tails, subsequences, Limits and Cauchy sequences of reals).
  2. There is a sequence (xk)(x_k) with all terms in A{c}A \setminus \{c\}, converging to cc, for which (f(xk))(f(x_k)) does not converge.

Only the choice-free half of the Heine criterion is used. The proof runs the implication from condition 1 to condition 2 of Heine criterion: limxcf(x)=L\lim_{x \to c} f(x) = L iff f(xk)Lf(x_k) \to L for every sequence in A{c}A \setminus \{c\} converging to cc, which is a theorem of ZF; no sequence is constructed here, both being supplied by the hypothesis. So this corollary, the workhorse for showing that a limit fails to exist, costs no choice principle at all.

Facts & Assumptions

[L1]

Heine criterion, the direction from the ε\varepsilon-δ\delta limit to sequences: if limxcf(x)=L\lim_{x \to c} f(x) = L then f(zk)Lf(z_k) \to L for every sequence (zk)(z_k) with all terms in A{c}A \setminus \{c\} converging to cc (Heine criterion: limxcf(x)=L\lim_{x \to c} f(x) = L iff f(xk)Lf(x_k) \to L for every sequence in A{c}A \setminus \{c\} converging to cc). That direction is proved without any choice principle.

[L2]

A sequence of reals has at most one limit, so two limits of the same sequence are equal, and a sequence with a limit converges (A sequence has at most one limit, Limits and Cauchy sequences of reals, Sequences of reals: bounded, eventually, frequently, tails, subsequences).

Proof

technique · contrapositive
1.1

Each of the two claims has the form "hypothesis \Rightarrow ff has no limit at cc"; we prove the contrapositive of each, namely that if some LRL \in \mathbb{R} satisfies limxcf(x)=L\lim_{x \to c} f(x) = L then neither hypothesis can hold.

contrapositive-reduce
1.2

Assume there is LRL \in \mathbb{R} with limxcf(x)=L\lim_{x \to c} f(x) = L.

assume-hyp
2.1

Let (zk)(z_k) be an arbitrary sequence with all terms in A{c}A \setminus \{c\} converging to cc. By [L1], (f(zk))(f(z_k)) converges, with limit LL.

step 1.2L1
3.1

Under hypothesis 1 this applies to (xk)(x_k) and to (yk)(y_k): f(xk)Pf(x_k) \to P and f(xk)Lf(x_k) \to L give P=LP = L by [L2], and likewise Q=LQ = L, so P=QP = Q; hypothesis 1, which asserts PQP \ne Q, therefore fails.

step 2.1L2
3.2

Under hypothesis 2 it applies to (xk)(x_k) and gives that (f(xk))(f(x_k)) converges; hypothesis 2, which asserts that it does not, therefore fails.

step 2.1L2
4.1

So the existence of a limit of ff at cc excludes both hypotheses; contrapositively, either hypothesis excludes the existence of a limit of ff at cc.

step 3.1step 3.2discharge-contrapositive

Remarks

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Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 63 results over 14 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.

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