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TheoremStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passverified 2026-08-09 (gpt-5.6-terra-codex-subscription)
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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

Statement

Let ARA \subseteq \mathbb{R}, let f:ARf : A \to \mathbb{R}, let cc be a limit point of AA (Limit point, isolated point, adherent point, derived set, and dense subset of R\mathbb{R}) and let LRL \in \mathbb{R}. The following are equivalent.

  1. 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).
  2. For every sequence (xk)kN(x_k)_{k \in \mathbb{N}} with xkAx_k \in A and xkcx_k \ne c for every kk, and xkcx_k \to c (Sequences of reals: bounded, eventually, frequently, tails, subsequences, Limits and Cauchy sequences of reals), the sequence (f(xk))kN(f(x_k))_{k \in \mathbb{N}} converges to LL.

The two directions do not cost the same. The implication from 1 to 2 is proved in ZF: the sequence is handed to the proof, and nothing is selected. The implication from 2 to 1, as proved below, invokes the axiom of countable choice (The Axiom of Countable Choice (ACω\mathrm{AC}_\omega)) exactly once, at step 3.2, to select one bad point from each of countably many nonempty sets. What this library does and does not claim about that cost is recorded in The sequence-to-ε\varepsilon direction of the Heine criterion uses countable choice for R\mathbb{R}, and where this library records that cost; the same asymmetry appears, for the same reason, in A point lies in the closure of ARA \subseteq \mathbb{R} iff some sequence in AA converges to it, so a subset of R\mathbb{R} is closed iff it is sequentially closed.

Because of this, the results on this page that can be proved directly from ε\varepsilon and δ\delta — the algebra of limits, order preservation, the squeeze theorem, composition — are proved that way, and not through this criterion. What the criterion is for is the transfer of sequential results to functions, and above all the negative use recorded in 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, which needs only the choice-free direction.

Facts & Assumptions

Given: A set ARA \subseteq \mathbb{R}, a function f:ARf : A \to \mathbb{R}, a limit point cc of AA and a real LL. Sequences are functions on N\mathbb{N}, and N\mathbb{N} contains 00 (Sequences of reals: bounded, eventually, frequently, tails, subsequences, The natural numbers N\mathbb{N} (von Neumann)), so the shrinking radii used below are 1/(k+1)1/(k+1) and never 1/k1/k.

[L1]

The function limit: limxcf(x)=L\lim_{x \to c} f(x) = L means that for every real ε>0\varepsilon > 0 there is a real δ>0\delta > 0 such that every xAx \in A with 0<xc<δ0 < |x - c| < \delta satisfies f(x)L<ε|f(x) - L| < \varepsilon (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).

[L2]

Sequential convergence: (yk)y(y_k) \to y means that for every rational ε>0\varepsilon > 0 there is KNK \in \mathbb{N} with yky<ε|y_k - y| < \varepsilon for all kKk \ge K (Limits and Cauchy sequences of reals, Sequences of reals: bounded, eventually, frequently, tails, subsequences). Testing instead against every positive REAL ε\varepsilon defines the same relation: every positive rational is a positive real, and below every positive real lies a positive rational (The rationals embed densely in the reals), which is the passage sanctioned in the remarks of Sequences of reals: bounded, eventually, frequently, tails, subsequences.

[L3]

Limit point: for every real δ>0\delta > 0 there is xAx \in A with 0<xc<δ0 < |x - c| < \delta (Limit point, isolated point, adherent point, derived set, and dense subset of R\mathbb{R}, The ε\varepsilon-neighbourhood and the punctured ε\varepsilon-neighbourhood of a point of R\mathbb{R}).

[L4]

Reciprocal Archimedean property: for every real ε>0\varepsilon > 0 there is a natural n1n \ge 1 with 1/n<ε1/n < \varepsilon (For every ε>0\varepsilon > 0 in a complete ordered field there is a natural n1n \ge 1 with 1/n<ε1/n < \varepsilon, Every complete ordered field is Archimedean); the canonical naturals satisfy n1R>0n \cdot 1_{\mathbb{R}} > 0 and are strictly increasing in nn (Canonical naturals are positive and strictly increasing); and 0<a<b0 < a < b gives 0<1/b<1/a0 < 1/b < 1/a (Inverses of positives are positive, and reciprocation reverses order).

[L5]

Countable choice: for every family (Xk)kN(X_k)_{k \in \mathbb{N}} of nonempty sets there is a function kxkk \mapsto x_k with xkXkx_k \in X_k for every kk (The Axiom of Countable Choice (ACω\mathrm{AC}_\omega)).

[L6]

Absolute value (Basic properties of the absolute value); and trichotomy, so the negation of u<ε|u| < \varepsilon is uε|u| \ge \varepsilon, and the negation of "for every ε\varepsilon there is δ\delta such that P" is "there is ε0\varepsilon_0 such that for every δ\delta, not P" (Ordered field).

Proof

technique · direct
1.1

Assume condition 1, let (xk)(x_k) be a sequence with xkAx_k \in A and xkcx_k \ne c for every kk and xkcx_k \to c, and let ε>0\varepsilon > 0 be an arbitrary real.

assume-hypL1L2L3
1.2

Assume condition 1 FAILS. Negating the quantifiers of [L1], there is a real ε0>0\varepsilon_0 > 0 such that for every real δ>0\delta > 0 some xAx \in A has 0<xc<δ0 < |x - c| < \delta and f(x)Lε0|f(x) - L| \ge \varepsilon_0.

assume-hypL1L6
2.1

By [L1] fix a real δ>0\delta > 0 such that every xAx \in A with 0<xc<δ0 < |x - c| < \delta satisfies f(x)L<ε|f(x) - L| < \varepsilon; and by [L2], δ\delta being a positive real, fix KNK \in \mathbb{N} with xkc<δ|x_k - c| < \delta for every kKk \ge K.

step 1.1L1L2choose
2.2

For kNk \in \mathbb{N} put Xk:={xA : 0<xc<1/(k+1)  and  f(x)Lε0}X_k := \{\, x \in A \ : \ 0 < |x - c| < 1/(k+1) \ \text{ and } \ |f(x) - L| \ge \varepsilon_0 \,\}. Each XkX_k is nonempty, since k+11k + 1 \ge 1 makes 1/(k+1)1/(k+1) a positive real and step 1.2 applies to that radius.

step 1.2L4L6
3.1

For every kKk \ge K we have xkAx_k \in A and xkcx_k \ne c, so 0<xkc<δ0 < |x_k - c| < \delta and hence f(xk)L<ε|f(x_k) - L| < \varepsilon. Since ε>0\varepsilon > 0 was an arbitrary real, f(xk)Lf(x_k) \to L; condition 1 therefore implies condition 2.

step 2.1L1L2L6
3.2

By countable choice applied to the family (Xk)kN(X_k)_{k \in \mathbb{N}}, fix a function kxkk \mapsto x_k with xkXkx_k \in X_k for every kNk \in \mathbb{N}.

step 2.2L5choose
4.1

That sequence has xkAx_k \in A and xkcx_k \ne c for every kk, and it converges to cc: given a real ε>0\varepsilon > 0, [L4] supplies a natural n1n \ge 1 with 1/n<ε1/n < \varepsilon, and every knk \ge n has k+1>n1k + 1 > n \ge 1, hence xkc<1/(k+1)<1/n<ε|x_k - c| < 1/(k+1) < 1/n < \varepsilon.

step 3.2L2L4L6
4.2

Yet (f(xk))(f(x_k)) does not converge to LL: every kk has f(xk)Lε0|f(x_k) - L| \ge \varepsilon_0, while a rational ε\varepsilon with 0<ε<ε00 < \varepsilon < \varepsilon_0 ([L2]) would require some KK with f(xk)L<ε<ε0|f(x_k) - L| < \varepsilon < \varepsilon_0 for all kKk \ge K.

step 3.2L2L6
5.1

So the failure of condition 1 produces a sequence witnessing the failure of condition 2; contrapositively, condition 2 implies condition 1, and with step 3.1 the two conditions are equivalent.

step 3.1step 4.1step 4.2

Remarks

Depends on

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