CounterexampleConstruction: AI-generatedVerification: AI-generatedprecheck passverified 2026-08-08 (gpt-5.6-terra-codex-subscription)

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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.
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Null times divergent has no rule: $x_{k} = 1 / k$ with $y_{k} = c k$ gives product limit $c$ , and with $y_{k} = k^{2}$ gives divergence

Statement refuted

That the product $0 \cdot (+ \infty)$ , left undefined by The extended real line $\overline{R} = R \cup {- \infty, + \infty}$ , its order, and the arithmetic that is left undefined, could be given a value compatible with limits: that there is $v \in \overline{R}$ such that for all sequences of reals with $x_{k} \to 0$ (Limits and Cauchy sequences of reals) and $y_{k} \to + \infty$ (Divergence to $+ \infty$ and to $- \infty$ ) the products $x_{k} y_{k}$ have the single limiting behaviour named by $v$ .

Equivalently: that knowing a factor is null and the other diverges to $+ \infty$ determines anything at all about the product. It does not, and the two undefined entries in the arithmetic of $\overline{R}$ are undefined for exactly this reason.

Facts & Assumptions

Given: The canonical naturals $ι (n) = n \cdot 1_{R}$ ; the sequence $x_{k} : = 1 / ι (k + 1)$ ; for a real $c > 0$ the sequence $y_{k}^{(c)} : = c ι (k + 1)$ ; and the sequence $z_{k} : = ι (k + 1) ι (k + 1)$ .

[L1]

Canonical naturals: $ι (n) > 0$ and invertible for $n \geq 1$ , $ι$ is strictly increasing, and $ι (n) \geq 1$ for $n \geq 1$ (Canonical naturals are positive and strictly increasing, Order on the natural numbers, $\leq$ is a linear order on $N$ ).

[L2]

Archimedean facts: for every real $η > 0$ there is a natural $p \geq 1$ with $1 / p < η$ , and for every real $M$ there is a natural $p \geq 1$ with $M < ι (p)$ ; and $0 < u < v$ gives $0 < 1 / v < 1 / u$ (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).

[L3]

Convergence to a real and divergence to $+ \infty$ ; to establish convergence it suffices to produce a threshold for every real $ε > 0$ ; a constant sequence converges to its value (Limits and Cauchy sequences of reals, Sequences of reals: bounded, eventually, frequently, tails, subsequences, Divergence to $+ \infty$ and to $- \infty$ ).

[L4]

A sequence diverging to $+ \infty$ is unbounded and therefore does not converge to any real (Divergence to $+ \infty$ and to $- \infty$ , Every convergent sequence is bounded); a limit, when it exists, is unique (A sequence has at most one limit).

[L5]

Order and field arithmetic: multiplying an inequality by a positive element preserves it; $0 < 1 < 2$ and $1 \neq 2$ ; $u \cdot (1 / u) = 1$ for $u \neq 0$ ; and the algebra of limits (Sign rules for products and monotonicity of multiplication, Order is preserved by adding a constant and by adding inequalities, The multiplicative identity is positive, Algebra of limits: sums, scalar multiples, products and quotients, Integer powers $a^{m}$ , Ordered field, Complete ordered field (least-upper-bound property)).

[L6]

The product $0 \cdot (\pm \infty)$ is left undefined in $\overline{R}$ (The extended real line $\overline{R} = R \cup {- \infty, + \infty}$ , its order, and the arithmetic that is left undefined).

Counterexample

technique · direct

1.1givenL1L2L3

The sequence $x_{k} = 1 / ι (k + 1)$ is well defined, positive, and converges to $0$ : given a real $ε > 0$ , take a natural $p \geq 1$ with $1 / p < ε$ ; for $k \geq p$ we have $ι (k + 1) > ι (p) > 0$ , hence $0 < x_{k} < 1 / p < ε$ .

1.2givenL1L2L3L5

For every real $c > 0$ the sequence $y_{k}^{(c)} = c ι (k + 1)$ diverges to $+ \infty$ : given a real $M$ , the quotient $M / c$ is real, so there is a natural $p \geq 1$ with $M / c < ι (p)$ , and for $k \geq p$ we get $ι (k + 1) > ι (p) > M / c$ , hence $y_{k}^{(c)} = c ι (k + 1) > M$ after multiplying by $c > 0$ .

1.3givenL1L2L3L5

The sequence $z_{k} = ι (k + 1) ι (k + 1)$ diverges to $+ \infty$ : given a real $M$ , take a natural $p \geq 1$ with $M < ι (p)$ ; for $k \geq p$ we have $ι (k + 1) \geq 1$ and $ι (k + 1) > ι (p) > M$ , so $z_{k} \geq ι (k + 1) > M$ .

2.1step 1.1step 1.2L3L5

For every real $c > 0$ the product sequence is constant: $x_{k} y_{k}^{(c)} = (1 / ι (k + 1)) c ι (k + 1) = c$ for every $k$ , so it converges to $c$ .

2.2step 1.1step 1.3L1L2L3L4L5

The product with $(z_{k})$ is $x_{k} z_{k} = (1 / ι (k + 1)) ι (k + 1) ι (k + 1) = ι (k + 1)$ , which diverges to $+ \infty$ by the argument of step 1.3 with the single factor, and therefore converges to no real number.

3.1step 2.1step 2.2L4L5L6∎

Now take the three pairs $(x, y^{(1)})$ , $(x, y^{(2)})$ and $(x, z)$ . In each, the first sequence is null and the second diverges to $+ \infty$ , so each pair satisfies the hypotheses of the refuted claim; but the three products converge to $1$ , converge to $2$ , and converge to $+ \infty$ in the extended sense. Since $1 \neq 2$ and limits are unique, no single $v \in \overline{R}$ describes all three, and the claim is false.

Remarks

This is why the entry is blank in the table. The extended real line $\overline{R} = R \cup {- \infty, + \infty}$ , its order, and the arithmetic that is left undefined leaves $0 \cdot (\pm \infty)$ undefined not out of caution but because any value assigned to it would make some instance of a product rule false, and the three pairs above already realise three different behaviours.
The same phenomenon rules out $(+ \infty) + (- \infty)$ . Taking $a_{k} = ι (k)$ and $b_{k} = - ι (k)$ gives a sum that is constantly $0$ , while $a_{k} = ι (k)$ and $b_{k} = - ι (k) - ι (k)$ gives a sum diverging to $- \infty$ ; both pairs have $a_{k} \to + \infty$ and $b_{k} \to - \infty$ .
Measure theory's convention is not a counterexample to this. Texts that set $0 \cdot \infty : = 0$ are fixing the value of a formula in a context where the factor $0$ is the measure of a null set, not asserting a limit rule; the distinction is spelled out in Which extended-real operations this library leaves undefined, and where each $lim sup$ statement needs the hypothesis.
Index range. The classical statement writes $x_{k} = 1 / k$ and $y_{k} = c k$ , which requires $k \geq 1$ . Written on $N$ , which contains $0$ (Sequences of reals: bounded, eventually, frequently, tails, subsequences), the same sequences are $1 / (k + 1)$ and $c (k + 1)$ , as above.

Null times divergent has no rule: $x_{k} = 1 / k$ with $y_{k} = c k$ gives product limit $c$ , and with $y_{k} = k^{2}$ gives divergence

Statement refuted

Facts & Assumptions

Counterexample

Remarks

Depends on

Dependency tree · 2 levels

Sources

Null times divergent has no rule: xk=1/kxk​=1/k with yk=ckyk​=ck gives product limit cc, and with yk=k2yk​=k2 gives divergence

Statement refuted

Facts & Assumptions

Counterexample

Remarks

Depends on

Dependency tree · 2 levels

Sources

Null times divergent has no rule: $x_{k} = 1 / k$ with $y_{k} = c k$ gives product limit $c$ , and with $y_{k} = k^{2}$ gives divergence