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.
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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

1.1

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 $0 < x_{k} < 1 / p < ε$ .

givenL1L2L3

1.2

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 $c > 0$ .

givenL1L2L3L5

1.3

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 $z_{k} \geq ι (k + 1) > M$ .

givenL1L2L3L5

2.1

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 $c$ .

step 1.1step 1.2L3L5

2.2

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.

step 1.1step 1.3L1L2L3L4L5

3.1

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.

step 2.1step 2.2L4L5L6∎

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.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

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Sources

Indeterminate form (Wikipedia) (standard reference, not scraped)
Extended real number line (Wikipedia) (standard reference, not scraped)
T. Tao, Analysis I, 3rd ed., §6.1 (standard reference, not scraped)

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