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A null sequence times an unbounded sequence need not be null: $(k + 1)^{- 1} \cdot (k + 1) = 1$

Statement refuted

Refuted claim: if $(x_{k})$ converges to $0$ , then the product $x_{k} y_{k}$ converges to $0$ for every sequence $(y_{k})$ of reals.

The witnesses are $x_{k} : = (k + 1)^{- 1}$ and $y_{k} : = k + 1$ , where $k + 1 = σ (k)$ is read in $R$ through the canonical-natural embedding (Canonical naturals are positive and strictly increasing): $(x_{k})$ is null (The sequence $1 / (k + 1)$ converges to $0$ ) and $(y_{k})$ diverges to $+ \infty$ , but $x_{k} y_{k} = 1$ for every $k$ , so the product does not converge to $0$ : its distance from $0$ is $1$ at every index, and the rational test value $ε = 1$ never passes. The boundedness hypothesis of A null sequence times a bounded sequence is null therefore cannot be dropped; it is the only hypothesis that lemma uses beyond convergence of the null factor.

Facts & Assumptions

Given: The sequences $x_{k} : = (k + 1)^{- 1}$ and $y_{k} : = k + 1$ , where $k + 1 = σ (k)$ is the successor of $k$ read in $R$ through the canonical-natural embedding (Sequences of reals: bounded, eventually, frequently, tails, subsequences, Canonical naturals are positive and strictly increasing).

[F1]

$(x_{k})$ converges to $0$ and every term $x_{k}$ is positive (The sequence $1 / (k + 1)$ converges to $0$ ).

[F2]

Divergence to $+ \infty$ : $y_{k} \to + \infty$ means that for every real $M$ there is $K$ with $y_{k} > M$ for all $k \geq K$ (Divergence to $+ \infty$ and to $- \infty$ ).

[F3]

Archimedean property: for every real $x$ there is a natural number $n \geq 1$ with $x < n \cdot 1_{R}$ (Every complete ordered field is Archimedean).

[F4]

Canonical naturals: for every $n \geq 1$ the element $n \cdot 1_{R}$ is positive, and $n \mapsto n \cdot 1_{R}$ is strictly increasing on ${1, 2, 3, \dots}$ , hence preserves non-strict order there (Canonical naturals are positive and strictly increasing).

[F5]

Order is compatible with addition in $N$ : for all naturals $m, n, k$ one has $m \leq n ⟺ m + k \leq n + k$ (Order is compatible with addition).

[F6]

Field arithmetic: every nonzero element $u$ has a multiplicative inverse $u^{- 1}$ with $u \cdot u^{- 1} = 1$ (Field).

[F7]

The boundedness hypothesis: if $(x_{k})$ converges to $0$ and $(y_{k})$ is bounded, then $(x_{k} y_{k})$ converges to $0$ (A null sequence times a bounded sequence is null).

[F8]

Absolute value: $∣ u ∣ = u$ whenever $u \geq 0$ (Absolute value in an ordered field).

[F9]

Convergence: a sequence $(z_{k})$ of reals converges to $z$ when for every rational $ε > 0$ there is $K$ with $∣ z_{k} - z ∣ < ε$ for all $k \geq K$ (Limits and Cauchy sequences of reals).

[F10]

Strict order: for all reals $u, v$ exactly one of $u < v$ , $u = v$ , $v < u$ holds, so $u < u$ is false (Ordered field).

[F11]

$0 < 1$ in $R$ (The multiplicative identity is positive).

[F12]

The map $q \mapsto \hat{q}$ is an embedding of ordered fields, so the rational $1$ is identified with the real $1$ (The rationals embed densely in the reals).

[F13]

Boundedness: $(z_{k})$ is bounded when there is a real $M$ with $∣ z_{k} ∣ \leq M$ for every $k \in N$ (Sequences of reals: bounded, eventually, frequently, tails, subsequences).

Counterexample

technique · direct

1.1givenF2F3F4F5

$(y_{k}) = (k + 1)$ diverges to $+ \infty$ : given a real $M$ , [F3] supplies a natural $N \geq 1$ with $M < N$ , and for every $k \geq N$ one has $N + 1 \leq k + 1$ by [F5], hence $N < N + 1 \leq k + 1$ by [F4], so $y_{k} = k + 1 > M$ ; this is exactly the condition of [F2] with $K = N$ .

1.2givenF1

$(x_{k})$ converges to $0$ and every term $x_{k}$ is positive by [F1].

1.3givenF4F6algebra

For every $k$ the factor $k + 1$ is nonzero, and field arithmetic [F6] gives $x_{k} y_{k} = (k + 1)^{- 1} (k + 1) = 1$ .

2.1step 1.3F8F9F10F11F12algebra

The product sequence is therefore constantly $1$ . At the rational test value $ε : = 1$ , identified with its real image by [F12] and positive by [F11], the convergence condition of [F9] would require $∣ x_{k} y_{k} - 0 ∣ < 1$ for all large $k$ ; but $∣ x_{k} y_{k} - 0 ∣ = ∣ 1 ∣ = 1$ by [F8] and [F11], and $1 < 1$ is false by [F10]. So no threshold satisfies the test: the product does not converge to $0$ .

3.1step 1.1step 1.2step 2.1F7F13∎

The refutation is complete: $(x_{k})$ is null by step 1.2, $(y_{k})$ diverges to $+ \infty$ by step 1.1, and the product fails to converge to $0$ by step 2.1. Moreover $(y_{k})$ is not bounded in the sense of [F13]: if $∣ y_{k} ∣ \leq M$ held for every $k$ , then for an index beyond the threshold belonging to the real $M$ of step 1.1 one would have both $y_{k} > M$ and $y_{k} \leq ∣ y_{k} ∣ \leq M$ . Since [F7] is true under the additional boundedness hypothesis with the same conclusion, the hypothesis is exactly what the counterexample removes, and it cannot be dropped.

Depends on

Used by

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Dependency tree · two levels

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Sources

John K. Hunter, An Introduction to Real Analysis (complete author-hosted lecture notes, 307 pp.) (standard reference, not scraped)
Jiri Lebl, Basic Analysis I: Introduction to Real Analysis, Volume I (author-hosted complete book PDF) (standard reference, not scraped)

A null sequence times an unbounded sequence need not be null: (k+1)−1⋅(k+1)=1(k+1)−1⋅(k+1)=1