False statementConstruction: AI-adaptedVerification: AI-adaptedprecheck passverified 2026-08-05 (gpt-5.6-sol-codex-subscription)

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FALSE: limits preserve strict inequalities

Statement

False claim: if $(x_{k})$ and $(y_{k})$ are convergent sequences of reals (Sequences of reals: bounded, eventually, frequently, tails, subsequences, Limits and Cauchy sequences of reals) with $x_{k} < y_{k}$ for every $k \in N$ , then

$\lim_{k} x_{k} < \lim_{k} y_{k} .$

The correct statement replaces both strict inequalities by non-strict ones and is Limits preserve non-strict inequalities. The claim above is refuted by $x_{k} = 0$ and $y_{k} = 1 / (k + 1)$ , whose limits are both $0$ .

Facts & Assumptions

Given: The constant sequence $x_{k} : = 0$ and the sequence $y_{k} : = ((k + 1) \cdot 1_{R})^{- 1}$ , where $n \cdot 1_{R}$ denotes the canonical natural of $R$ (Canonical naturals are positive and strictly increasing, Sequences of reals: bounded, eventually, frequently, tails, subsequences).

[L1]

$(x_{k})$ converges to $x$ when for every rational $ε > 0$ there is $K \in N$ with $∣ x_{k} - x ∣ < \hat{ε}$ for all $k \geq K$ (Limits and Cauchy sequences of reals); a sequence of reals is a function $N \to R$ (Sequences of reals: bounded, eventually, frequently, tails, subsequences), so a constant sequence converges to its value, $∣ x - x ∣ = 0 < \hat{ε}$ holding at every index.

[L2]

Archimedean property: for every $z \in R$ there is a natural $N \geq 1$ with $z < N \cdot 1_{R}$ (Every complete ordered field is Archimedean).

[L3]

Canonical naturals: $n \cdot 1_{R} > 0$ for every $n \geq 1$ , and $n \mapsto n \cdot 1_{R}$ is strictly increasing on ${1, 2, 3, \dots}$ (Canonical naturals are positive and strictly increasing).

[L4]

Inverses and order: $a > 0$ implies $a^{- 1} > 0$ ; $0 < a < b$ implies $0 < b^{- 1} < a^{- 1}$ ; and $(u^{- 1})^{- 1} = u$ for $u \neq 0$ (Inverses of positives are positive, and reciprocation reverses order, Field).

[L5]

Absolute value: $∣ u ∣ = u$ when $u \geq 0$ , and $∣ u - 0 ∣ = ∣ u ∣$ (Basic properties of the absolute value, Order on the reals).

[L6]

Order arithmetic: transitivity and trichotomy in $R$ (Complete ordered field (least-upper-bound property), Ordered field). On $N$ , $m < n$ if and only if $σ (m) \leq n$ , so $σ (k) = k + 1$ is the immediate successor of $k$ (Discreteness: $σ (n)$ is the immediate successor); transitivity of the linear order therefore gives $k \geq N \Rightarrow k + 1 > N$ ( $\leq$ is a linear order on $N$ ).

[L7]

If sequences of reals $(x_{k})$ and $(y_{k})$ converge to $x$ and $y$ and $x_{k} \leq y_{k}$ eventually, then $x \leq y$ (Limits preserve non-strict inequalities).

[L8]

A sequence of reals has at most one limit (A sequence has at most one limit), so the symbols $\lim_{k} x_{k}$ and $\lim_{k} y_{k}$ appearing in the false claim and below denote.

Refutation

technique · direct

1.1

For every $k$ the canonical natural $(k + 1) \cdot 1_{R}$ is positive by [L3], hence invertible with positive inverse by [L4]; so $y_{k} > 0 = x_{k}$ , that is $x_{k} < y_{k}$ for every $k \in N$ .

L3L4

1.2

The constant sequence $(x_{k}) = (0)$ converges to $0$ .

2.1

The sequence $(y_{k})$ converges to $0$ . Let $ε > 0$ be rational; then $ε^{- 1} > 0$ by [L4], so [L2] supplies a natural $N \geq 1$ with $ε^{- 1} < N \cdot 1_{R}$ , and [L4] applied to $0 < ε^{- 1} < N \cdot 1_{R}$ gives $0 < (N \cdot 1_{R})^{- 1} < ε$ . For $k \geq N$ we have $k + 1 > N$ by [L6], hence $(k + 1) \cdot 1_{R} > N \cdot 1_{R} > 0$ by [L3], hence $0 < y_{k} < (N \cdot 1_{R})^{- 1} < ε$ by [L4], and therefore $∣ y_{k} - 0 ∣ = y_{k} < ε$ by [L5].

step 1.1L1L2L3L4L5L6

3.1

Both sequences converge, and their limits are unique by [L8], so $\lim_{k} x_{k} = 0 = \lim_{k} y_{k}$ ; the conclusion $\lim_{k} x_{k} < \lim_{k} y_{k}$ therefore fails by trichotomy, although the hypothesis $x_{k} < y_{k}$ holds at every single index. The claim is therefore false.

step 1.1step 1.2step 2.1L6L8

4.1

What survives is the non-strict statement [L7]: from $x_{k} \leq y_{k}$ eventually one may conclude $\lim_{k} x_{k} \leq \lim_{k} y_{k}$ , and here that conclusion holds with equality.

step 3.1L7∎

Remarks

The reason is structural rather than accidental. A strict inequality between two sequences is a statement about each index separately, and a gap that is positive at every index may shrink towards $0$ ; the limit records only what is left after the shrinking. Non-strict inequalities survive precisely because " $\geq 0$ " is stable under this shrinking, which is the content of Limits preserve non-strict inequalities.
Strictness at every index is never enough by itself, and the failure has nothing to do with the limit being $0$ . The witness may be shifted: for any real $a$ , the sequences $x_{k} : = a$ and $y_{k} : = a + 1 / (k + 1)$ again satisfy $x_{k} < y_{k}$ at every index, and both converge to $a$ by the sum rule applied to a constant sequence and a null sequence (Algebra of limits: sums, scalar multiples, products and quotients), so no value of the common limit is exceptional. What does repair the claim is a quantitative strengthening of the hypothesis, for instance a uniform gap $y_{k} - x_{k} \geq c$ for a fixed real $c > 0$ : then $(y_{k} - x_{k})$ converges to $\lim_{k} y_{k} - \lim_{k} x_{k}$ (Algebra of limits: sums, scalar multiples, products and quotients) and Limits preserve non-strict inequalities, applied to the constant sequence $c$ and to $(y_{k} - x_{k})$ , gives $\lim_{k} y_{k} - \lim_{k} x_{k} \geq c > 0$ . The moral is that $x_{k} < y_{k}$ carries no lower bound on the gap, not that hypotheses on the sequences are powerless.
The sequence $1 / (k + 1)$ used here is the standard witness that the Archimedean property is what makes $R$ have no infinitesimals (Every complete ordered field is Archimedean); by For positive terms, null and divergence to $+ \infty$ are reciprocal its reciprocals diverge to $+ \infty$ .

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

39 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

J. K. Hunter, An Introduction to Real Analysis, Ch. 3 (standard reference, not scraped)
Limit of a sequence (Wikipedia) (standard reference, not scraped)
T. Tao, Analysis I, 3rd ed., §6.4 (standard reference, not scraped)
W. Rudin, Principles of Mathematical Analysis, 3rd ed., Ch. 3 (standard reference, not scraped)