False statementConstruction: AI-adaptedVerification: AI-adaptedSession-authored (Fable 5 assisted)precheck 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 $k \in \mathbb{N}$ , 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 $y_k := \bigl((k+1) \cdot 1_{\mathbb{R}}\bigr)^{-1}$ , where $n \cdot 1_{\mathbb{R}}$ denotes the canonical natural of $\mathbb{R}$ (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 $\varepsilon > 0$ there is $K \in \mathbb{N}$ with $|x_k - x| < \hat\varepsilon$ for all $k \ge K$ (Limits and Cauchy sequences of reals); a sequence of reals is a function $\mathbb{N} \to \mathbb{R}$ (Sequences of reals: bounded, eventually, frequently, tails, subsequences), so a constant sequence converges to its value, $|x - x| = 0 < \hat\varepsilon$ holding at every index.

[L2]

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

[L3]

Canonical naturals: $n \cdot 1_{\mathbb{R}} > 0$ for every $n \ge 1$ , and $n \mapsto n \cdot 1_{\mathbb{R}}$ 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 $u \ne 0$ (Inverses of positives are positive, and reciprocation reverses order, Field).

[L5]

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

[L6]

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

[L7]

If sequences of reals $(x_k)$ and $(y_k)$ converge to $x$ and $y$ and $x_k \le y_k$ eventually, then $x \le y$ (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 $(k+1) \cdot 1_{\mathbb{R}}$ 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 $k \in \mathbb{N}$ .

L3L4

1.2

The constant sequence $(x_k) = (0)$ converges to $0$ .

2.1

The sequence $(y_k)$ converges to $0$ . Let $\varepsilon > 0$ be rational; then $\varepsilon^{-1} > 0$ by [L4], so [L2] supplies a natural $N \ge 1$ with $\varepsilon^{-1} < N \cdot 1_{\mathbb{R}}$ , and [L4] applied to $0 < \varepsilon^{-1} < N \cdot 1_{\mathbb{R}}$ gives $0 < (N \cdot 1_{\mathbb{R}})^{-1} < \varepsilon$ . For $k \ge N$ we have $k + 1 > N$ by [L6], hence $(k+1) \cdot 1_{\mathbb{R}} > N \cdot 1_{\mathbb{R}} > 0$ by [L3], hence $0 < y_k < (N \cdot 1_{\mathbb{R}})^{-1} < \varepsilon$ by [L4], and therefore $|y_k - 0| = y_k < \varepsilon$ 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 $x_k \le y_k$ eventually one may conclude $\lim_k x_k \le \lim_k y_k$ , 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 " $\ge 0$ " 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 $y_k - x_k \ge c$ 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 $\lim_k y_k - \lim_k x_k \ge c > 0$ . 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 $\mathbb{R}$ 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

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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)