LemmaStatement: AI-adaptedProof: AI-adaptedprecheck passjudge pass (z-ai/glm-5.2)verified 2026-07-26 (claude-opus-5)

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The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

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.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

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Limits preserve non-strict inequalities

Statement

Let $(x_{k})$ and $(y_{k})$ be sequences of reals (Sequences of reals: bounded, eventually, frequently, tails, subsequences) converging to $x$ and $y$ respectively (Limits and Cauchy sequences of reals). If $x_{k} \leq y_{k}$ eventually, that is for all $k$ from some index on, then

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

In particular, if $x_{k} \geq 0$ eventually then $\lim_{k} x_{k} \geq 0$ , and if $a \leq x_{k} \leq b$ eventually then $a \leq \lim_{k} x_{k} \leq b$ .

The conclusion is not strict, and cannot be made strict; see the remarks below and the false statement at the end of this page.

Facts & Assumptions

Given: Sequences $(x_{k})$ , $(y_{k})$ of reals with $(x_{k})$ converging to $x$ , $(y_{k})$ converging to $y$ , and an index $K_{0} \in N$ with $x_{k} \leq y_{k}$ for every $k \geq K_{0}$ (Sequences of reals: bounded, eventually, frequently, tails, subsequences, Limits and Cauchy sequences of reals). Write $z_{k} : = y_{k} - x_{k}$ and $z : = y - x$ .

[L1]

Convergence, quantified over rational $ε > 0$ (Limits and Cauchy sequences of reals).

[L2]

Difference rule: $(z_{k}) = (y_{k} - x_{k})$ converges to $y - x = z$ (Algebra of limits: sums, scalar multiples, products and quotients).

[L3]

Small rationals: for every real $η > 0$ there is a rational $ε$ with $0 < ε < η$ , by density (The rationals embed densely in the reals) or by the Archimedean property (Every complete ordered field is Archimedean) applied to $1 / η$ (Inverses of positives are positive, and reciprocation reverses order).

[L4]

Absolute value: $∣ u ∣ < c$ if and only if $- c < u < c$ , for $c > 0$ (Basic properties of the absolute value).

[L5]

Order arithmetic in $R$ : adding a constant preserves $<$ and $\leq$ ; $u \leq v$ and $v < w$ give $u < w$ ; trichotomy, so exactly one of $z < 0$ , $z = 0$ , $z > 0$ holds and the negation of $z < 0$ is $z \geq 0$ ; $u \leq v$ if and only if $v - u \geq 0$ ; and $u < u$ is impossible (Order is preserved by adding a constant and by adding inequalities, Complete ordered field (least-upper-bound property), Ordered field).

[L6]

The order on $N$ is total, so any two indices admit a common upper bound ( $\leq$ is a linear order on $N$ ).

[L7]

For $c \in R$ the constant sequence $(c)$ converges to $c$ (Sequences of reals: bounded, eventually, frequently, tails, subsequences, Limits and Cauchy sequences of reals).

[L8]

A sequence of reals has at most one limit (A sequence has at most one limit), which is what licenses writing $\lim_{k} x_{k}$ and $\lim_{k} y_{k}$ for the limits named in the statement; without it those symbols would not denote.

Proof

technique · contradiction

1.1givenL2L5

By [L2] the sequence $(z_{k})$ converges to $z = y - x$ , and by hypothesis $z_{k} = y_{k} - x_{k} \geq 0$ for every $k \geq K_{0}$ .

1.2assume-contra

Suppose, for contradiction, that $z < 0$ .

2.1step 1.2L3L5choose

Then $- z > 0$ , so by [L3] we may choose a rational $ε$ with $0 < ε < - z$ .

3.1step 1.1step 2.1L1L4L5

Applying convergence of $(z_{k})$ to this $ε$ gives $K_{1} \in N$ with $∣ z_{k} - z ∣ < ε$ for all $k \geq K_{1}$ , hence $z_{k} - z < ε$ and so $z_{k} < z + ε$ for all such $k$ .

4.1step 1.1step 2.1step 3.1L5L6

Fix an index $k$ with $k \geq K_{0}$ and $k \geq K_{1}$ . Then $0 \leq z_{k} < z + ε < z + (- z) = 0$ , so $0 < 0$ , which is impossible.

5.1step 4.1L5L7L8discharge-contradiction∎

The assumption $z < 0$ is therefore untenable; by trichotomy $z \geq 0$ , that is $y - x \geq 0$ , that is $x \leq y$ ; since $x$ and $y$ are the unique limits of $(x_{k})$ and $(y_{k})$ by [L8], that is exactly $\lim_{k} x_{k} \leq \lim_{k} y_{k}$ . Since $(x_{k})$ and $(y_{k})$ were an arbitrary pair satisfying the hypotheses, the conclusion applies to every such pair, and the two stated special cases are instances of it. Let $(w_{k})$ be convergent. If $w_{k} \geq 0$ from some index on, apply the conclusion to the pair consisting of the constant sequence $0$ , which converges to $0$ by [L7], and of $(w_{k})$ : it gives $0 \leq \lim_{k} w_{k}$ . If $a \leq w_{k} \leq b$ from some index on, apply it first to the constant sequence $a$ and $(w_{k})$ , then to $(w_{k})$ and the constant sequence $b$ : it gives $a \leq \lim_{k} w_{k}$ and $\lim_{k} w_{k} \leq b$ .

Remarks

The two special cases are instances of the main claim, discharged in step 5.1 by taking one of the two sequences constant; that a constant sequence converges to its value (Sequences of reals: bounded, eventually, frequently, tails, subsequences) is the only extra ingredient they need.
The inequality does not become strict. From $x_{k} < y_{k}$ for every $k$ one may conclude only $\lim x_{k} \leq \lim y_{k}$ ; the witness $x_{k} = 0 < 1 / (k + 1) = y_{k}$ has equal limits (FALSE: limits preserve strict inequalities). Intuitively, the order relation $<$ is not preserved by passage to a limit because a strict gap may shrink to nothing, while $\leq$ is preserved because it is closed under that shrinking.
The proof routes through the single sequence $(z_{k}) = (y_{k} - x_{k})$ and the difference rule of Algebra of limits: sums, scalar multiples, products and quotients. That is not an economy of writing only: it isolates the one thing being proved, namely that a sequence eventually $\geq 0$ cannot have a negative limit.

Limits preserve non-strict inequalities

Statement

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 2 levels

Sources