LemmaStatement: AI-adaptedProof: AI-adaptedprecheck passjudge pass (openai/gpt-5.4)audited 2026-07-24

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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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A non-null Cauchy sequence is eventually bounded away from zero, with constant sign

Statement

If $(a_{n})$ is Cauchy but not null, there are a rational $δ > 0$ and an index $N_{0}$ such that $∣ a_{n} ∣ > δ$ for all $n \geq N_{0}$ ; moreover either $a_{n} > δ$ for all $n \geq N_{0}$ , or $a_{n} < - δ$ for all $n \geq N_{0}$ .

Facts & Assumptions

Given: A Cauchy sequence $(a_{n})$ that is not null.

[A1]

Negation of null (Null sequence): there is a rational $ε_{0} > 0$ such that for every $N$ some $n \geq N$ has $∣ a_{n} ∣ \geq ε_{0}$ .

[A2]

Cauchy (Cauchy sequence of rationals).

[L1]

Ordered-field arithmetic: $ε_{0} / 3 > 0$ , $ε_{0} - ε_{0} / 3 > ε_{0} / 3$ (The rationals form a totally ordered field).

[L2]

Triangle inequality, in the form $∣ u ∣ \geq ∣ v ∣ - ∣ v - u ∣$ (Absolute value and the triangle inequality).

Proof

technique · direct

1.1A1

Fix $ε_{0} > 0$ witnessing that $(a_{n})$ is not null.

1.2A2L1

Fix $N_{0}$ with $∣ a_{m} - a_{n} ∣ < ε_{0} / 3$ for all $m, n \geq N_{0}$ .

2.1step 1.1step 1.2

Pick $n_{0} \geq N_{0}$ with $∣ a_{n_{0}} ∣ \geq ε_{0}$ .

3.1step 2.1step 1.2L2L1

For every $n \geq N_{0}$ : $∣ a_{n} ∣ \geq ∣ a_{n_{0}} ∣ - ∣ a_{n_{0}} - a_{n} ∣ > ε_{0} - ε_{0} / 3 > ε_{0} / 3 = : δ$ ; so $∣ a_{n} ∣ > δ$ for all $n \geq N_{0}$ .

4.1step 3.1step 1.2L1∎

Sign stability: if some $a_{m} > δ$ and some $a_{n} < - δ$ with $m, n \geq N_{0}$ , then $∣ a_{m} - a_{n} ∣ > 2 δ = 2 ε_{0} / 3 > ε_{0} / 3$ , impossible by step 1.2; so beyond $N_{0}$ all terms have one sign, and by step 3.1 either $a_{n} > δ$ for all $n \geq N_{0}$ or $a_{n} < - δ$ for all $n \geq N_{0}$ .

A non-null Cauchy sequence is eventually bounded away from zero, with constant sign

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources