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

How statement and proof provenance work

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.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

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

1.1

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

1.2

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

A2L1

2.1

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

step 1.1step 1.2

3.1

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 $n \geq N_{0}$ .

step 2.1step 1.2L2L1

4.1

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 $n \geq N_{0}$ .

step 3.1step 1.2L1∎

Depends on

Used by

Free tail ultrafilters and bounded real ultralimit calculus Lemma
The null ideal is maximal Lemma
The reals form a totally ordered field Theorem

Dependency tree · two levels

9 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

T. Tao, Analysis I, 3rd ed., §5.4 (standard reference, not scraped)
L. S. Krapp, Constructions of the real numbers: a set theoretical approach (Oxford, 2014) (standard reference, not scraped)