Alphabeta Math
LemmaStatement: AI-adaptedProof: AI-adaptedprecheck passjudge pass (openai/gpt-5.4)audited 2026-07-24
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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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Every Cauchy sequence of rationals is bounded

Statement

Every Cauchy sequence (an)n≥1 of rational numbers (Cauchy sequence of rationals) is bounded: there exists a rational M≥0 such that ∣an∣≤M for all n≥1.

Facts & Assumptions

Given: A Cauchy sequence (an)n≥1 of rational numbers.

[A1]

For every rational ε>0 there exists N∈N with ∣am−an∣<ε for all m,n≥N.

[L1]

Triangle inequality on Q: ∣x∣≤∣x−y∣+∣y∣ (Absolute value and the triangle inequality).

Proof

technique · direct
1.1

Apply [A1] with ε=1: fix N∈N such that ∣am−an∣<1 for all m,n≥N.

A1
1.2

For every n≥N: ∣an∣≤∣an−aN∣+∣aN∣.

L1
2.1

For every n≥N: ∣an∣<∣aN∣+1.

step 1.1step 1.2algebra
2.2

Define M=max⁡{∣a1∣,∣a2∣,…,∣aN−1∣,∣aN∣+1}, a maximum of finitely many rationals, hence rational and ≥0.

step 1.1choose
3.1

For every n<N: ∣an∣≤M, since ∣an∣ appears in the maximum.

step 2.2algebra
3.2

For every n≥N: ∣an∣<∣aN∣+1≤M.

step 2.1step 2.2algebra
4.1

For every n≥1: ∣an∣≤M, so (an) is bounded.

step 3.1step 3.2∎

Depends on

Used by

Dependency tree · two levels

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Sources