Alphabeta Math
LemmaStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck 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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Null sequences form an ideal

Statement

The set N\mathcal{N} of null sequences is an ideal of the ring C\mathcal{C} of Cauchy sequences (Cauchy sequences form a commutative ring): it is a subgroup under addition, and czNc \cdot z \in \mathcal{N} whenever cCc \in \mathcal{C} and zNz \in \mathcal{N}.

Facts & Assumptions

Given: Null sequences (zn),(wn)(z_n), (w_n), a Cauchy sequence (cn)(c_n), and a rational ε>0\varepsilon > 0.

[A1]

Null: beyond some index, zn|z_n| is smaller than any prescribed positive rational.

[L1]

Ordered-field arithmetic in Q\mathbb{Q} (The rationals form a totally ordered field).

[L2]

Triangle inequality and multiplicativity of |\cdot| (Absolute value and the triangle inequality).

[L3]

Cauchy sequences are bounded (Every Cauchy sequence of rationals is bounded).

[L4]

Null sequences are Cauchy, so NC\mathcal{N} \subseteq \mathcal{C} (Null sequences are Cauchy).

Proof

technique · direct
1.1

Fix Nz,NwN_z, N_w with zn<ε/2|z_n| < \varepsilon/2 for nNzn \ge N_z and wn<ε/2|w_n| < \varepsilon/2 for nNwn \ge N_w.

A1L1
1.2

Fix M0M \ge 0 with cnM|c_n| \le M ([L3]) and set B=max(M,1)1B = \max(M, 1) \ge 1, so cnB|c_n| \le B for all nn.

L3
2.1

Sum: for nmax(Nz,Nw)n \ge \max(N_z, N_w), zn+wnzn+wn<ε|z_n + w_n| \le |z_n| + |w_n| < \varepsilon; negation: zn=zn|-z_n| = |z_n|. So N\mathcal{N} is a subgroup under addition.

step 1.1L2
2.2

Fix NN with zn<ε/B|z_n| < \varepsilon/B for nNn \ge N.

step 1.2A1L1
3.1

Product: for nNn \ge N, cnzn=cnznBzn<ε|c_n z_n| = |c_n|\,|z_n| \le B\,|z_n| < \varepsilon; so (cnzn)(c_n z_n) is null.

step 1.2step 2.2L2
4.1

N\mathcal{N} is a nonempty additive subgroup of C\mathcal{C} absorbing multiplication by C\mathcal{C}: an ideal.

step 2.1step 3.1L4

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 28 results over 14 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources