LemmaStatement: AI-adaptedProof: AI-adaptedprecheck passverified 2026-09-09 (gpt-6-astra)

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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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The null ideal is maximal

Statement

If $(a_{n}) \in C ∖ N$ , then the ideal of $C$ generated by $N$ and $(a_{n})$ is all of $C$ . Hence $N$ is a maximal ideal.

Facts & Assumptions

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

[L1]

Away-from-zero: there are $δ > 0$ and $N_{0}$ with $∣ a_{n} ∣ > δ$ for all $n \geq N_{0}$ (A non-null Cauchy sequence is eventually bounded away from zero, with constant sign).

[L2]

Triangle inequality and $∣ u v ∣ = ∣ u ∣ ∣ v ∣$ (Absolute value and the triangle inequality).

[L3]

Cauchy definition and field arithmetic in $Q$ : $ε δ^{2} > 0$ for $ε > 0$ (The rationals form a totally ordered field).

[L4]

Null definition; a sequence that is $0$ from some index on is null (Null sequence).

[L5]

Ideal arithmetic in $C$ : an ideal containing $1$ is the whole ring (Cauchy sequences form a commutative ring, Null sequences form an ideal).

Proof

technique · direct

1.1L1

Fix $δ > 0$ and $N_{0}$ with $∣ a_{n} ∣ > δ$ for all $n \geq N_{0}$ ; in particular $a_{n} \neq 0$ there.

2.1step 1.1choose

Define $b_{n} = 1$ for $n < N_{0}$ and $b_{n} = 1 / a_{n}$ for $n \geq N_{0}$ .

3.1step 2.1step 1.1L2L3

$(b_{n})$ is Cauchy. Given $ε > 0$ , choose a Cauchy index $N_{1}$ for $(a_{n})$ at $ε δ^{2}$ and put $N = \max (N_{0}, N_{1})$ . For $m, n \geq N$ , $∣ b_{m} - b_{n} ∣ = \frac{∣ a_{n} - a_{m} ∣}{∣ a_{m} ∣ ∣ a_{n} ∣} \leq \frac{∣ a_{m} - a_{n} ∣}{δ^{2}} < ε$ . The non-strict comparison includes $a_{m} = a_{n}$ .

3.2step 2.1L4

For $n \geq N_{0}$ , $a_{n} b_{n} = 1$ , so the sequence $(a_{n} b_{n} - 1)$ is $0$ from $N_{0}$ on, hence null.

4.1step 3.1step 3.2L3L4L5∎

Therefore $1_{C} = (a_{n}) (b_{n}) - ((a_{n} b_{n}) - 1)$ lies in the ideal generated by $(a_{n})$ and $N$ , so that ideal is all of $C$ . The constant sequence $1$ is not null (take $ε = 1 / 2$ ), so $N$ is proper. Any ideal strictly containing $N$ contains a non-null element and thus equals $C$ : $N$ is maximal.

The null ideal is maximal

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources