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

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.

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

1.1

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

2.1

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

step 1.1choose

3.1

$(b_{n})$ 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 $a_{m} = a_{n}$ .

step 2.1step 1.1L2L3

3.2

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.

step 2.1L4

4.1

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.

step 3.1step 3.2L3L4L5∎

Depends on

Used by

Dependency tree · two levels

14 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

W. Aitken, Constructing the Real Numbers, Chapter 9, Section 6, Lemmas 17–18 (standard reference, not scraped)
T. Tao, Analysis I, 3rd ed., §5.3 (standard reference, not scraped)
L. S. Krapp, Constructions of the real numbers: a set theoretical approach (Oxford, 2014) (standard reference, not scraped)