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

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Cauchy sequences form a commutative ring

Statement

The set $C$ of Cauchy sequences of rationals, with termwise addition and multiplication and the constant sequences $0$ and $1$ , is a commutative ring with identity.

Facts & Assumptions

Given: Cauchy sequences $(a_{n}), (b_{n})$ and a rational $ε > 0$ .

[A1]

Cauchy: for every rational $δ > 0$ there is an index beyond which any two terms differ by less than $δ$ .

[L1]

$Q$ is a field; ring axioms hold termwise (The rationals form a field).

[L2]

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

[L3]

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

[L4]

Ordered-field arithmetic: $ε / 2$ , $ε / (A + B)$ are positive rationals for $A, B \geq 1$ (The rationals form a totally ordered field).

Proof

technique · direct

1.1L1

Termwise, all ring axioms (associativity, commutativity, identities, inverses for $+$ , distributivity) hold in $C$ because they hold in $Q$ at every index; the constant sequences $0, 1$ are Cauchy. What needs proof is closure of $C$ under $+$ , negation, and $\cdot$ .

1.2A1L4

Fix $N_{a}$ with $∣ a_{m} - a_{n} ∣ < ε / 2$ for $m, n \geq N_{a}$ and $N_{b}$ likewise for $(b_{n})$ .

1.3L3

Fix rational bounds $A, B \geq 1$ with $∣ a_{n} ∣ \leq A$ and $∣ b_{n} ∣ \leq B$ for all $n$ .

2.1step 1.2L2

Sum: for $m, n \geq \max (N_{a}, N_{b})$ , $∣ (a_{m} + b_{m}) - (a_{n} + b_{n}) ∣ \leq ∣ a_{m} - a_{n} ∣ + ∣ b_{m} - b_{n} ∣ < ε$ ; so $(a_{n} + b_{n})$ is Cauchy. Negation likewise: $∣ (- a_{m}) - (- a_{n}) ∣ = ∣ a_{m} - a_{n} ∣$ .

2.2step 1.3A1L4

Set $δ = ε / (A + B) > 0$ and fix $M_{a}, M_{b}$ with $∣ a_{m} - a_{n} ∣ < δ$ for $m, n \geq M_{a}$ and $∣ b_{m} - b_{n} ∣ < δ$ for $m, n \geq M_{b}$ .

3.1step 1.3step 2.2L2

Product: for $m, n \geq \max (M_{a}, M_{b})$ , $∣ a_{m} b_{m} - a_{n} b_{n} ∣ = ∣ a_{m} (b_{m} - b_{n}) + b_{n} (a_{m} - a_{n}) ∣ \leq A ∣ b_{m} - b_{n} ∣ + B ∣ a_{m} - a_{n} ∣ < (A + B) δ = ε$ ; so $(a_{n} b_{n})$ is Cauchy.

4.1step 1.1step 2.1step 3.1∎

$C$ is closed under addition, negation, and multiplication, contains $0$ and $1$ , and satisfies the ring axioms termwise: a commutative ring with identity.

Cauchy sequences form a commutative ring

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources