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

1.1

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 $\cdot$ .

1.2

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

A1L4

1.3

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

2.1

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: $∣ (- a_{m}) - (- a_{n}) ∣ = ∣ a_{m} - a_{n} ∣$ .

step 1.2L2

2.2

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 $m, n \geq M_{b}$ .

step 1.3A1L4

3.1

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.

step 1.3step 2.2L2

4.1

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

step 1.1step 2.1step 3.1∎

Depends on

Used by

Dependency tree · two levels

12 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

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)