The Cauchy sequences of rationals form a commutative ring that is not an integral domain: two eventually-constant sequences with disjoint supports multiply to zero

Let $X := \{\, n \in \mathbb{N} : n \ge 1 \,\}$ (The natural numbers $\mathbb{N}$ (von Neumann), Order on the natural numbers) and let $\mathcal{C}$ be the set of Cauchy sequences of rationals, that is, the set of those functions $X \to \mathbb{Q}$ that satisfy the condition of Cauchy sequence of rationals; that definition indexes its sequences $(a_n)_{n \ge 1}$, so $X$, and not $\mathbb{N}$, is the index set. Give $\mathcal{C}$ the termwise operations and the constant sequences $0$ and $1$. Then:

1. $\mathcal{C}$ is a commutative ring in the sense of Ring: an abelian group under addition and a monoid under multiplication, with multiplication distributing over addition on both sides and Commutative ring, with $1 \ne 0$;

2. $\mathcal{C}$ is a subring (Subring: a subset containing $1_R$ and closed under addition, additive inverses and multiplication) of the ring $\mathbb{Q}^{X}$ of all functions $X \to \mathbb{Q}$ with pointwise operations (The ring $R^{X}$ of all functions from a set $X$ into a ring, with pointwise operations);

3. $\mathcal{C}$ is not an integral domain (Zero divisor, and integral domain: a commutative ring with $1 \ne 0$ and no zero divisors): the sequences

   $a = (1,0,0,0,\dots), \qquad b = (0,1,1,1,\dots)$

   defined by $a_1 = 1$, $a_n = 0$ for $n \ge 2$, and $b_1 = 0$, $b_n = 1$ for $n \ge 2$, are both Cauchy, both nonzero, and satisfy $ab = 0$; so each is a zero divisor.

• The index set is stated because it is not $\mathbb{N}$. Cauchy sequence of rationals indexes its sequences from $n = 1$, while The ring $R^{X}$ of all functions from a set $X$ into a ring, with pointwise operations takes an arbitrary index set; the ambient ring in claim 2 is therefore $\mathbb{Q}^{X}$ with $X = \{\, n \in \mathbb{N} : n \ge 1 \,\}$, not $\mathbb{Q}^{\mathbb{N}}$. Since $\mathbb{N}$ contains $0$ (The natural numbers $\mathbb{N}$ (von Neumann)) the two are different sets of functions, and the subring claim would be false as stated about the second.

• Why this matters for the construction of $\mathbb{R}$. The real numbers are defined as the quotient $\mathcal{C}/\mathcal{N}$ by the null sequences (The real numbers), and The reals form a field proves that quotient is a field. The present example shows the field property cannot come from $\mathcal{C}$ alone: $\mathcal{C}$ is not even a domain. What The reals form a field actually uses is The null ideal is maximal, a property of $\mathcal{N}$ inside $\mathcal{C}$. The two zero divisors above cause no trouble in the quotient: $a$ is a null sequence (Null sequence), so its class is $0$, and $b$ differs from the constant sequence $1$ by a null sequence, so its class is $1$.

• The witnesses are chosen to keep the verification short. Both are eventually constant, hence Cauchy with no $\varepsilon$ bookkeeping, and their supports are disjoint, which makes the product zero at every index.