Cauchy sequences form a commutative ring

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

Fix $N_a$ with $|a_m - a_n| < \varepsilon/2$ for $m,n \ge N_a$ and $N_b$ likewise for $(b_n)$.

Fix rational bounds $A, B \ge 1$ with $|a_n| \le A$ and $|b_n| \le B$ for all $n$.

Sum: for $m,n \ge \max(N_a, N_b)$, $|(a_m + b_m) - (a_n + b_n)| \le |a_m - a_n| + |b_m - b_n| < \varepsilon$; so $(a_n + b_n)$ is Cauchy. Negation likewise: $|(-a_m) - (-a_n)| = |a_m - a_n|$.

Set $\delta = \varepsilon/(A+B) > 0$ and fix $M_a, M_b$ with $|a_m - a_n| < \delta$ for $m,n \ge M_a$ and $|b_m - b_n| < \delta$ for $m,n \ge M_b$.

Product: for $m,n \ge \max(M_a, M_b)$, $|a_m b_m - a_n b_n| = |a_m(b_m - b_n) + b_n(a_m - a_n)| \le A|b_m - b_n| + B|a_m - a_n| < (A+B)\delta = \varepsilon$; so $(a_n b_n)$ is Cauchy.

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