Is $R^\infty$ a ring?

Question

Let $(R,+,\cdot,0,1)$ be a ring, and consider the set $$R^\infty=\left\{\{a_n\}_{n=1}^\infty:a_k\in R\text{ for all } k \in \mathbb{Z}_{>0}\right\}$$ with operations $\oplus$ and $\odot$ on $R^\infty$ defined by $$\begin{split} \{a_n\}_{n=1}^\infty\oplus\{b_n\}_{n=1}^\infty&=\{a_n+b_n\}_{n=1}^\infty,\\ \{a_n\}_{n=1}^\infty\odot\{b_n\}_{n=1}^\infty&=\{a_n\cdot b_n\}_{n=1}^\infty. \end{split}$$ Is it then correct to conclude that $(R^\infty,\oplus,\odot,(0,0,\ldots),(1,1,\ldots))$ is a ring? I know the proof for the case $R^n$ for some finite natural number $n$ (which just comes down to verifying axioms), but I'm not completely sure if this proof can be extended to the case $n=\infty$.

Answer 1

More generally, if $I$ is any set and $(R_i)_{i\in I}$ is a family of rings, then the product $$ \prod_{i\in I} R_i := \left\{ \, (r_i)_{i\in I} \,\Big|\, \forall i\in I: r_i \in R_i \, \right\} $$ with addition and multiplication defined component-wise is a a ring. The proof is the same as the proof that $R^n$ is a ring, just with more heavy notation.

For $I=\{1,\dots,n\}$ and all $R_i=R$ you get $R^n$ and for for $I=\mathbb N$ and again all $R_i=R$ you get what you called $R^\infty$ in the question.

Answer 2

Well, you are correct. I'd consider the ring $R^\infty$ as the set of all functions $f:{\Bbb Z}_{>0}\rightarrow {\Bbb R}$. Then for two such functions $f,g$, $$f+g : {\Bbb Z}_{>0}\rightarrow {\Bbb R}: a\mapsto f(a)+g(a)$$ and $$f\cdot g : {\Bbb Z}_{>0}\rightarrow {\Bbb R}: a\mapsto f(a)\cdot g(a).$$ The operations are componentwise and the properties depend on the properties of the codomain, here ${\Bbb R}$.