I'm reading Torsten Wedhorn's notes on Adic Spaces (link here), and I've come across the following definition (page 41):
Let $A$ be a non-archimedean topological ring, let $I$ be an index set and fix a family $T=(T_i)_{i\in I}$ of subsets $T_i$ of $A$ such that for all $i\in I,m\in\Bbb N$ and for every neighborhood $U$ of $0$ in $A$ the subgroup $T_i^mU$ is a neighborhood of $0$. For $v\in\Bbb N_0^{(I)}$, we set $$T^v:=\prod_{i\in I}T_i^{v(i)}.$$ Then $T^vU$ is a neighborhood of $0$ for all $v$ and for all neighborhoods $U$ of $0$.
My confusion lies with the last line. If $I$ is finite then this is certainly true because we would just have $T^v U=T_1^{v_1}(T_2^{v_2}\cdots (T_n^{v_n}(U)))$ for some $n$, and there's a clear induction argument. However, how can we be sure this is true if $I$ is infinite?
It may have something to do with the fact that $A$ is non-archimedean. In case this may be useful, I'll include the definition:
Definition A topological ring $A$ is called non-archimedean if it has a fundamental system of neighborhoods of $0$ consisting of subgroups of $(A,+)$.
I would assume that the notation $\mathbb{N}_0^{(I)}$ means the set of all finite-support functions $I\to\mathbb{N}_0$. So $v(i)=0$ for all but finitely many $i$, and you don't have to worry about an infinite product. (Indeed, it's not even clear to me what such an infinite product would mean.)