Theorem 7.14 Let I be an index set and, for each $\iota \in I $ suppose that $(X_{\iota},\mathcal{T}_{\iota}) $ is a topological space. Let $ \mathcal B $ be the collection of all subsets of $\Pi_{\iota \in I} X_{\iota} $ obtained in the following way. let $F \subset I $ and, for each $\iota \in F $, let $U_{\iota} $ be in $ \mathcal T_{\iota}$.
Define: $$ U(F,(U_{\iota })_{\iota \in F } = \{(x_{\iota})_{\iota \in I} \in \Pi_{\iota \in I} X_{\iota}|x_{\iota}\in U_{\iota}, \forall \iota \in F\}$$
Let $\mathcal B $ be the collection of all subsets of $\Pi_{\iota \in I} X_{\iota}$ obtained in the way. then $\mathcal B $ is a basis for a topology on $\Pi_{\iota \in I} X_{\iota}$.
I have two questions:
1) Honestly i am not entirely sure what this is even saying but would like to work through a proof of it in hopes i can understand the theorem.
2) I believe in the special case that $I=\Bbb N $ this appears to be the same basis as $\mathcal B'$ below; how could i show this is true or prove its not?
Suppose $(X_n,\mathcal T_n ), n \in \Bbb N,$ is a sequence of topological spaces.
Let $\mathcal B'$ consist of all subsets of $\Pi_{n=1}^{\infty} X_n$ which are of the form $$U_1 \times \cdots \times U_N \times X_{N+1} \times \cdots= \{(x_n)_{n=1}^{\infty}| x_n \in U_n, 1\leq n \leq N \} $$ Where $N\geq 1 $ and $U_n $ is in $ \mathcal T_n $ for all $ 1\leq n \leq N $. the collection $\mathcal B' $ is a basis for a topology on $\Pi_{n=1}^{\infty} X_n$ called the product topology.
Edit: it turns out that this is actually wrong because F may be countable this cannot be the product topology when I is countable this is the box topology!