Theorem 7.8 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 topolgy.
I am studying for an exam and would like to know how to prove the above theorem; though my prof says we need not know the proofs of theorems for our exam i think learning the proof of this would be a meaningful learning experience and would like to know how to prove this and our textbook does not do so. A reference to where this is proved our proof would be much appreciated.
Edit: do i just need to prove that $\mathcal B'$ is a basis? or do i need something more than that?
Since the statement of the theorem is that that family forms a basis for the topology, then yes, to prove the theorem all you need is to show it forms a basis.
It is clear that $\mathcal B'$ covers $X = \prod_{n \in \mathbb N}X_n$: actually $X \in \mathcal B'$ (just take $U_n = X_n$, for all $n$).
Now let $U, V \in \mathcal B'$. Then there exist $m,n \in \mathbb N$, and $U_i, V_i \in \mathcal T_i$ such that $$U = U_1 \times \cdots \times U_n \times X_{n+1} \times X_{n+2} \times \cdots$$ and $$V = V_1 \times \cdots \times V_m \times X_{m+1} \times X_{m+2} \times \cdots.$$ Suppose, wlog, $n \leq m$ and then define $U_{n+1} = X_{n+1}, \ldots, U_m = X_m$.
Then $W_i = U_i \cap V_i \in \mathcal T_i$ and $U \cap V = W$.
This is enough to prove that $\mathcal B'$ is a basis for a topology on $X$ (notice the statement by the end of the second paragraph after definition: "A sufficient but not necessary condition for..."), and it is indeed the product topology.