Neighborhood basis in product topology 
Let $x = (x_1, x_2, \dots) \in \{0,1\}^{\mathbb{N}}$. Show that the sets $$B_n(x)=\{(y_1, y_2, \dots) \mid y_i = x_i \text{ for all } i =1,2,\dots,n\}$$ form a neighborhood base at $x$.

Let $O$ be a neighborhood of $X$, then $O$ contains a basic open set $U = \prod_{n \in \Bbb N} U_n$ such that $x \in U \subset O$.
Now for $U$ we have that $U_n = \{0,1\}$ except for finitely many indices say $\{n_1, n_2, \dots, n_k\}$.
I'm wondering how I should use these indices for $B_n(x)$. If I let $m = \min\{n_1, n_2, \dots, n_k\}$, then do I have $B_m(x) \subset U \subset O$?
Alternative way for this would be to express $$U = \{f : \Bbb N \to \{0,1\} | f(n_1) \in U_{n_1}, \dots, f_{n_k} \in U_{n_k} \}$$ and $B_n(x) = \{f \in \Bbb N \to \{0,1\} \mid f(x_1) \in\{y_1\}, \dots ,f(x_n) \in \{y_n\} \}$ but it isn't that helpful.

Using your notation,what you want to show is that for every generic neighbourhood $O$ of your point $x=(x_1,x_2...)$, there is a $B_n()$ contained in it. Let
$U=\prod_{n\in{\Bbb N}}U_n$ the open set contained in O by definition of neighborhood. Since $x\in U$, calling F= {$n_1,...,n_k$}, $U_j=x_j$ $\forall j\in F$. So lets take $M=max{F}$ then $x\in B_M(x)$ and $B_M(x)\subset U \subset O$ .
