Topological basis of infinite product Let $X$ be the infinite product of the space $\{0,1\}$, where $\{0,1\}$ has the discrete topology and $X$ the product topology.
Let $A(a_0,...,a_n)=\{(x_n)\in X:x_0=a_0,...,x_n=a_n\}$ for $a_0,...,a_n\in \{0,1\}$.

How do I prove that these sets $A(a_0,...,a_n)$ form a basis for the product topology on $X$?

What I thought:
Per definition $$\{U\times\{0,1\} \times\{0,1\}\times...|U\text{ open in}\{0,1\}\}\cup\{\{0,1\} \times V\times\{0,1\}\times...|V\text{ open in}\{0,1\}\}\cup...$$
is a subbasis of the product topology. That means that the set of finite intersections of elements herein form a basis. But all these finite intersections are exactly of the form of $A(a_0,..,a_n)$.
Is my reasoning correct?
 A: To quote my own answer here: a base for the product $X = \prod_{i=1}^\infty X_i$ are all sets of the form 
$$\prod_{i=0}^n U_i \times \prod_{i=n+1}^\infty X_i$$
where all $U_i$ are open in $X_i$.
Now for $X_i = \{0,1\}$ the only non-empty and non-$X_i$ open sets are of the form
$\{0\}$ and $\{1\}$ and picking a product with $U_i = \{a_i\}$ is the same as demanding $x_i = a_i$
So then when $U_i = \{a_i\}, a_i =0,1$: 
$$\prod_{i=0}^n U_i \times \prod_{i=n+1}^\infty X_i = \{(x_i)_i \in X: x_0 = a_0,\ldots,x_n = a_n\}$$
So your question is a special case of this more general fact.
A direct proof an be given as well:
Suppose $p = (p_i) \in O$ where $O$ is product open. All sets of the form $p_n^{-1}[V_n]$, where $V_n \subset \{0,1\}$ is open, form a subbase as you rightly state. If we we take $V_n = \emptyset$ we get $\emptyset$, or $V_n =\{0,1\}$ we get the whole product, so we can assume that $V_n = \{0\}$ or $V_n = \{1\}$, for every non-trivial subbasic set. The finite intersections of subbasic elements form a base, so we have $n_0, \ldots, n_k \in \mathbb{N}$ (in increasing order) and $i_0,\ldots i_k \in \{0,1\}$ such that:
$$p \in \bigcap_{j=0}^k p_{n_j}^{-1}[\{i_j\}] \subset O$$
But then certainly:
$p \in A:= \{(x_i) \in X: x_0 = q_0, x_1 = q_1,\ldots x_{n_k} = q_{n_k}\}$
where $q_i = i_{n_j}$ if $i = n_j \in \{n_0, \ldots ,n_k\}$ and $q_i = p_i$ otherwise. And this set lies inside $\bigcap_{j=0}^k p_{n_j}^{-1}[\{i_j\}]$ by definition (and so $p \in A \subset O$). So sets like $A$ (fixed start sets) form a base for the product topology. 
