Construction of Basis for Product Topology In a book I read the following:

Let $(E_i, \mathcal{T}_i)_{i\in I}$ be a family of topological spaces, and let $E = \prod_{i\in I} E_i$ be the Cartesian product of the $E_i$'s. Denote by $\pi_i$ the natural projection from $E$ into $E_i$, defined by $\pi_i((e_j)_{j\in I} = e_i$. The product topology on E is the topology generated by the basis consisting of the finite intersections of sets of the form $\pi^{-1}_i(X_i)$ where $X_i$ is an open set of $E_i$. These sets are nothing else than the products $\prod_{i\in I} X_i$ where the $X_i$'s are open sets of $E_i$ and where $X_i = E_i$ except for a finite number of indices.

I don't understand why the sets of finite intersections of the $\pi_i^{-1}(X_i)$ are of the form $\prod_{i\in I} X_i$ with $X_i = E_i$ cofinitely often. Because for example the set $\pi_i^{-1}(X_i)$ is in the set of finite intersection, but could well contain an element $X = \prod_{j\in I} X_j$ were all $X_j \ne E_j$ as long as $\pi_i(X) = X_i$. So why does it follows that $X_j = E_j$ infinitely often?

Think about what each of the generating elements $\pi_i^{-1} (X_i)$ look like in the case each of your $E_i = \Bbb R$ and $X_i$ is an open interval.  Your open sets look like things of the form $$ \Bbb R \times \Bbb R \times \dots \times (a,b) \times \dots \times \Bbb R \times \Bbb R \times \dots$$
Finite intersections of such sets will only ever get you a finite number of intervals anywhere but the majority of the sets in the product will still be $\Bbb R$.
