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?
Best Answer
$$\pi_i^{-1}(X_i)=E_1\times E_2\times ... \times X_i\times E_{i+1}\times...$$ If we intersect finitely many of these, say with indices $1..k$, for simplicity to writing up, we get $$\pi_1^{-1}(X_1)\cap \pi_2^{-1}(X_2)\cap..\cap \pi_k^{-1}(X_k)=X_1\times..\times X_k\times E_{k+1}\times E_{k+2}\times...$$ (It's also needed that, for the same index, we have $\pi_i^{-1}(X_i)\cap\pi_i^{-1}(Y_i)=\pi_i^{-1}(X_i\cap Y_i)$.)