Let $X = \prod_{i\in I}X_i$ be a product space with the product topology. Then $X$ has a basis consisting of elements of the form $U = \prod_{i \in I} V_i$ where $V_i \ne X_i$ for only finitely many $i$.
This is how the basis for the product topology is defined, but it can also be defined by the projection maps $pr_i : X \to X_i$ as $$\left\{\bigcap_{i \in K}pr_i^{-1}(V_i) \mid V_i \text{ open in } X_i, K \subset I \text{ finite} \right\}$$
Now what I want to show is that $$\bigcap_{i \in K}pr_i^{-1}(V_i) = \prod_{i \in I }V_i$$ where the rhs satisfies that $V_i \ne X_i$ for only finitely many $i \in I$.
I know that $$pr_i^{-1}(V_i)=\{(x_1, \dots,x_i, \dots) \mid pr_i(x_1, \dots,x_i, \dots) =x_i \in V_i\} = X_1 \times \dots \times X_{i-1} \times U_i \times X_{i+1} \times \dots$$
So that $$\bigcap_{i \in K}pr_i^{-1}(V_i) = (V_1 \times X_2 \times \dots) \cap (X_1 \times V_2 \times \dots) \cap(X_1 \times X_2 \times V_3 \times \dots) \cap \dots$$
But I don't know if this rhs is same as $\prod_{i \in I}V_i$ because of the repitition of all the $X_i$'s and products?
To be precise, we have to show that $$\bigcap_{i \in K}pr_i^{-1}(V_i) = \prod_{i \in I }V_i$$ where the rhs satisfies $V_i = X_i$ for $i \in I \setminus K$. Note that therefore $V_i \ne X_i$ is possible only for $i \in K$, i.e. only for finitely many $i$.
But we have $x = (x_i) \in \bigcap_{i \in K}pr_i^{-1}(V_i)$ iff $x_i = pr_i(x) \in V_i$ for $i \in K$ iff $x = (x_i) \in \prod_{i \in I }V_i$ where $V_i = X_i$ for $i \in I \setminus K$.