Suppose each $X_i$ is a topological space with basis $\mathbb{B}_i$.
Set $\mathbb{B}=\{ \prod_iB_i $ $:$ $B_i\in \mathbb{B}_i$ for finitely many $i$ and $B_i=X_i$ for remaining indices. $\}$
Then $\mathbb{B}$ is a basis for the product topology.
My attempt:
- I must show that $\prod_iB_i$ is open. Let $x=(x_i)_{i\in I} \in \prod_iB_i$, so $x_i\in B_i$ for all $i\in I$. Suppose $(x_{i_1},....x_{i_n})\in B_{i_1}\times....\times B_{i_n}$ and $x_j\in X_j$ for $j\notin \{ i_1,....,i_n \}$. Then, observe:
$\prod_iB_i=\bigcup_{k=1}^n\pi^{-1}_{i_k}(B_{i_k})$ $\cup \bigcup_{j\neq i_1....,i_n }\pi_j^{-1}(X_j)$ which is the union of open sets, hence open.
- Let $U$ be open in $\prod_i X_i$. So, for each $x\in U$, there exists $\bigcap_{j=1}^n\pi_{i_j}^{-1}(U_{i_j})$ containing $x$, and contained within $U$.
this means that $x_{i_1}\in U_{i_1}, x_{i_2}\in U_{i_2},.... x_{i_n} \in U_{i_n}$ and therefore:
$x\in \prod_{i\in I}B_i\subseteq U$ where $x_{i_k} \in B_{i_k}\subseteq U_{i_k}$ and $B_j=X_j$ for $j\neq i_m$. $(B_{i_m}$ is a basis member in $X_{i_m}$).
Is this correct?
No, $\prod_i B_i = \bigcap_{j=1}^n \pi_{i_j}^{-1}[B_{i_j}]$, a finite intersection of (subbasic) open sets, not a union.
Such sets exactly form the finite intersections of the subbasic elements $\pi_i^{-1}[O]$ where $O \subset X_i$ is open and $i \in I$, so it's exactly the base needed to generate the minimal topology that makes all projections continuous, which is the definition of the product topology. So point 2 I don't see the point of. It seems tautological: your definition already says that finite intersections of inverse images of open sets under projections form a base? In that case only the intersection remark (that I made above) is all that's needed.