# basis for product topology finitely many

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:

1. 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.

1. 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.