Given the subbasis $\mathcal{S}_i$ of $X_i$, how may we construct a subbasis of $\prod_i X_i$ (in the product topology).

Question

Let $(X_i,\tau_i)$ be topological spaces and the product topology $(X,\tau)$ we may construct, given the bases $\mathcal{B}_i$ of $\tau_i$, a basis $\mathcal{B}$ of $\tau$ as follows: any element $B_i\in \mathcal{B}$ is of the form $$\prod_j B_j,i$$ where $B_{j,i}\ne X_j$ for finitely many $j$, in which case $B_{j,i}\in \mathcal{B}_i$.

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I've been trying to come up with a similar result for subbasis. In particular, if we are given the bases $\mathcal{S}_i$ of $\tau_i$, how may we construct a subbasis $\mathcal{S}$ of $\tau$?

Answer 1

Let us consider $\pi_j:\Pi_{i\in \Lambda}X_i\to X_j$ defined by $$\pi_j ((x_i)_{i\in \Lambda} =x_j$$

Let $S_j=\{\pi_j^{-1}(B_j):B_j\in\mathcal{B}_j\}$

Then $\mathcal{S}=\bigcup_{i\in \Lambda}\mathcal{S}_j$ is the subbasis of the product topology. (Verify)

(Product topology is the coarset topology for which every projection map is continuous )