Let $\prod_iX_i$ be equipped with the product topology. Suppose each $X_i$ has sub basis $\S_i$.
Show that $S=\{ \pi_i^{-1}(S_i) : S_i\in \S_i \forall i \}$ is a subbasis for $\prod_iX_i$.
My attempt:
It suffices to show that the topology generated by $S$ is $\prod_iX_i$.
The topology is generated by the basis $\{$ $\bigcap_{j=1}^n: \pi_{i_j}^{-1}(S_{i_j})$ $:$ $S_{i_j}\in \S_{i_j} \forall i_j \}$
Let $U$ be open in $\prod_iX_i$. Let $x\in U$. So, there exists basis member $x\in \bigcap_{j=1}^n\ \pi_{i_j}^{-1}(U_{i_j})\subseteq U$.
Now, $U_{i_j}$, being open in $X_{i_j}$ means that $U_{i_j}=\bigcup_{m\in I}\bigcap_{k=1}^m(S_{i_k})$.
But, I'm not sure if all this is correct so far. May I have help?
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**EDIT**
Okay, to make things less complicated. Observe the lemma:
Lemma: Let $\prod_iX_i$ be a space equipped with the product topology. Suppose each $X_i$ is a topological space with basis $\mathbb{B}_i$.
Put $\mathbb{B}$ $=$ $\{$ $\bigcap_{j=1}^{n}\pi_{i_j}^{-1}(B_{i_j})$ $:$ $B_{i_j}\in \mathbb{B}_{i_j}$, $n\in \mathbb{N}$, $i_j\in I$ $\}$.
Then $\mathbb{B}$ is a basis for $\prod_i X_i$.
Now, I will use the above lemma to answer the original problem.
It suffices to show that $\mathbb{B}$ is the basis generated by $S$. i.e.
I must show that $\mathbb{B}= \{$ $\bigcap_{j=1}^n \pi_{i_j}^{-1}(S_{i_j})$ $:$ $S_{i_j}\in \S_{i_j} \forall i_j \}$.
Since each $S_{i_j}\in \mathbb{B_{i_j}}$, we get the reverse inclusion.
Observe, given a basis element $B\in \mathbb{B}_i$, $B= \bigcap_{j=1}^nS_j$ where $S_j\in \S_i$ . Hence, $\bigcap_{j=1}^n\pi_{i_j}^{-1}(B_{i_j})$ $=$ $\bigcap_{k=1}^p \pi_{k}^{-1}(S_k)$.
You can avoid using or mentioning bases completely: I use the definition that $\mathcal{S}_i$ is a subbase for $(X_i, \mathcal{T}_i)$ iff $\mathcal{T}_i$ is the smallest topology that contains $\mathcal{S}_i$. And the definition for the product topology $\mathcal{T}_p$ on $X=\prod_{i \in I} (X_i, \mathcal{T}_i)$ is the smallest topology that makes all projections $\pi_i: X \to X_i$ continuous. It's this combination of minimality that will tell us that $\mathcal{T}_p$ is the smallest topology that contains $$\mathcal{S} = \{\pi_i^{-1}[S]: i \in I, S \in \mathcal{S}_i\}$$ which thus means that $\mathcal{S}$ is a subbase for $\mathcal{T}_p$. We will also need the standard fact that for any space $(Z,\mathcal{T})$ a function $f: (Z, \mathcal{T}) \to (X_i, \mathcal{T}_i)$ is continuous iff $$\forall S \in \mathcal{S}_i: f^{-1}[S] \in \mathcal{T}$$ so we can "test" continuity on subbases.
Main proof:
Suppose that $\mathcal{T}$ is a topology on $X$ that contains $\mathcal{S}$. For any $i$, and any $S \in \mathcal{S}_i$, $\pi_i^{-1}[S]\in \mathcal{S} \subseteq \mathcal{T}$ so the criterion for continuity from above tells us that $\pi_i$ is continuous. As this holds for all $\pi_i$, all $\pi_i$ are continuous, and by the above definition of the product topology $\mathcal{T}_p \subseteq \mathcal{T}$. It follows (As this holds for all such $\mathcal{T}$) that if $\mathcal{T}(\mathcal{S})$ is the minimal topology that contains $\mathcal{S}$, then $\mathcal{T}_p \subseteq \mathcal{T}(\mathcal{S})$
On the other hand, $\mathcal{T}_p$ is a topology that makes all $\pi$ continuous, and as all $\mathcal{S}_i \subseteq \mathcal{T}_i$, it follows immediately that $\mathcal{S} \subseteq \mathcal{T}_p$ and so $\mathcal{T}_p$ is some topology that contains $\mathcal{S}$ and so $\mathcal{T}(\mathcal{S}) \subseteq \mathcal{T}_p$ by minimality again. So we have two inclusions and
$$\mathcal{T}(\mathcal{S}) = \mathcal{T}_p$$ which exactly says that te product topology has $\mathcal{S}$ as a subbase.