Given a locally compact Hausdorff space $X$, let $S:= X^\mathbb{N}$ the corresponding product topology. I am trying to prove the following claim : $\sigma(S) = \sigma(p_i)$ where $\sigma(S)$ is the sigma algebra generated by the open sets in $S$, and $\sigma(p_i)$ is the sigma algebra generated by all projections (smallest sigma algebra making all projections measurable).
The argument I am trying to make is the following:
1. $S \subseteq \sigma(p_i)$.
2. $\{p_i^{-1}(E_j) :E_j \text{ is open in } X_i\} \subseteq \sigma(S)$.
In order to show (1), I am trying to understand how an element of $S$ looks like. I know that the following set is a base topology for $S$: $B =
\{ \Pi_{\alpha\in I} U_\alpha : \text{there are only finite number of } \alpha \text{ such that } X_\alpha \neq U_\alpha \}$, and proving that an element of $B$ is in $\sigma(p_i)$ can be done in the following manner:
Let $b \in B$, and the denote by $J$ the index of sets $U_i$ such that $U_i \neq X_i$. Then $b = \cap_{j \in J} p_j^{-1}(U_j)$, which is a finite (thus countable) intersection and therefor in $\sigma(p_i)$.
However, as far as I understand, a topology is any union of base elements, not necessarily countable. So I am not convinced that in order to show that $S \subseteq\sigma (p_i)$ it's enough to show that $B \subseteq \sigma(p_i)$ since sigma algebra is generated only by a countable union. Should I try to characterize a general element of $S$? Any tips about the next step?
BTW - I suspect that (2) is very easy: it is known that $p_i$ are continuous, so $p^{-1}_i(E)$ is open if $E$ is open, therefore in $\sigma(S)$. Is it correct?
Note that $\sigma(S)$ is a $\sigma$-algebra that makes all projections measurable, as $p_i^{-1}[O]$ for $O$ open, is open in the product topology and hence in $\sigma(S)$. So by minimality, $\sigma(p_i) \subseteq \sigma(S)$.
On the other hand, for all projections $p_i$ and all open $O$ in $X$, $p_i^{-1}[O] \in \sigma(p_i)$ as $p_i$ is measurable for $\sigma(p_i)$ by definition. So $\sigma(p_i)$ contains all open sets (as these are finite intersections of such sets) and so by minimality again, $\sigma(S) \subseteq \sigma(p_i)$.
Hence equality ensues.