Jtdylktuy

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 $S$: $B =
{ Pi_{alphain I} U_alpha : text{there are only finite number of } alpha text{ such that } X_alpha neq U_alpha }$, 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 $S subseteqsigma (p_i)$ 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.