Let $S$ be a measurable space and $T$ an uncountable index set. For a subset $I\in T$, we write $\pi_1:S^T\to S^I$ for the natural projection. Show that $S^{\otimes T}=\underset{I \,is\, a\, countable\, subset\, of\, T}{\cup}\,\,\,\pi_1^{-1}(S^{\otimes I})$.
Consequently, if $S$ is a topological space with a nontrivial topology, then $B(S^T)\supset B(S)^{\otimes T} $, where $B$ refers to the Borel algebra.
Intuitively, I see why $S^{\otimes T}$ should be the limit of the $S^{\otimes I}$ taken over all the countable subsets of $T$, but I'm not sure how to prove this. I took an element of the LHS. It should belong to the RHS, which is a union, so we need to show that it belongs to one of the elements of the union, but which one? I think I can prove that $RHS\subseteq LHS$. This is because $S^{\otimes I}$ can naturally be identified with a subset of $S^{\otimes T}$ (just let the indices not in $ I$ map to $0$).