Consider the space $[0,1]^A$(the product space indexed by $A$ rather than the space of all functions from $A$ to $[0,1]$). Let $B \subset A$ be a finite set, can we have an open set of the form $\pi_B^{-1}(V)$ where $V \subset [0,1]^B$ is not necessarily open? Here $\pi_B: [0,1]^A \to [0,1]^B$ is the projection map.
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I don't think so, because $\pi_B$ is open, and $\pi_B(\pi_B^{-1}(V)) = V$. I'm not fully sure I've understood your question - I don't understand why you're talking about the product $\sigma$-algebra when the question seems to be only about topology. – Izaak van Dongen Nov 05 '23 at 01:01