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Of course, every open set in every topological space has the Baire property and belongs to the Baire property $\sigma$-algebra, which is why we have to be careful to distinguish these notions, all arising from generalizing results in Réné Baire's thesis.
Perhaps some background information would be helpful. An open set is not necessarily a Baire set (a set whose indicator function is a Baire function, or equivalently belongs to the Baire $\sigma$-algebra, generated by the zero sets). A compact Baire set is $G_\delta$, so in a compact Hausdorff space, the Baire open sets are the $F_\sigma$ open sets. Therefore if $X$ is uncountable, the complement of a point in the product space $2^X$ is a non-Baire open set. In a perfectly normal space, e.g. a metrizable space, every open set is a Baire set.
@DanielAsimov See my comment to another answer to this same question. "Positive reals" means reals $\geq 0$ in functional analysis, those $> 0$ are "strictly positive reals".
We can't use open sets, as indicated by fedja. We can find a $G_\delta$ subset of $\mathbb{R}^3$ whose projection onto $\mathbb{R}^2$ is a complete analytic set (other sense, obviously), and therefore not Borel. The projection map $\mathbb{R}^3 \rightarrow \mathbb{R}^2$ is linear, and therefore real-analytic. I think this is essentially the original non-Borel subset.
@DanielLoughran When I read the question, I assumed that a connected groupoid was required to be non-empty, like a connected topological space or a connected graph, for the same reason that 1 is not a prime number. So non-emptiness of $\mathcal{G}(\overline{k})$ is not missing (but maybe Niels uses a different definition of connectedness of groupoids).
Although product is surjective, interestingly, the squaring map is not surjective. The function $x^3 e^{-x^2}$ is not a square, because it vanishes to order 3 at 0, so it isn't the square of a smooth function, because 3 is not even. The $n$th power map is not surjective for similar reasons.
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@IosifPinelis I mean "What ChatGPT is" in a very literal sense. As in how it works. So neither of your two options. I emphasize again that I have not downvoted you.
You're never going to get all compact Hausdorff spaces as a limit of finite spaces because Stone spaces are closed under limits (because the product of Stone spaces is Stone and a closed subspace of a Stone space is Stone).
See this earlier question. I intend to expand my comments to an answer but I've been quite busy lately. In short, the category of compact Hausdorff spaces is the free $\aleph_1$-cofiltered completion of the category of compact metrizable spaces (or closed subsets of $[0,1]^{\mathbb{N}}$ if you need a small rather than just essentially small category when you take a completion). This is best viewed under Gelfand duality - a commutative unital C$^*$-algebra is an $\aleph_1$-filtered colimit of separable commutative unital C$^*$-algebras.
You can drop "complete". However, on a complete metric space having separable support is equivalent to being Radon, and so the existence of a real-valued measurable cardinal is also equivalent to there being a non-Radon measure on a complete metric space. (You can prove that there is a non-Radon measure on an incomplete separable metric space only using ZFC, but it requires the C.)
When F. Burton Jones, the student of Moore, proved that $2^{\aleph_0} < 2^{\aleph_1}$ implies every separable normal Moore space is metrizable, he included the hypothesis explicitly (see Theorem 5, page 6 of the pdf). He does say "The author has tried for some time without success to prove $2^{\aleph_1} > 2^{\aleph_0}$", but forcing wouldn't be invented for 26 years, so it would have seemed reasonable at the time (especially since Moore didn't let students consult the literature).
