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Given a family of sets $G\subset P(X)$, can one characterize by "closure properties" alone whether or not $G$ arises as the family of all $G_\delta$ for some topology on $X$? some Polish space topology on $X$?

Of course I'm not interested in the trivial answer, namely: if there exists a system of sets $T$ closed under unions and finite intersections such that every $A\in G$ is a countable intersection of sets in $T$.

I don't feel sure about the best way to bar this non-answer! Feel free to improve my question if you have an answer.

I suppose I'm looking for something like this: (positive) statements with only universal quantifiers over combinatorially-restricted families of sets in $G$. Combinatorially restricted can allows for restrictions on the cardinality of the family, disjointness, nestedness, etc.

This is the easiest question of this type where I don't know the answer. So how about arbitrary levels of the Borel hierarchy?

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A minor correction: your trivial answer isn't quite stated correctly, since you also need the other direction of your requirement, so that $A\in G$ if and only if $A$ is the intersection of countably many sets in $T$. – Joel David Hamkins Oct 9 '12 at 2:24
Hmm, why can't we say that a collection of subsets $G$ is the $G_{\delta}$ if and only if the sets are closed under unions and countable intersections? If this condition is satisfied then we can let $G$ be a topology, which gives a $P$-space, i.e. the $G_{\delta}$ sets are precisely the open sets. – Gjergji Zaimi Oct 9 '12 at 2:39
Gjergji, the $G_\delta$ sets in $\mathbb{R}$ include all singletons, and so are not closed under arbitrary unions. – Joel David Hamkins Oct 9 '12 at 2:53
Ah, of course. That was silly of me. – Gjergji Zaimi Oct 9 '12 at 2:55
$G_\delta$'s aren't even closed under countable disjoint unions or countable nested unions since ${\Bbb Q}$ is not $G_\delta$ in ${\Bbb R}$. – David Feldman Oct 9 '12 at 4:11

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