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https://math.stackexchange.com/a/61110/1248511 I just saw this proof for showing that the rationals are not a Gδ set(without using baire's theorem) and while I'm not really sure about what exactly I'm asking, I wanted to know if there's any relation between the uncountability of the irrationals and the fact that they can be expressed as the intersection of open sets

NoetherBoy
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    Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking. – Community Nov 10 '23 at 10:23
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    I think you talk about countability rather than commensurability. Yes, no countable dense set is $G_\delta$, but there are a lot of other subsets that also are not. – mihaild Nov 10 '23 at 10:32
  • @mihalid Yes, thank you, I mixed up the terms. Does this mean that every uncountable, dense set is Gδ? – NoetherBoy Nov 10 '23 at 10:37
  • @NoetherBoy no, there are uncountable dense sets that are not even Borel (Bernstein sets for example) – Alessandro Codenotti Nov 10 '23 at 10:42
  • @mihalid Ah i see. But there could never be a countable dense set Gδ, solely because of its countablity property? – NoetherBoy Nov 10 '23 at 10:51
  • @AlessandroCodenotti Ah ok, does this mean that every countable dense set is not Gδ just because of it's countability alone? – NoetherBoy Nov 10 '23 at 11:03
  • @NoetherBoy the proof you linked works for any countable dense set. This doesn't say anything about uncounatble or non-dense sets - some of them are $G_\delta$, some are not. – mihaild Nov 10 '23 at 13:58

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