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I understand that a $G_\delta$ set is a set which is a countable intersection of open sets. My question is: Is there any other characterization for $G_\delta$ sets (on $\mathbb{R}$)? For example, can I say that the interior of this sets is not empty? or that they are dense somewhere (means that they are not nowhere dense)? or any other topological characterization? Somehow I find it difficult to imagine those sets.

Also, How can I prove that the $\mathbb{Q}$ is not a $G_\delta$ set.

Thank you, Shir

topsi
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    A point is a $G_\delta$ set which is nowhere dense and has empty interior. So is the empty set, for that matter. – MartianInvader Jan 02 '14 at 17:13
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    $G_{\delta}$ sets are also Borel sets, if that helps. – Wintermute Jan 02 '14 at 17:16
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    It probably won't help with your visualization, but $E \subseteq {\mathbb R}$ is a $G_{\delta}$ set if and only if there exists a function $f:{\mathbb R} \rightarrow {\mathbb R}$ that is continuous at each point in $E$ and discontinuous at each point not in $E.$ – Dave L. Renfro Jan 02 '14 at 17:39
  • Is there a set which is Borel and not $G_\delta$, or a set which is $G_\delta$ but not Borel? – topsi Jan 02 '14 at 17:39
  • Dave, I can't se the consistency with the argument that a point is a $G_\delta$ set. How can a function on reals be continuous on just one point from the domain? – topsi Jan 02 '14 at 17:44
  • Is the cantor set $G_\delta$? – topsi Jan 02 '14 at 17:50
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    Shir $\Bbb Q$ is Borel, as a countable union of singletons (which are closed); but it is not $G_\delta$. To your second question, the Cantor set is closed and so $G_\delta$ (every closed set is a $G_\delta$ set). – Asaf Karagila Jan 02 '14 at 17:58
  • Yes I see now, ok thank you – topsi Jan 02 '14 at 18:08
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    Shir, the function defined by $f(x) = x$ for rational $x$ and $f(x)=-x$ for irrational $x$ is continuous only at $x=0$. – Santiago Canez Jan 02 '14 at 18:19
  • See http://math.stackexchange.com/q/622739 – GEdgar Jan 02 '14 at 19:42
  • For your question about $\mathbb Q$ not being $G_\delta$, see here: http://math.stackexchange.com/questions/69451/how-to-show-that-mathbbq-is-not-g-delta – Martin Sleziak Jan 10 '14 at 07:41

2 Answers2

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For the question of why $\mathbb{Q}$ is not $G_{\delta}$, you may want to use the following two results:

  1. A non-empty countable complete metric space has an isolation point (a pretty straight forward corollary of Baire category theorem).

  2. A subspace of a complete metric space is completely metrisable if and only if it is a $G_{\delta}$ subset.

Use $1.$ to infer that $\mathbb{Q}$ is not completely metrisable, and $2.$ to conclude that it thus can not be a $G_{\delta}$ subset of the complete metric space $\mathbb{R}$.

T. Eskin
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In the case of the $\Bbb R$, $G_\delta$ subsets do not have many restrictions in terms of interior or denseness. But you can prove that every closed subset of $\Bbb R$ is also $G_\delta$. An equivalent, perhaps easier, thing to prove is that every open subset of $\Bbb R$ is $F_\sigma$. You can certainly prove that an open interval is $F_\sigma$. If you know in addition that every open subset of $\Bbb R$ is a countable disjoint union of open intervals and that a countable union of $F_\sigma$ subsets is still $F_\sigma$, you can prove that result.

The best way to see that $\Bbb Q$ is not a $G_\delta$ will often use the Baire Category Theorem. If you know that $\Bbb R$ is a Baire space, then you can show $\Bbb Q$ is not $G_\delta$ easily with this result that you should do as an exercise: A dense $G_\delta$ subset in any topological space is comeager. Why does this finish it? Because $\Bbb Q$ is already meager as it is a countable union of singletons. If $\Bbb Q$ were comeager, $\Bbb R$ would be the union of two meager subsets.