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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

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    $\begingroup$ A point is a $G_\delta$ set which is nowhere dense and has empty interior. So is the empty set, for that matter. $\endgroup$ – MartianInvader Jan 2 '14 at 17:13
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    $\begingroup$ $G_{\delta}$ sets are also Borel sets, if that helps. $\endgroup$ – Wintermute Jan 2 '14 at 17:16
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    $\begingroup$ 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 $E.$ $\endgroup$ – Dave L. Renfro Jan 2 '14 at 17:39
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    $\begingroup$ 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). $\endgroup$ – Asaf Karagila Jan 2 '14 at 17:58
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    $\begingroup$ Shir, the function defined by $f(x) = x$ for rational $x$ and $f(x)=-x$ for irrational $x$ is continuous only at $x=0$. $\endgroup$ – Santiago Canez Jan 2 '14 at 18:19
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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 $\mathbb{R}$.

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  • $\begingroup$ I see. ok. I'll go over these two statements. Thank you!! $\endgroup$ – topsi Jan 5 '14 at 9:54
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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.

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