A set is of first category if it is the union of nowhere dense sets and otherwise it is of second category.
How can we prove that irrational numbers are of second category and the rationals are of of first category?
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1Do you mean a countable union? – T. Eskin Dec 24 '12 at 09:55
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$\mathbb Q = \bigcup_{q \in \mathbb Q} \{ q \}$ hence the rationals are a countable union of nowhere dense sets.
Assume the irrationals are also a countable union of nowhere dense sets: $I = \bigcup_{n \in \mathbb N} U_n$. Then $\mathbb R = \bigcup_{q \in \mathbb Q} \{ q \} \cup \bigcup_{n \in \mathbb N} U_n$ is also a countable union of nowhere dense sets.
Rudy the Reindeer
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Recall that $\mathbb{R\setminus Q}=\bigcap_{q\in\mathbb Q}\mathbb R\setminus\{q\}$.
This is a countable intersection of dense open sets, and by Baire's category theorem the result is dense, i.e. second-category.
Asaf Karagila
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How do you conclude that since the intersection is dense, then it is of second category. Isn't it the other way around ? – Svetoslav Feb 09 '16 at 16:26
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2The countable union of first-category sets is first-category. And since $\Bbb R$ is not the union of two sets of first-category you get the result. But it is also often phrased in terms of intersection of dense open sets being of second-category (which means necessarily that the complement is first-category, at least in the case of complete metric spaces). – Asaf Karagila Feb 09 '16 at 16:32
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Thanks for the fast response. Since you are online, I would like to ask you about this example http://math.stackexchange.com/a/459856/254733 – Svetoslav Feb 09 '16 at 16:40
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It is clear that the intersection is dense in $\mathbb R$ since $\mathbb R$ is a Baire space, or just by noting that it contains all the rationals. But why is this set uncountable ? The argument is that it is of second category. I can not see why it is of second category. – Svetoslav Feb 09 '16 at 16:42
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