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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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Do you mean a countable union? –  Thomas E. Dec 24 '12 at 9:55

2 Answers 2

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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.

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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.

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