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The rational number are countable. Let $U_i=\Bbb R \setminus \{x_i\}$ where $x_i \in \Bbb Q$. Can we say $U_i$ is dense subset of $\Bbb R$, and then, by Baire category theorem, $\bigcap_{i=1}U_i$ is also dense subset of $\Bbb R$. And $\bigcap_{i=1} U_i$ is the set of irrational number?

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    $\begingroup$ Yes you can say that since $U_n$ are open and dense in $\Bbb{R}$ and their intersection is the set of irrationals..but in general,proving the density of irrationals with Baire is like hunting a rabbit with hydrogen bombs ;p $\endgroup$ – Marios Gretsas Dec 10 '19 at 1:10
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    $\begingroup$ More interesting is the fact (with the same proof) that $\mathbb R$ is uncountable. $\endgroup$ – GEdgar Dec 10 '19 at 2:17
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Yes, that proof is valid, but the use of Baire is unnecessary and overkill (IMHO), if you just want to show $\Bbb P$ (the irrationals) are dense.

Georg Cantor used this argument (essentially; he used completeness directly, not a derived fact like Baire's theorem, which wasn't formulated as such at that time) for his first proof that $\Bbb R$ is uncountable, which is a also a consequence: if $\Bbb R$ were countable, we could use $\Bbb R\setminus \{x\}$ where $x \in \Bbb R$, etc.

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