# Can we use the Baire category theorem to prove the denseness of irrational number?

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?

• 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 – Marios Gretsas Dec 10 '19 at 1:10
• More interesting is the fact (with the same proof) that $\mathbb R$ is uncountable. – GEdgar Dec 10 '19 at 2:17

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.