# Problem on Baire's category theorem.

I'd like to show that the set of irrational numbers in $$[0,1]$$ cannot be represented as a countable union of closed sets.

The hint says to use Baire's category theorem. I know two versions of such theorem:

1. Baire's category theorem for complete metric spaces. If $$(X,d)$$ is a complete metric space, then $$X$$ is a Baire space.
2. Baire's category theorem for locally compact spaces. If $$(X, \mathcal{T})$$ is a locally compact Hausdorff space, then $$X$$ is a Baire space.

Let $$A$$ be the set of irrationals in $$[0,1]$$. Suppose $$A=\bigcup_n F_n$$, where each $$F_n$$ is closed. Note that $$\overset{\circ}{F_n} \subset \overset{\,\,\circ}A =\emptyset$$ so that each $$F_n$$ has empty interior.
On the other hand, putting $$B=\mathbb{Q}\cap [0,1]$$, we obviously have $$B=\bigcup_{b\in B}\{b\}$$. Singletons are closed and have empty interior. Note also that $$B$$ is countable. Write
$$[0,1]=A\cup B=\bigcup_n F_n \cup \bigcup_{b\in B}\{b\}$$ This shows that $$[0,1]$$ can be written as a union of countably many closed sets with empty interior. As $$[0,1]$$ is a complete metric space, the Baire Category Theorem implies that $$[0,1]$$ has empty interior. But this is a contradiction.
Suponer que existen conjuntos cerrados $$C_i$$ tales que $$[0,1]\setminus \mathbb Q\cap[0,1]=\bigcup_{i =1}^{\infty}C_i$$. Entonces, $$[0,1] = \bigcup_{i =1}^{\infty} C_{i} \cup \bigcup_{q \in \mathbb{Q}\cap[0,1]} \{q\}$$ y al aplicar Baire, concluimos que uno de los $$C_i$$ tiene interior no nulo. Pero se sabe que el interior del conjunto $$[0,1]\setminus \mathbb Q\cap[0,1]$$ es nulo, así que llegamos a una contradicción.