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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.
Martin Sleziak
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YENI LAURACIO QUISPE
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2 Answers2

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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.

Antonio Maria Di Mauro
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Reveillark
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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.

Matematleta
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