1
$\begingroup$

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.
$\endgroup$
1
3
$\begingroup$

Let $A$ be the set of irrationals in $[0,1]$. Suppose $A=\bigcup_n F_n$, where each $F_n$ is closed. Note that $$ {F_n}^\circ \subset A^\circ =\emptyset $$ 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.

$\endgroup$
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.