Is $\mathbb{R}\setminus\mathbb{Q}$ a union of countable family of closed sets?

Question

Can we represent set of irrational numbers as union of countable family of closed sets?

Answer 1

The answer is NO.

If $\mathbb R\!\smallsetminus\!\mathbb Q=\bigcup_{n\in\mathbb N}F_n$, where $F_n$, $n\in\mathbb N$, are closed, then $$ \mathbb Q=\mathbb R\smallsetminus\!\bigcup_{n\in\mathbb N}F_n =\bigcap_{n\in\mathbb N} (\mathbb R\!\smallsetminus\! F_n)=\bigcap_{n\in\mathbb N} U_n $$ where $U_n=\mathbb R\!\smallsetminus\! F_n$ open. Clearly, each of these $U_n$'s is also dense, as their intersection is dense.

If we now write $\mathbb Q$ as $\mathbb Q=\{q_n\}_{n\in\mathbb Q}$, as set $V_n =U_n\!\smallsetminus\!\{q_n\}$, then these $V_n$'s are also open and dense and $$ \bigcap_{n\in\mathbb N}V_n=\varnothing, $$ which contradicts Baire's Theorem.

Answer 2

No, this is not possible, for if it was, ${\mathbb R}$ would be the union of countably many nowhere dense closed subsets, contradicting Baire's category theorem.