Baire's Theorem and Irrationals I am asked to show that the irrational numbers are not a countable union of closed subsets of $\mathbb{R}$ given that if a complete metric space is the countable union of of closed subsets then at least one of them has a nonempty interior.
So far: I assume that the irrationals are the countable union of closed subsets of  $\mathbb{R}$. I know that the irrationals are uncountable, so we cannot have this union consist of all singleton subsets. Therefore these closed sets must consist of intervals of irrational numbers. 
At this point I cannot see how these intervals must all have a nonempty interior. Since each of these sets is closed, its closure is itself, and and so the closure minus the boundary of the interval should be nonempty given that any interval no matter how small contains infinitely many rationals and irrationals. 
Can someone please point me in the right direction and/or point out the flaws in my reasoning?
Thanks
 A: One statement of the Baire category theorem is that a complete non empty metric space cannot be written as a countable collection of nowhere dense sets.
(A set $N$ is nowhere dense if $(\overline{N})^\circ = \emptyset$.)
(Reminds me of a joke: While discussing the double negative, an english professor announced that there is no double positive in the English language  to which a smart alec in the back row shouted 'Yea, right'.)
So, suppose $\mathbb{Q}^c = \cup_k C_k$, where the $C_k$ are closed and nowhere dense. Since we can write $\mathbb{Q} = \cup_k \{q_k\}$, where the 
$\{q_k\}$ are obviously nowhere dense, then we could write
$\mathbb{R} = \mathbb{Q}^c \cup \mathbb{Q} = \cup_k (C_k \cup \{q_k\})$, which
would contradict the completeness of $\mathbb{R}$.
Hence, if $\mathbb{Q}^c = \cup_k C_k$, at least one $C_k$ must not be nowhere dense, hence $C_k^\circ \neq \emptyset$ and hence contains some non
empty interval $(a,b)$. But since this must contain a rational, we have
a contradiction.
