$\mathbb{Q}$ is not open, is not closed, but is the countable union of closed sets. I want to prove that $\mathbb{Q}$ (the set of rational numbers) is not open, is not closed, but is the countable union of closed sets. I tried to show that $\mathbb{Q}$ doesn't contain all of its limit points which would imply not closed. However, I am not able to prove the other two things. I need little help to prove this. Thanks.
 A: $\Bbb Q$ is not closed because it is dense, and if a set is both dense and closed then it is equal to the whole space (in this case, $\Bbb R$).
$\Bbb Q$ is not open either because open sets are either empty, or contain an interval which makes them uncountable; but $\Bbb Q$ is countably infinite so it is neither empty nor uncountable.
Lastly $\Bbb Q$ is the countable union of singletons, all of which are closed.
Interestingly enough, $\Bbb Q$ cannot be written as an intersection of countably many open sets.
A: Hint
$$r\in\mathbb{Q},\quad\forall \epsilon>0, (r-\epsilon,r+\epsilon)\not\subset\mathbb{Q}$$
$$x\in\mathbb{R}\setminus\mathbb{Q},\quad\forall \epsilon>0, (x-\epsilon,x+\epsilon)\not\subset\mathbb{R}\setminus\mathbb{Q}$$
$$\displaystyle\mathbb{Q}=\bigcup_{r\in\mathbb{Q}}\{r\}$$
A: Since any neighborhood $(q - \epsilon,  q + \epsilon)$ of a rational $q$ contains irrationals, $\mathbb{Q}$ has
no internal points. This implies that $\mathbb{Q}$ is not open. Since every irrational number  is the limit of a sequence of rationals,  $\mathbb{Q}$ is not closed (for a set to be closed it should contain all of its limit points). Since every one-point-set $\{x\} \subset \mathbb{R}$ is closed, and since $\mathbb{Q}$ is countable, we have that
$$\mathbb{Q} = \bigcup_{p\in \mathbb{Q}} \{p\}$$
is a  countable union of closed sets.
