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.
[Math] $\mathbb{Q}$ is not open, is not closed, but is the countable union of closed sets.
general-topology
Best Answer
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.