The question has actually 2 parts. I know the question has been asked before, but I want to know if my reasoning makes sense.I've done the first part but am unsure about the second. It goes as follows:
$\text{1) Show that} \ (\mathbb{R} - \mathbb{Q}) \ \text{is not the countable union of of closed sets} \\ \text{2) Show that} \ (\mathbb{Q}) \ \text{is not the countable intersection of open sets}$
First I should ask: Can I just argue that as the countable union of closed sets is closed and since $(\mathbb{R} - \mathbb{Q})$ is neither open nor closed the statement would make no sense? (an analogous case could be used in the $\mathbb{Q}$ part.
My second question is: For the second part, not using the argument above but using the first part of the problem, is it correct to say that $$ (\mathbb{R} - \mathbb{Q}) \neq \bigcup_ {n\in\mathbb{N}} C_n \text{ , where $C_n$ is a closed set $\Rightarrow$ }$$ $$(\mathbb{R} - \mathbb{Q})^c \neq (\bigcup_ {n\in\mathbb{N} } C_n)^c \text{ , where little c is the complement}\Rightarrow $$ $$(\mathbb{Q}) \neq \bigcap_ {n\in\mathbb{N} } (C_n)^c =\bigcap_ {n\in\mathbb{N}} O_n \text{ , where $O_n$ is an open set, which would prove part 2?} $$
