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:
- Show that $\mathbb{R} – \mathbb{Q}$ is not the countable union of closed sets.
- Show that $\mathbb{Q}$ is not the countable intersection of open sets.
My question is: For the second part, using the first part of the problem, is it correct to say that
$$ (\mathbb{R} – \mathbb{Q}) \neq \bigcup_ {n\in\mathbb{N}} C_n \text{ , where $C_n$ is a closed set $\Rightarrow$ }$$
$$(\mathbb{R} – \mathbb{Q})^c \neq (\bigcup_ {n\in\mathbb{N} } C_n)^c \text{ , where little c is the complement}\Rightarrow $$
$$(\mathbb{Q}) \neq \bigcap_ {n\in\mathbb{N} } (C_n)^c =\bigcap_ {n\in\mathbb{N}} O_n \text{ , where $O_n$ is an open set} $$
Best Answer
I follow what you have done, but I don't think the logic quite works. I think you successfully show that $Q$ is not the intersection of some specific open sets, $(C_n)^c$ formed from a specific choice of closed sets $C_n$.
The standard proof would be by contradiction starting from the other end......
Assume that $O_n$ is any countable collection of open sets with $Q = \cap O_n$
Then $ \mathbb R \setminus \mathbb Q = \mathbb R \setminus \cap O_n = \cup \mathbb R \setminus O_n$
And it is the case that the complement of any open set is closed, so that with $C_n = R \setminus O_n$ then $\mathbb R \setminus \mathbb Q = \cup C_n$ would be the countable union of closed sets, which contradicts the first part.