Why is $\mathbb{Q}$ not a countable intersection of open sets in $\mathbb{R}$?
The hint shown in the book is to recall the Baire Category Theorem: A countable intersection of open dense sets in $\mathbb{R}$ is again dense.
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Do you see why, if such an intersection existed, all of the sets being intersected must be dense? – Mark Schultz-Wu Jul 18 '18 at 19:25
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Q is already dense, so what is the contradiction? – Shen Chong Jul 18 '18 at 19:28
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Assume $\mathbb{Q} = \bigcap_{n=1}^\infty U_n$ where $U_n$ are open. $\mathbb{Q} \subseteq U_n$ so $U_n$ are dense.
We have
$$\emptyset = (\mathbb{R} \setminus \mathbb{Q}) \cap \mathbb{Q} = \left(\bigcap_{q \in \mathbb{Q}} \mathbb{R} \setminus \{q\}\right) \cap \left( \bigcap_{n=1}^\infty U_n\right)$$
so $\emptyset$ is a countable intersection of dense open sets. This is a contradiction with your hint.
mechanodroid
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