Prove that $\mathbb{Q}\cap[0,1]$ is not a $G_{\delta}$ set. 
Prove that $\mathbb{Q}\cap[0,1]$ is not a $G_{\delta}$ set.

I know I need to make use of the Baire category theorem, but I don't see how. I also assume a proof by contradiction would be used here, i.e. assume $\mathbb{Q}\cap[0,1]$ is a $G_{\delta}$ set, then there is a countable collection of open sets $\{\mathcal{O}_n\}$ for which
$$
\mathbb{Q}\cap[0,1]=\bigcap_{n=1}^{\infty}\mathcal{O}_n.
$$
And I take it this would lead to some contradiction about the Baire category theorem, which says that the intersection of any countable open dense sets is dense.
 A: Hint :
$$\emptyset = \mathbb{Q} \cap \left( \mathbb{R} \setminus\mathbb{Q}\right)= \mathbb{Q} \cap \left( \bigcap_{q \in \mathbb{Q}} \mathbb{R} \setminus \lbrace q \rbrace\right)$$
If $\mathbb{Q}$ was a $G_{\delta}$, then $\emptyset$ would be a countable intersection of dense $G_{\delta}$'s. Can you see the contradiction ?
A: $[0, 1]$ is a complete metric space so we can apply the Baire category theorem to it. If we had such a countable intersection $[0, 1] \cap \mathbb Q = \bigcap U_i$ then each $V_i = U_i \cap [0, 1]$ would be an open subset of $[0, 1]$ containing $\mathbb Q$. Hence, each $V_i$ is open and dense in $[0, 1]$. Furthermore, we'd still have $[0, 1] \cap \mathbb Q = \bigcap V_i$
On the other hand, let $\{q_i\}_{i \in \mathbb N}$ be an enumeration of $\mathbb Q \cap [0, 1]$. Then $V_i - \{q_i\}$ will still be open and dense in $[0, 1] \cap \mathbb Q$, so the intersection $\bigcap (V_i - \{q_i\})$ must be dense in $[0, 1]$ by the Baire category theorem. However, $\bigcap (V_i - \{q_i\}) = (\bigcap V_i) - [0, 1] \cap \mathbb Q = \emptyset$. But the empty set isn't dense in $[0, 1]$ - contradiction. Hence, $\mathbb Q \cap [0, 1]$ is not a $G_\delta$.
