$\mathbb{Q}$ is not $G_{\delta}$ on the Michael Line

Let $M=\mathbb{R}$ and $\tau_M=\{U\cup A: U$ open in $\mathbb{R}, A\subset \mathbb{R} \setminus \mathbb{Q}\}$. Then $(M,\tau_M)$ is a topological space called the Michael Line.

So how do I show that $\mathbb{Q}$ is not $G_\delta$ on the Michael Line.

I did show that $\mathbb{Q}$ is not $G_\delta$ in $\mathbb{R}$ with its usual topology using Baire Catagory Theorem. So how does it follow from this?

Thank you for any help or idea!

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Suppose otherwise, that $\mathbb{Q} = \bigcap_{n \in \mathbb{N}} ( U_n \cup A_n )$. Since each $A_n$ is disjoint from $\mathbb{Q}$ it must be that $\mathbb{Q} \subseteq U_n$ for all $n$, and therefore $\mathbb{Q} = \bigcap_{n \in \mathbb{N}} U_n$.
If $\Bbb Q$ were a $G_\delta$ set, then (by definition) there would be some countable collection $\{U_n\cup A_n\}_{n\in\Bbb N},$ with each $U_n$ open in $\Bbb R$ and each $A_n\subseteq\Bbb R\smallsetminus\Bbb Q$ such that $$\Bbb Q=\bigcap_{n\in\Bbb N}(U_n\cup A_n).$$ Since we necessarily have $\Bbb Q\cap A_n=\emptyset$ for all $n\in\Bbb N,$ then $\Bbb Q\subseteq U_n$ for all $n\in\Bbb N$. Hence, $\Bbb Q$ would be a countable intersection of (standard) open sets, which you've proved is not the case.
Thank you very much. Oh, is it right if I conclude that the Michael Line is also not metrizable because there is no metric that generates $\tau_M$? Can I say it is because every ball is closed in M? – Akaichan Feb 22 '13 at 5:07
Thank you for the link. It's great. But on result #5, the writer says $\mathbb{Q}$ is a closed set. I believe that $\mathbb{Q}$ is neither open or close so am I missing something? Sorry for so many questions. – Akaichan Feb 22 '13 at 5:26
$\Bbb Q$ is readily closed in the Michael line, since $U=\emptyset$ is open in the usual topology and so $\Bbb R\smallsetminus\Bbb Q$ is open in the Michael line. – Cameron Buie Feb 22 '13 at 5:31