# $\mathbf{Q}$ is not a countable intersection of open sets $\rightarrow$?

I need help understanding a statement. I have been told that :

$\mathbf{Q}\subset\mathbf{R}$ is not a countable intersection of open sets. In other words, $\mathbf{Q}\neq\bigcap\limits_{i=1}^{\infty}U_i$ (where each $U_i$ is open). This implies that there is no function $f:\mathbf{R}\rightarrow\mathbf{R}$ such that $f$ is continuous at every $x\in\mathbf{Q}$ and is discontinuous at every $x\notin\mathbf{Q}$

I am not sure how the first two sentences imply the last one.

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See here math.stackexchange.com/questions/518632/… for a proof that the set of points of continuity is $G_{\delta}$. –  Deven Ware Nov 5 '13 at 5:36
I am sorry, but I haven't learned anything regarding $G_\delta$. Does it mean that it is Borel measurable? How would I prove this statement without using $G_\delta$? –  user97554 Nov 5 '13 at 5:41
Nevermind: en.wikipedia.org/wiki/G%CE%B4_set –  user97554 Nov 5 '13 at 5:43
@user97554 $G_\delta$ set is countable intersection of open sets. You gives $\Bbb{Q}$ is not $G_\delta$ already. –  tetori Nov 5 '13 at 5:45

A subset of $\mathbb R$ is called a $G_{\delta}$-set if it is the intersection of countably many open sets.

Theorem. For any function $f:\mathbb R\to\mathbb R$, the set of all points at which $f$ is continuous is a $G_{\delta}$-set.

Proof. Let a function $f:\mathbb R\to\mathbb R$ be given. For each $n\in\mathbb N$, let $U_n$ be the union of all open sets $D$ such that $|f(x)-f(y)|\lt\frac1n$ for all $x,y\in D$, and let $A=\bigcap_{n=1}^{\infty}U_n$. Clearly $A$ is a $G_{\delta}$-set, being the intersection of the open sets $U_n$. It is easy to verify that $f$ is continuous at a point $x_0\in\mathbb R$ if and only if $x_0\in A$.

Theorem. A countable dense subset of $\mathbb R$ cannot be a $G_{\delta}$-set.

Proof. Let $A=\{a_n:n\in\mathbb N\}$ be a dense subset of $\mathbb R$, and assume for a contradiction that $A=\bigcap_{n=1}^{\infty}U_n$ where each $U_n$ is an open set. Since $A$ is dense, each $U_n$ is dense. Let $I_0$ be some closed interval (understood to be of the form $[a_0,b_0]$ with $a_0\lt b_0$). Inductively define for each $n\in\mathbb N$ a closed interval $I_n\subseteq I_{n-1}$ so that $a_n\notin I_n$ and $I_n\subseteq U_n$. The intersection of the nested closed intervals $I_n$ contains a point $x$, which must be in $A$ since it is in each $U_n$, but is not equal to any $a_n$. This contradiction proves the theorem.

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If such function exists, then we can consider

$$A=\{x\in\Bbb{R}:f(x)\text{ is continuous at }x\}.$$

But it is known that the set of points of continuity of a real function is $G_\delta$.

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