# Countable Dense Set not a $G_{\delta}$ set? Where's the contradiction?

Prove the following theorem:

Theorem: If $D$ is a countable dense subset of $\Bbb{R}$, there is no function $f : \Bbb{R} \to \Bbb{R}$ that is continuous precisely at the points of $D$.

(a) Show that if $f : \Bbb{R} \to \Bbb{R}$, then the set $C$ of points at which $f$ is continuous is a $G_{\delta}$ set.

(b) Show that $D$ is not a $G_{\delta}$ set in $\Bbb{R}$ [Hint: Suppose $D = \cap W_n$, where $W_n$ is open in $\Bbb{R}$. For $d \in D$, set $V_d = \Bbb{R}-\{d\}$. Show that $W_n$ and $V_d$ are dense in $\Bbb{R}$.]

I already worked on part (a), but I am having trouble with part (b). Presumably, I am asked to prove the following: If $D \subseteq \Bbb{R}$ is a countable dense set, then $D$ cannot be a $G_{\delta}$ set. Clearly $V_d$ is dense, the countable intersection of them is dense:

$$\Bbb{R} = \overline{\bigcap_{d \in D} V_d} = \overline{\bigcap_{d \in D} (\Bbb{R} - \{d\}} = \overline{\Bbb{R} - D}$$

Moreover, since $D \subseteq W_n$ for every $n$, clearly the $W_n$ are dense...But what was the point of showing this? To conclude that $\bigcap W_n$ is dense? We already know this as $D$ was assumed to be dense. Also, where is the contradiction in all of this? I honestly don't see any contradiction

On the one hand, $\left ( \bigcap_{n \in \mathbb{N}} W_n \right ) \cap \left ( \bigcap_{d \in D} V_d \right )=\emptyset$. On the other hand, since $W_n$ and $V_d$ are both open and dense, the Baire category theorem would imply that this intersection should be dense again. This is a contradiction.