Every dense $G_\delta$ subset of $\Bbb R$ is uncountable Every dense $G_\delta$ subset of $\Bbb R$ is uncountable. I know that I have to use Baire's Theorem but I don't know how.
Thanks!
 A: Let $A$ be a dense $G_\delta$. Then $A =\bigcap\limits_{i=1}^\infty O_i$ where each $O_i$ is open. So, $A^C=\bigcup\limits_{i=1}^\infty O_i^C  $. 
Each $O_i^C$ is closed. 
Show, using the denseness of $A$, that each $O_i^C$ is nowhere dense. 
Once you do this, it follows by definition that $A^C$ is of first category. It then follows that $A$ must be uncountable. Otherwise, $\Bbb R=A\cup A^C$ would be of first category (this is where Baire comes into play).
A: Let $A$ be a dense $G_{\delta}$ set. Let $A=\cap_{n\in \Bbb N}U_n$ where each $U_n$ is open. Each $U_n$ is dense because $U_n\supset A$ and $A$ is dense.  
Let $B=\{b_n:n\in \Bbb N\}$ be any non-empty countable subset of $\Bbb R.$ (It does not matter whether $b_m=b_n$ for some distinct $m,n.$) Let $V_n=U_n\setminus \{b_n\}$. Then $V_n$ is dense and open. 
We have $$A \setminus B=\cap_{n\in \Bbb N}(\;U_n\setminus \{b_n\}\;)= \cap_{n\in\Bbb N}V_n.$$ By the Baire Category Theorem,  $A\setminus B$ is dense. Therefore $A\not \subset B.$ In particular $A\ne B.$ 
