A question about uncountable, dense sets in R If you have an uncountable subset of $\mathbb{R}$, which is dense in $\mathbb{R}$, is its complement countable? 
I'm trying to see if you can take the set of irrationals, remove a countable amount, and still have a dense subset in $\mathbb{R}$.If this is the case, this sets complement would still be countable.
My goal is to determine if you can have an uncountable subset of $\mathbb{R}$, which is dense in $\mathbb{R}$, yet its complement is uncountable.
If somebody could just let me know if this is possible or not I would appreciate it. Thanks.

No: Try considering a countable dense subset, and attach an interval.

 Specifically, we let  $$A = \Bbb{Q} \cup [0, 1]$$ It is dense in $\Bbb{R}$ because it contains $\Bbb{Q}$, it is uncountable because it contains $[0, 1]$, and it has uncountable complement.

