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.

If you want to partition $\Bbb R$ into two uncountable dense subsets, you can take the positive rationals and negative irrationals as one set and the non-positive rationals and positive irrationals as the other.  You can partition $\Bbb R$ into uncountably many disjoint countable dense subsets, by taking the quotient $\Bbb {R/Q}$
