TITLE: A question about uncountable, dense sets in R
QUESTION [5 upvotes]: 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.

REPLY [5 votes]: The Cantor set $C$ is a nowhere dense uncountable closed set. Let $X=C+Q=\{c+q:c\in C,\;q\in Q\}$ where $Q$ is the set of all rational numbers; i.e., the union of all rational translates of $C$. Then $X$ is an uncountable dense set, in fact, $X$ meets each interval of $\mathbb R$ in a set of cardinality $2^{\aleph_0}$. Since $X$ is meager, its complement also meets each interval of $\mathbb R$ in a set of cardinality $2^{\aleph_0}$.