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.
 A: 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.

A: 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 $\Bbb {R/Q}$
A: 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 $2^{\aleph_0}$.
A: To follow up on your idea, the following statement is true: the irrationals $\mathcal I$ minus any countable set $C=\{c_1,c_2,\ldots\}$ is still dense in $\mathbb R$.
To see this, let $a,b$ be two reals numbers, $a<b$. Then it is well-known $\mathcal I\cap (a,b)$ is uncountable. Hence $(\mathcal I\setminus C)\cap (a,b)$ is nonempty. So there is an element of $\mathcal I\setminus C$ between any two reals, so that set is dense.
Edit: I see I am answering the statement in the question: "I'm trying to see if you can take the set of irrationals, remove a countable amount, and still have a dense subset in $\mathbb R$," which is completely different from the next statement: "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." I'm leaving this answer since it does answer part of the question body!
