TITLE: Is complement of a dense set in $\mathbb{R}$ dense in $\mathbb{R}$?
QUESTION [5 upvotes]: $\mathbb{Q}$ is dense in $\mathbb{R}$. Also, its complement, $\mathbb{R-Q}$, is dense in $\mathbb{R}$. I know that we can proof denseness of $\mathbb{Q}$ and $\mathbb{R-Q}$ separately for each of them. 
Is it true that complement of EVERY dense set is dense, as well? 
Thank you.
PS My knowledge is too elementary; to me, akech's proof is understandable.

REPLY [9 votes]: No, this is not at all true. An extreme example was given in the comments, with $\mathbb{R}$ dense in $\mathbb{R}$. 
However there are plenty more: $\mathbb{R} \setminus \mathbb{Z}$ is dense, yet $\mathbb{Z}$ is not. $\mathbb{R} \setminus F$ is dense for every finite $F$, yet $F$ is not dense.
(This answers assume the topology used is the  usual on.)

REPLY [6 votes]: No.  Quick counterexample: $\bigcup_{k\in \mathbb Z}(k,k+1)$ is dense in $\mathbb R$ since its closure is $\mathbb R$, but its complement is $\mathbb Z$, which is extremely not dense.