# Is complement of a dense set in $\mathbb{R}$ dense in $\mathbb{R}$?

$\mathbb{Q}$ 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.

• certainly not. $\mathbb R$ is dense in $\mathbb R$ but its complement is not. – Ittay Weiss Jan 27 '15 at 0:56
• Take the set of all real numbers except for $1$. That is a dense subset. What is its complement? – KCd Jan 27 '15 at 1:03

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.
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.