5

$\mathbb{Q}$ 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.

Community
  • 1

2 Answers2

12

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.)

quid
  • 42,135
6

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.

Jon
  • 706
  • For all the counterexamples given above one set among X or its complement is CLOSED. So the question maybe, assume that X is a dense subset of the reals such that X and its compolement are NOT CLOSED. Is it necessary that its complement is dense too? – Idris Addou Jan 27 '15 at 04:34
  • @Idris I'm not sure how to prove such a result in general. I do know the sets of algebraic and transcendental numbers are neither open nor closed dense complements of each other, as a second example. But I haven't come up with a counterexample, either... – Jon Jan 27 '15 at 06:10
  • 1
    Thank you for your answer. I think I have a counterexample which prove that the answer to my question is negative. Consider X the complement of the set of all the points of the sequence (1/n). X contains zero, so X and its complement are not closed. However, X is dense but not its complement. – Idris Addou Jan 27 '15 at 11:55