$\mathbb{Q}$ 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.

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

(This answers assume the topology used is the usual on.)


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.

  • $\begingroup$ 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? $\endgroup$ – Idris Jan 27 '15 at 4:34
  • $\begingroup$ @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... $\endgroup$ – Jon Jan 27 '15 at 6:10
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    $\begingroup$ 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. $\endgroup$ – Idris Jan 27 '15 at 11:55

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