Dense and nowhere dense

Question

Let $X$ be a topological space and $A$ be a non-empty subset of $X$. Then one can conclude that if $X\setminus A$ is nowhere dense in $X$, $A$ is dense in $X$,

Is the above statement true in general? I know if $A$ is open then the result is true, but I am not sure otherwise.

In case it could be of use, I gave (for metric spaces) some equivalences of "has empty interior", and also of "is nowhere dense", in my answer to Is saying a set is nowhere dense the same as saying a set has no interior?. — Dave L. Renfro, Jun 25 '18 at 18:29

score 5 · Accepted Answer · answered Jun 25 '18 at 17:33

5

No, that is not true in general.

Consider $X = \mathbb{R}$. Clearly $A = \mathbb{Q}$ is dense in $\mathbb{R}$, but $\mathbb{R} \setminus \mathbb{Q}$ is also known to be dense as well.

answered Jun 25 '18 at 17:33

Thank you so much – Golam biswas Jun 25 '18 at 17:40

score 1 · Answer 2 · answered Jun 25 '18 at 21:02

1

No, as others pointed out already. Sets can be dense and co-dense, like the rationals and the irrationals in the real numbers.

But $A$ is nowhere dense iff $X\setminus \overline{A}$ is dense. So something close to it is true.

answered Jun 25 '18 at 21:02

Henno Brandsma

242,131

score 0 · Answer 3 · answered Apr 22 '20 at 16:44

0

$A$ is dense $\Rightarrow A^-=X\Rightarrow A^e=\emptyset \Rightarrow A^{c\circ}=\emptyset$. $A^{c\circ}=\emptyset$ does not always imply $A^{c-\circ}=\emptyset$, since $A^c\subseteq A^{c-}$.

answered Apr 22 '20 at 16:44

yushang

23

Dense and nowhere dense

3 Answers3