3
$\begingroup$

Let $A$ be a subset of a topological space $X$.

(1) $A$ is nowhere dense if the interior of its closure is empty;

(2) $A$ has dense interior if the closure of its interior is $X$;

(3) $A$ has empty interior if its interior is empty;

(4) $A$ is dense if its closure is $X$.

Now, to have dense interior (= to be dense and open) is stronger than to be dense.

I have a doubt: Is to be nowhere dense stronger than to have empty interior?

$\endgroup$
1
  • 4
    $\begingroup$ The rational have empty interior but they are dense. $\endgroup$
    – EQJ
    Jan 18, 2020 at 18:30

3 Answers 3

3
$\begingroup$

Yes it is. The interior of the closure of a nowhere dense set is empty and the interior of a set is contained in the interior of the closure.

Edit: it is indeed strictly stronger. Consider the set of rational numbers, this set is dense but it has empty interior.

$\endgroup$
7
  • 1
    $\begingroup$ Perhaps you mean to say the the interior of the closure of a nowhere dense set is empty. What you wrote is incorrect. $\endgroup$
    – MPW
    Jan 18, 2020 at 18:48
  • $\begingroup$ That is. Thanks. $\endgroup$
    – EQJ
    Jan 18, 2020 at 18:49
  • $\begingroup$ Better now, +1 ${}$ $\endgroup$
    – MPW
    Jan 18, 2020 at 18:51
  • 1
    $\begingroup$ So, just to be sure, to be nowhere dense is a stronger requirement than to have empty interior. $\endgroup$
    – Ben
    Jan 18, 2020 at 18:53
  • 1
    $\begingroup$ Now add the example from your comment to show that it is strictly stronger. $\endgroup$
    – GEdgar
    Jan 18, 2020 at 18:53
1
$\begingroup$

$A$ is nowhere dense iff $int (\overline A)=\emptyset.$ Since $B\supset C\implies int(B)\supset int(C),$ we have $int(\overline A)=\emptyset\implies int(A)=\emptyset.$

$\endgroup$
0
$\begingroup$

Here's another way to compare the two:

  • Empty interior is equivalent to: every nonempty open set of $X$ contains at least one point of $X - A$.
  • Nowhere dense is equivalent to: every nonempty open set of $X$ contains a nonempty open set that intersects $A$ trivially, i.e., contains an entire open set of points that lie in $X - A$.

We also have:

  • $A$ has empty interior $\iff$ $X-A$ is dense.
  • $A$ is nowhere dense $\iff$ $X-A$ has dense interior.

In particular, a dense subset is never nowhere dense (unless $X = \varnothing$) because its complement has empty interior. So a dense subset with empty interior is an example that shows that to be nowhere dense is a stronger property than to have empty interior. Example: $\mathbb Q \subset \mathbb R$.

$\endgroup$
7
  • $\begingroup$ I think your definition of nowhere dense should say "nonempty open set that intersects $X-A$ non-trivially" rather than "nonempty open set that intersects $A$ trivially" $\endgroup$
    – Ben
    Aug 22, 2021 at 16:27
  • $\begingroup$ The statement "$A$ has empty interior $\Leftrightarrow X-A$ is dense." contradicts the statements "$\mathbb{R}-\mathbb{Q}$ is dense" and "$\mathbb{Q}\subset\mathbb{R}$ has nonempty interior" $\endgroup$
    – Ben
    Aug 22, 2021 at 17:21
  • $\begingroup$ @Ben 1. No. $\operatorname{int}(\operatorname{cl}(A)) = \varnothing$ is equivalent to $\operatorname{cl}(\operatorname{int}(A)) = X$, because closure becomes interior and vice versa upon taking complements. So this means that the complement has dense interior, and this is precisely what the second bullet says. 2. The statement "$\mathbb Q \subset \mathbb R$ has nonempty interior" is false. There is no contradiction. $\endgroup$ Aug 26, 2021 at 15:19
  • $\begingroup$ but you state "It is now clear that $\mathbb{Q}\subset\mathbb{R}$ has nonempty interior" just after the 2nd bullet point. $\endgroup$
    – Ben
    Aug 26, 2021 at 15:25
  • $\begingroup$ @Ben Right, sorry. I fixed that sentence. Thanks! $\endgroup$ Aug 26, 2021 at 19:02

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .