# Is to be nowhere dense stronger than to have empty interior?

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?

• The rational have empty interior but they are dense. – YTS Jan 18 at 18:30

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.

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

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

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 $$X - A$$.

It is now clear that $$\mathbb Q \subset \mathbb R$$ has nonempty interior, but is not nowhere dense.

We also have:

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

It is clear that $$\mathbb R - \mathbb Q$$ is dense, but its interior (which is empty) is not.