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?
 A: 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.
A: $A$ is nowhere dense iff $int (\overline A)=\emptyset.$ Since $B\supset C\implies int(B)\supset int(C),$ we have $int(\overline A)=\emptyset\implies int(A)=\emptyset.$
A: 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: $\mathbb Q \subset \mathbb R$.
