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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?

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    $\begingroup$ The rational have empty interior but they are dense. $\endgroup$ – YTS Jan 18 at 18:30
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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.

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    $\begingroup$ 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 at 18:48
  • $\begingroup$ That is. Thanks. $\endgroup$ – YTS Jan 18 at 18:49
  • $\begingroup$ Better now, +1 ${}$ $\endgroup$ – MPW Jan 18 at 18:51
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    $\begingroup$ So, just to be sure, to be nowhere dense is a stronger requirement than to have empty interior. $\endgroup$ – Ben Jan 18 at 18:53
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    $\begingroup$ Now add the example from your comment to show that it is strictly stronger. $\endgroup$ – GEdgar Jan 18 at 18:53
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$A$ is nowhere dense iff $int (\overline A)=\emptyset.$ Since $B\supset C\implies int(B)\supset int(C),$ we have $int(\overline A)=\emptyset\implies int(A)=\emptyset.$

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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.

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