Are the rationals a nowhere dense set? I thought a nowhere dense set was a set where the closure of the complement is the whole space but surely the complement of the rationals is the irrationals who's closure is the reals so surely it's nowhere dense?
 A: No they are not: Wikipedia and Wolfram MathWorld indicate that a "nowhere dense set" is one whose closure has empty interior.  Since $\bar{\mathbb{Q}}=\mathbb{R}$ in this case, the rationals are not nowhere dense.
It appears you are taking the closure of the complement, as opposed to the interior of the complement
A: You have stumbled onto something that might seem confusing.  There are (at least) three different notions concerning density in topological spaces.


*

*$A \subseteq X$ is called dense if $\overline{A} = X$.

*$A \subseteq X$ is called co-dense if $X \setminus A$ is dense, i.e., if $\overline{ X \setminus A} = X$, or, equivalently, if $\mathrm{Int} (A) = \emptyset$.

*$A \subseteq X$ is called nowhere dense if $\overline{A}$ is co-dense, i.e., if $\mathrm{Int} ( \overline{A} ) = \emptyset$.


While the complement of every dense set is co-dense (by definition) this does not mean that the complement cannot be dense itself.  The rationals $\mathbb{Q}$ and its complement, $\mathbb{R} \setminus \mathbb{Q}$, the irrationals, are an example of this phenomenon in the real line.
Nowhere denseness is a much stronger version of co-denseness.  Clearly, every nowhere dense set is co-dense (as every subset of a co-dense set is co-dense).  But the complements of nowhere dense sets cannot be co-dense themselves, that is, nowhere dense sets cannot be dense.
A: A nowhere dense set is a set whose closure has empty interior. Equivalently, the interior of the complement is dense. So you neglected to take the interior of the complement before the closure.
