We recall the definition of a nowhere dense subset of a metric space:
"A subset $A$ of a metric space $(X,d)$ is nowhere dense if $Int(\bar A)=\emptyset$"
I don't understand how it is that $\mathbb Z\subset \mathbb R$ is nowhere dense; how can the interior of its closure be the empty set?
Best Answer
The set $\Bbb Z$ consists entirely of isolated points, so it has no limit points. Hence, its closure is $\Bbb Z$.
Also because $\overline{\Bbb Z}=\Bbb Z$ consists of isolated points, it has no interior points, meaning its interior is empty.
To see any $k \in \Bbb Z$ is an isolated point, note that the only integer in the open interval $\left( k- \frac 12, k+\frac 12 \right)$ is $k$.