Interior and closure of $n^3$ in the topology on $\mathbb Z$ generated by bi-infinite arithmetic progressions Consider the topology on the integers $\mathbb{Z}$ defined by the collection of all bi-infinite arithmetic progressions
$$\mathscr{B} = \{A_{a,d} \mid d\in \mathbb{N}, 0 \le a < d\}$$
$$A_{a,d} = \{a + dn \mid n \in \mathbb{Z}\}$$
Compute the closure and interior of $C = \{ n^3 \mid n \in \mathbb{Z}\}$
So... (I think) the set $C$ is neither open or closed. And I can't think of any open set (arithmetic progression) that would lie inside $0, 1, 8, 27, 64$, etc, so is the interior empty? I'm not confident about that....
I have no idea about the closure, but I know that all closed sets are open in this topology, so is it asking the intersection of all arithmetic progressions that contain the integers cubed? 
 A: First of all, in this topology, it is not true that all closed sets are open in this topology. For example, every open set is infinite, but the singleton set $\{0\}$ is closed, as it is the complement of the open set
$$
\bigcup_{d=1}^\infty \bigcup_{a=1}^{d-1} A_{a,d} = \mathbb Z\setminus \{0\}.
$$
In fact, every finite set is closed.
Second, the set of cubes is $\{ \dots, -64, -27, -8, -1, 0, 1, 8, 27, 64, \dots\}$ - you left out the negative ones.
The set of cubes does in fact have empty interior. Showing this is equivalent to showing: every arithmetic progression contains a non-cube. This can be done in several ways; perhaps the easiest way is noting that the cubes get farther and farther apart, so eventually they can't contain two consecutive integers at distance $d$ from each other.
I believe the set of cubes is a closed set. Showing this is equivalent to showing that its complement is open, which is equivalent to: given any non-cube $k$, there exists $d>0$ such that none of the numbers in $A_{k,d}$ are cubes. In other words, if $k$ is a non-cube, then there exists a $d$ such that $k$ is not a cube modulo $d$. I think this is true, but it's not trivial as far as I can tell right now.
A: A small supplement to Greg Martin's answer. As is shown in that answer, to prove that the set of cubes is closed is the same as to prove the following: for each non-cube $k$ there exists $d > 0$ such that $A_{k,d}$ contains no cubes.
This is quite simple. Since $k$ is not a cube, there exists a prime $p$ such that $k = lp^m$, where $p$ doesn't divide $l$ and $3$ doesn't divide $m$. Now just take $d = p^{m+1}$.
Consider an arbitrary number $x = k + nd$ from set $A_{k,d}$. Note that
$$
x = lp^m + np^{m+1} = p^m(l + np).
$$
$p^m$ divides $x$, but $p^{m+1}$ doesn't divide $x$. Since $m$ is not a multiple of $3$, it follows that $x$ is not a cube. Done.
