TITLE: Is $\mathbb{Q}$ the continuous image of a Golomb-like space, or vice versa?
QUESTION [5 upvotes]: The set $${\cal B} = \big\{\emptyset\big\}\cup\big\{\{a + bn: n\in\omega\}: a\in\omega, b\in(\omega\setminus\{0\})\big\}$$ is a basis for a topology $\tau$ on $\omega$. Is there a surjective continuous map from $\mathbb{Q}$ with the Euclidean topology onto $(\omega,\tau)$, or the other way round, or in neither direction?
Note. These two spaces are not homeomorphic, since $\mathbb{Q}$ is homogeneous, and $(\omega,\tau)$ is not.

REPLY [6 votes]: Your topology is by definition the profinite topology of $\mathbf{Z}$, restricted to the nonnegative numbers $\mathbf{N}$. Hence it's a nonempty countable metrizable space. By a classical theorem of Sierpiński (Dasgupta - Countable metric spaces without isolated points, pdf), this implies that it is homeomorphic to $\mathbf{Q}$, in contradiction with your claim.
Golomb's space is defined as yours, but restricting to positive numbers and to coprime arithmetic progressions: precisely this subtlety drastically changes the topology (which is then connected—removing 0 is necessary to have it Hausdorff).