# Is $\mathbb{Q}$ the continuous image of a Golomb-like space, or vice versa?

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.

• And this is not the definition of Golomb space: in the Golomb space $0$ is excluded, and $a,b$ are coprime. – YCor Mar 30 '18 at 6:30
• $a\mapsto\lfloor a+\sqrt{2}\rfloor$ is a continuous surjection from $\mathbb Q$ to a discrete countable space. Feel free to take it from there. – Wojowu Mar 30 '18 at 6:48
• Note: this topology is the topology induced by inclusion of $\omega$ (the natural numbers) into $\widehat{\mathbf{Z}}$ (the profinite completion of $\mathbf{Z}$). So this topology is actually homeomorphic to $\mathbf{Q}$, and is far from "Golomb-like", much more standard. (See math.stackexchange.com/questions/355779/… for this topological characterization) – YCor Mar 30 '18 at 12:33

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.