Is $\Bbb{Z}$ compact under the evenly spaced integer topology?

Question

Let us consider $\Bbb{Z}$ with the evenly spaced integer topology

The topology is generated by the basis $U_{a,b} = a \Bbb{Z} + b$, where $a\ne0$.

$\{U_{a,x}\}_{x = 1\dots a}$ covers $\Bbb{Z}$ for any $a \in \Bbb{Z}$ finitely since $x$ ranges over $1..a$.

Does this somehow imply that any open cover of $\Bbb{Z}$ contains a finite subcover?

Answer 1

The collection $$p\mathbb{Z}$$ as $p$ ranges over primes covers all but one and minus one. If I then include $$ 5\mathbb{Z}+1$$ and $$5\mathbb{Z}-1$$ has that a finite subcover?

Answer 2

Start with $A_1 = 2\mathbb{Z},$ then $A_2 = 4\mathbb{Z}+1,$ then $A_3 = 8\mathbb{Z}-1,$ and in general $A_n = 2^n\mathbb{Z}+a_n$ where $a_n$ is an element of minimal magnitude in $\mathbb{Z} \setminus \bigcup_{m=1}^{n-1} A_m.$ (when you must choose between two elements for $a_n,$ choose the positive value)

Then $\{A_n\}_{n=1}^\infty$ is a partition of $\mathbb{Z}$ by nonempty sets $A_n = U_{2^n,a_n}$

---

To see this, just note that $\mathbb{Z}\setminus \bigcup_{m=1}^{n-1} A_m$ is a set of the form $2^{n-1}\mathbb{Z}+b_n,$ and every step we're just removing every other remaining element.