# Show that $\mathbb{Z}$ with the Furstenberg's topology is compact

Let $a,b \in \mathbb{Z}, b \neq 0$ and $S(a,b):= \{a + nb \in \mathbb{Z}: n\in \mathbb{Z} \} \subseteq \mathbb{Z}$. We define the Furstenberg's topology as following: \begin{align*} W \subseteq \mathbb{Z} \, \, \, \text{open iff} \, \, \, W = \emptyset \, \, \text{or} \, \, \forall \, a\in W \, \, \exists \, b \neq 0 \, \text{such that} \, \, S(a,b) \subseteq W \end{align*} I have to show that $\mathbb{Z}$ with this topology is not compact.

I showed that is not connected but I can't see how to do with the compactness.

• This topology should not be credited to Furstenberg. It is the pro-goup topology on $\mathbb{Z}$, a much older definition. – J.-E. Pin Apr 18 '18 at 9:22

Consider these subsets of $\mathbb Z$:
• the odd numbers (numbers of the form $2a+1$);
• the multiples of $3$ (numbers of the form $3a$);
• even numbers which are not multiples of $4$ (numbers of the form $4a+2$);
• multiples of $4$ which are not multiples of $8$ (numbers of the form $8a+4$)
and so on. They form an open cover of $\mathbb Z$, but this open cover has no finite subcover. In fact, no finite number of elements of this open cover contain every power of $2$.