Furstenberg Topology is Topology

Theorem
Let $\struct {\Z, \tau}$ be the topological space formed by the Furstenberg topology on the set of integers $\Z$.

Then $\tau$ is indeed a topology on $\Z$.

Proof
In view of Union from Synthetic Basis is Topology it suffices to show that $\BB$ is a synthetic basis on $\Z$.

Recall the definition of synthetic basis:

$(\text B 1)$
$\BB$ is trivially a cover of $\Z$, since $\Z \in \BB$.

$(\text B 2)$
Let $a_1 \Z + b_1, a_2 \Z + b_2 \in \BB$.

If:
 * $\paren {a_1 \Z + b_1} \cap \paren {a_2 \Z + b_2} = \O$

then it is done, since $\O = \bigcup \O$ and $\O \subseteq \BB$.

Now, suppose that:
 * $\exists x \in \paren {a_1 \Z + b_1} \cap \paren {a_2 \Z + b_2}$

Let $\lcm \set {a_1, a_2}$ be the lowest common multiple of $a_1$ and $a_2$.

Then:

That is:
 * $\paren {a_1 \Z + b_1} \cap \paren {a_2 \Z + b_2} = \lcm \set {a_1, a_2} \Z + x$

This concludes the proof, since:
 * $\lcm \set {a_1, a_2} \Z + x = \bigcup \AA$

where:
 * $\AA := \set {\lcm \set {a_1, a_2} \Z + x} \subseteq \BB$

Also see

 * Definition:Furstenberg Topology