Furstenberg Topology is Topology

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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

Recall the definition of the Furstenberg topology:

Let:

$\BB := \set {a \Z + b : a,b \in \Z, a \ne 0}$

where:

$a \Z + b := \set {a k + b : k \in \Z}$

Let:

$\tau := \set {\bigcup \AA : \AA \subseteq \BB}$


Then $\tau$ is called Furstenberg topology on $\Z$.

$\Box$


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:

A synthetic basis on $S$ is a subset $\BB \subseteq \powerset S$ of the power set of $S$ such that:

\((\text B 1)\)   $:$   $\BB$ is a cover for $S$      
\((\text B 2)\)   $:$     \(\ds \forall U, V \in \BB:\) $\exists \AA \subseteq \BB: U \cap V = \bigcup \AA$      

That is, the intersection of any pair of elements of $\BB$ is a union of sets of $\BB$.


$(\text B 1)$

$\BB$ is trivially a cover of $\Z$, since $\Z \in \BB$.

$\Box$


$(\text B 2)$

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:

\(\ds y\) \(\in\) \(\ds \paren {a_1 \Z + b_1} \cap \paren {a_2 \Z + b_2}\)
\(\ds \leadstoandfrom \ \ \) \(\ds \forall i \in \set {1,2}: \, \) \(\ds y\) \(\in\) \(\ds a_i \Z + b_i\) Definition of Set Intersection
\(\ds \leadstoandfrom \ \ \) \(\ds \forall i \in \set {1,2}: \, \) \(\ds y - x\) \(\in\) \(\ds a_i \Z + b_i - x\)
\(\ds \leadstoandfrom \ \ \) \(\ds \forall i \in \set {1,2}: \, \) \(\ds y - x\) \(\in\) \(\ds a_i \Z\) as $b_i - x \in a_i \Z$
\(\ds \leadstoandfrom \ \ \) \(\ds y - x\) \(\in\) \(\ds a_1 \Z \cap a_2 \Z\) Definition of Set Intersection
\(\ds \leadstoandfrom \ \ \) \(\ds y - x\) \(\in\) \(\ds \lcm \set {a_1, a_2} \Z\) Intersection of Integer Ideals is Lowest Common Multiple
\(\ds \leadstoandfrom \ \ \) \(\ds y\) \(\in\) \(\ds \lcm \set {a_1, a_2} \Z + x\)

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, because:

$\lcm \set {a_1, a_2} \Z + x = \bigcup \AA$

where:

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

$\blacksquare$


Also see