# Arithmetic progressions form infinite basis on $\mathbb{Z}$

Let $B(a,b) = \{ax+b: a,b \in \mathbb{Z}, a \neq 0, x \in \mathbb{Z}\}$ be a so called arithmetic progression

I am required to show that that $\mathcal{B} = \{B(a,b) | a,b \in \mathbb{Z}\}$ is a basis on $\mathbb{Z}$ that generates the Furstenberg Topology, and each $B(a,b)$ is an infinite set hence all open sets in $\tau_F$ are infinite

If $\mathcal{B}$ is a basis, then:

1. $\mathcal{B}$ covers $\mathbb{Z}$
2. If $z \in B_1\cap B_2$, then there exists $B_3 \in \mathcal{B}$ such that $z \in B_3 \subseteq B_1 \cap B_2$

We know $1$ is true, since $B(1,0) = \mathbb{Z} \in \mathcal{B}$

I am not sure how to prove $2$, this seems very involved.

Case 1: Let $B_1 = \{a_1x+b_1\}$, $B_2 = \{a_2x+b_2\}$, then $B_1 \cap B_2 = \varnothing$ if $a_1 \neq a_2$...But $\varnothing$ is not in the basis!

Case 2: Let $B_1 = \{a_1x+b_1\}$, $B_2 = \{a_2x+b_2\}$, then $B_1 \cap B_2 = \varnothing$ if $a_1 = a_2, b_1 \neq b_2$

Case 3: Let $B_1 = \{a_1x+b_1\}$, $B_2 = \{a_2x+b_2\}$, then $B_1 \cap B_2 = \{b_1\} = \{b_2\}$ if $a_1 \neq a_2, b_1 = b_2$

Case 4: Let $B_1 = \{a_1x+b_1\}$, $B_2 = \{a_2x+b_2\}$, then $B_1 \cap B_2 = B_1 = B_2$ if $a_1 = a_2, b_1 = b_2$

It would seem that only case $4$ do we have definition for a basis...but then it is trivial

Also I need to show that all open sets are infinite. Obviously each basic open set is infinite, because it is an affine map, hence a bijection. Union of infinite set are infinite.

In the notation of your post, we want to show that if the arithmetic sequences $B_1$ and $B_2$ have a point $z$ in common, then they have a whole infinite arithmetic sequence $B_3$ in common.
Since $B_1$ and $B_2$ have $z$ in common, there are integers $x_1$ and $x_2$ such that $z=a_1x_1+b_1=a_2x_2+b_2$.
Let $B_3$ be the set of all numbers of the shape $a_1a_2x+z$, where $x$ ranges over the integers. Then $B_3$ is a subsequence of $B_1$. For note that $$a_1a_2x+z=a_1a_2 x+a_1x_1+b_1=a_1(a_2x+x_1)+b_1.$$ A similar argument shows that $B_3$ is a subsequence of $B_2$, so we are finished.
• @MSEisadatingsite: Note that in the calculation we could replace $a_1a_2x+z$ by $Mx+z$ where $M$ is the lcm of $a_1$ and $a_2$. But this is not needed to prove the required result. As to the deity part, there are many hundreds, perhaps many thousands, of MSE users who could have written a solution. – André Nicolas Jun 22 '16 at 19:19