TITLE: Confirming something is a basis for a topology
QUESTION [1 upvotes]: I am brand new to topology and had some questions regarding topological bases. 
I am currently working through Furstenburg's proof of the infinitude of primes. I am stuck trying to understand this:
Consider A defined in the following way:
$A(n,a) = \{a+nq : q\in \mathbb{Z}\}$
Now consider $B$ defined in the following way:
$B=\{A(n,a):a,n \in \mathbb{Z}, n\neq 0  \}$
Now I know (because it says in the proof) that B is a basis for a topology on $\mathbb{Z}$ . Apparently the proof of this is very simple but I just don't see exactly what I am supposed to do to prove the following two conditions are true:
1) For each $x \in \mathbb{Z}$, there is at least one basis element containing $x$.
2) If $x$ belongs to the intersection of two basis elements, there is a third basis element containing $x$ that is in the intersection of the first 2 basis elements.
It is mostly this second part I am having trouble understanding. Again it seems very simple but if someone could just show what I need to do to prove that this is a basis, that would be greatly appreciated. Thanks a lot.

REPLY [1 votes]: $A(0,1) =\mathbb{Z}$ so condition 1. is easy. (or $a \in A(a,q)$ for any $q$).
If $a_3 \in A(a_1,q_1) \cap A(a_2,q_2)$, then check that  $$a_3 \in A(a_3, \gcd(q_1, q_2)) \subseteq A(a_1,q_1) \cap A(a_2, q_2)$$
for the second condition.
Facts: $a_3 \in A(a_1, q_1)$ iff $A(a_3, q_1) = A(a_1, q_1)$, and $A(a,q_1) \cap A(a,q_2) = A(a, \gcd(q_1, q_2))$. So I think you have $=$ instead of $\subseteq$ above, even.
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