Arithmetic progression topology I'm being asked to prove that $\mathcal{A}=\left\lbrace A_{a,b}:a,b\in\mathbb{Z}\right\rbrace$ is a basis for a topology on $\mathbb{Z}$, being:
$$A_{a,b}=\left\lbrace a+nb:n\in\mathbb{Z}\right\rbrace=\left\lbrace ...,a-2b,a-b,a,a+b,a+2b,...\right\rbrace$$
Obviously every $a\in \mathbb{Z}$ belongs to a basis element, for example $A_{a,b}$, for any $b$. I'm having difficulties though, proving that if two of those sets $A_{a,b}$ and $A_{a',b'}$ share an element $x$, then there exists another $A_{c,d}$, such that $x\in A_{c,d}$ and $A_{c,d}\subset A_{a,b},A_{a',b'}$.
Any hint?
 A: Hint: Try $d = \operatorname{lcm}(b,b')$ or $d = bb'$ - either should work fine. The idea being that we're looking for things that are in a list of step size of $b$ (so to say) and in another list of step size $b'$.
A: Hint: Chinese Remainder Theorem.
If $\rm x\equiv a\bmod b,\bar{a}\bmod\bar{b}$, then pick $\rm d:=lcm(b,\bar{b})$ and $\rm c$ a residue of $\rm x$ mod $\rm d$.
A: As a hint, sets with large values of the second argument are smaller, as they include fewer numbers.
A: If $z\in D_1= A_{a_1,b_1}\in A$ and $z\in D_2=a_{a_2,b_2}\in A,$ then $D_1=A_{z,b_1}$ and $D_2=A_{z,b_2}$ so $z\in D_3\subset (D_1\cap D_2)$ where $D_3=A_{z,(b_1 b_2)}\in A.$
A: To avoid the space becoming discrete, we must not allow $b = 0$. Also, $A_{a,b} = A_{a,-b}$. Hence it is sufficient to take $b \in \mathbb{Z}^+$. Finally, if $a > b$, then by taking $c = a \mod b$, we see that $A_{a,b} = A_{c,b}$. Thus the basic open sets are of the form $A_{a,b}$, $(b \in \mathbb{Z}^+, 0 \le a < b)$. Or in other words, the basic open sets are of the form $b\mathbb{Z} + a$.
Next given $A_{a,b}, A_{a',b'}$, we have to find $A_{c,d} \subset  A_{a,b} \cap A_{a',b'}$. Let $z,w \in A_{c,d}$. Then $z,w \in A_{a,b} \Rightarrow b \mid (z-w)$ and similarly, $ b' \mid (z-w)$. Thus $lcm(b,b') \mid (z-w)$. So, we take $d = lcm(b,b')$.
Next we have to find $c$. As noted we shall find $c$ such that $0 \le c <d$. Let $x$ be the smallest nonnegative integer in $A_{c,d}$. Then $x \in A_{a,b} \Rightarrow b \mid (x - a)$. This gives $x \cong a \mod b$. Similarly, $x \cong a' \mod b'$. Using the Chinese remainder theorem, we get $c$. Note that in Chinese remainder theorem, either there is no solution or there are infinitely many solutions. Here our requirement of proving that the given sets form an open base, forces that we have to find the solution, only when the two sets intersect. Thus the solution is guaranteed.
