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I want to show that set $\mathcal{B} := \{S_{a,b}\}_{a \in \mathbb{Z},b \in \mathbb{Z}^{+}}$ where $S_{a,b} = a + b\mathbb{Z}$ is the basis for the topology on $\mathbb{Z}$. I am struggling to prove that: if $ x \in S_{a_1, b_1} \cap S_{a_2, b_2}$ then there exists $S_{a_3,b_3}$ such that $x \in S_{a_3,b_3} \subset S_{a_1, b_1} \cap S_{a_2, b_2}$. My main question is how to find those $a_3, b_3$?

I tried to take some specific values of $a_1,b_1,a_2$ and $b_2$ and tried to find $a_3,b_3$ in terms of $(a_1,b_1)$ and $(a_2,b_2)$ such that $S_{a_3,b_3}$ is contained in both $S_{a_1,b_1}$ and $S_{a_2,b_2}$ but couldn't find any specific relation.
Any hint on how to proceed will help!

not_a_math_guy
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    Your phrasing is off. If you ask how to show that $\mathcal B$ is the basis for *the* topology, one has to respond: For *what* topology? What it looks like you are asking is to show that $\mathcal B$ is *a* basis for *a* topology. – Lee Mosher Jan 17 '25 at 13:40
  • Also, you should perhaps show us what specific numerical values of $x,a_1,b_2,a_2,b_2$ you tried, and where you got stuck finding $a_3,b_3$. – Lee Mosher Jan 17 '25 at 13:43
  • @LeeMosher Actually I want to show that $\mathcal{B}$ is a topological basis on $\mathbb{Z}$. For this, I believe I just need to show it follows the definition of basis and from there I can find the topology it generates. I am not able to see how does what topology matters here? Am I missing something? – not_a_math_guy Jan 17 '25 at 13:54
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    Hint: use the Chinese remainder theorem. – Kolakoski54 Jan 17 '25 at 14:43
  • Duplicate question. Use site search "Fürstenberg topology integers". – Martin Brandenburg Jan 17 '25 at 15:43
  • To further explain my first comment, by standard English semantics, your use of the definite article *the* in the phrase "$\mathcal B$ ... is the basis for the topology on $\mathbb Z$" implies that a topology on $\mathbb Z$ has already been given preceding that phrase. Hence my question: *what* topology? As you have indicated in your comment (and as I already suspected), you did not intend to make that implication. To clarify your intention, I therefore suggested that you hit the edit button and replace those two definite articles *the* with indefinite articles *a*. – Lee Mosher Jan 17 '25 at 20:27

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I'll do the rest from the comment of @Kolakosli54

$S_{a,b}$ is the class $x\equiv a (\text{mod }b)$. Thus, we need to find the intersection of the classes $x\equiv a_1 (\text{mod }b_1)$ and $x\equiv a_2 (\text{mod }b_2)$. This means we are on the conditions of the generalized Chinese Remainder Theorem, therefore, if the intersection is nonempty, there exist a unique $a_3\in \mathbb{Z}$ such that $x\equiv a_3 (\text{mod }\text{lcm}(b_1,b_2))$, i.e. $S_{a_3,b_3}\subseteq S_{a_1,b_1}\cap S_{a_2,b_2}$ where $b_3=\text{lcm}(b_1,b_2)$.

Emo
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