I'm trying to solve an exercise from Dugundji's Topology textbook and it says:
In $\mathbb{Z}$, let $p$ be a fixed prime. For each $a\in\mathbb{Z}^+$, define
$$U_{a}(n) = \{n+\lambda p^a ; \lambda\in\mathbb{Z}\}.$$
Show that the collection $B=\{U_{a}(n)\}$ is basis for some topology.
Well, I know $B$ satisfies the first property of basis trivially, because for each $n\in\mathbb{Z}$, $n\in U_{a}(n)$ for any $a$ (taking $\lambda =0$).
Now for the second property, I take $n\in\mathbb{Z}$ and $U_{a_1}(s)$,$U_{a_2}(t)\in B$ such that $n\in U_{a_1}(s)\cap U_{a_2}(t)$. I know I have to construct some $U_{a}(m)\in B$ such that $n\in U_{a}(m) \subseteq U_{a_1}(s)\cap U_{a_2}(t)$ but I just don't see how to do it. I mean, I know that from the fact that $n\in U_{a_1}(s)\cap U_{a_2}(t)$, there exist $\lambda_1,\lambda_2\in\mathbb{Z}$ such that
$$n = s+\lambda_1 p^{a_1},$$ $$n = t+\lambda_2 p^{a_2};$$
But I don't know how to use that for what I need to proof. Can someone help with that, please?
Hint
$$k \in U_a(n) \Rightarrow k =n+\beta p^a$$ As long as $a$ satisfies an obvious equality, you then have $$k = n+\beta p^a= s+\lambda_1 p^{a_1}+\beta p^a= s+ p^{a_1} \left( \mbox{ some integer } \right) \in U_{a_1}(s) \\ k = n+\beta p^a= t+\lambda_2 p^{a_2}+\beta p^a= t+ p^{a_2} \left( \mbox{ some integer } \right) \in U_{a_2}(t) \\$$