For each $a, d \in \Bbb N$, let $B(a,d) = \{a+nd \mid n \in \mathbb{N}\cup\{0\} \, \}$. I seek to show that $\mathscr{B} = \{B(a,d)\}$ is a basis for a topology on $\mathbb{N}$.
I have already shown that $\displaystyle\bigcup \mathscr{B} = \mathbb{N}$. At this point, I need only show that for every $B_1,B_2 \in \mathscr{B}$ with $m \in B_1 \cap B_2$ , then there exists $B_3$ satisfying $m \in B_3 \subset B_1 \cap B_2$. But I struggle to finish the proof.
Proof.
Suppose $B(a_1,d_1)$ and $B(a_2,d_2)$ satisfy $m \in B(a_1,d_1) \cap B(a_2,d_2)$ for some $m$. Since $m \in B(a_1,d_1) \cap B(a_2,d_2)$, we have
$$m = a_1 +n_1d_1 \hspace{1cm} m = a_2 +n_2d_2$$
Without knowing anything about $a_1,a_2,n_1,n_2,d_1,d_2$, it's hard to figure out how to come up with a third way to write $m$ and thus come up with a third set $B(a_3,d_3)$. Does anyone have any suggestions?
Since $m \in B(a_1, d_1) \cap B(a_2, d_2)$, $m$ is a solution of the system $$ x \equiv a_1 \bmod d_1 \qquad x \equiv a_2 \bmod d_2 $$ Setting $d = \text{lcm}(d_1, d_2)$, it follows from the Chinese remainder theorem that $m \in B(m, d)$ and $B(m,d) \subseteq B(a_1,d_1) \cap B(a_2,d_2)$.
Nota Bene. This topology is the profinite topology on the monoid $\Bbb N$.