Definition. $[a]^*_m = [a]_m \cap [a+m, \infty)$ is an arithmetic progression endowed with the corresponding infimum.
Theorem. If $\Sigma_B = \{[ax+b]^*_{(cx+d)}:x \in \mathbb{N}_{\ge 1}\}$ forms the base of a topology, $d = 1$.
Proof. If $\Sigma_B$ forms the base of a topology, then $(\Sigma_B, \cap)$ is a semigroup.
Let $S_i = [ai+b]^*_{(ci+d)}$, $S_j = [aj+b]^*_{(cj+d)}$, and $S_k = S_i \cap S_j = [ak+b]^*_{(ck+d)}$, so that $S_i, S_j \in \Sigma_B \implies S_k \in \Sigma_B$.
$S_k \subseteq S_i$ and $S_k \subseteq S_j$
$ak+b \equiv ai+b \pmod{ci+d} \implies k \equiv i \pmod{ci+d}$.
Similarly, $k \equiv j \pmod{cj+d}$.
$\therefore k \in \{[i]^*_{(ci+d)} \cap [j]^*_{(cj+d)}\}$.
$k = i + n(ci+d) = j + m(cj +d) = i + cni + nd = j + cmj +md$.
Both $i$ and $j$ range over the positive integers so the following holds when $n = j$ and $m = i$. $k = i + j(ci + d) = j + i(cj +d) = i + cij +dj = j + cij +di$
If $n = i$, $k = i(ci + d + 1) = j + i(cj + d) \implies$ $ci^2 + di + i = cij + dj + i$ such that $j = i$ and the domain of $k$ is the diagonal of $i$ and $j$. Letting $m = j$ has a similar implication, such that both $i$ and $j$ fail to range over the postive integers through the diagonal $i = j$.
As a result, $d = 1$ and $i + cij + j = j + cij + i$.
Do I have to prove the first statement or is it understood by definition? Is there a better way to justify letting n = j and m = i? Have I misused the term "diagonal"?