Basis for a topology on $\Bbb{Z}$ using $U_a = \{an^2 - 1: n \in \Bbb{Z}\}$.

Question

Let $U_a = \{ a n^2 - 1 : n \in \Bbb{Z}\}$. Then the $\{U_a\}_{a\in\Bbb{Z}}$ is a covering family of $\Bbb{Z}$.

For $x \in \Bbb{Z}, x = an^2 -1$ whenever $a = x + 1$ and $n = 1$.

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$U_a \cap U_b = \{ z \in \Bbb{Z} : z + 1 = an^2 = bm^2 $ for some $m,n \in \Bbb{Z}\}$.

Can we conclude that there is some $z \in U_d \subset U_a \cap U_b$ and thus $\{U_a\}_{a\in \Bbb{Z}}$ is a basis for a topology on $\Bbb{Z}$?

Answer 1

Let $z \in U_a \cap U_b.$ Therefore $z+1 = an^2 = bm^2$ for some $m,n \in \mathbb{Z}.$ Since $z+1 = (an^2)\cdot 1^2,$ we see that $z \in U_{an^2}$.

Also, if $w \in U_{an^2},$ then $w+1 = (an^2)\cdot k^2 = a(nk)^2$ for some $k \in \mathbb{Z},$ so $nk\in \mathbb{Z}$ and $w \in U_a,$ so $U_{an^2} \subseteq U_a.$ Since $an^2 = bm^2,$ we can similarly see $U_{an^2} = U_{bm^2} \subseteq U_b,$ so $U_{an^2} \subseteq U_a \cap U_b.$