Let $X=\Bbb{N} \setminus \lbrace 1 \rbrace$, $A_n=\lbrace d\in X : d|n \rbrace$, for $n\in \Bbb{N}$. Is $\tau=\lbrace A_n : n\in \Bbb{N}\}$ a topology?
My attempt: 1) It is obvious that $\emptyset, X \in \tau$, since $\emptyset = A_1 \in \tau$, $X=A_n\in \tau$.
2) Let $A_i, A_j \in \tau$
$A_i \cap A_j =A_{(i,j)}$, where $(i,j)$ is a greatest common divisor of $i$ and $j$.
3) Let $A_i \in \tau$, we have to show that $\bigcup A_i \in \tau$ or $ \bigcup A_i \notin \tau $.
I know that, this union is of every family of open sets, but I took $A_2, A_3$ to understand. And I found that $A_2 \cap A_3 \notin \tau$, since there is no set in $\tau$ equals $\lbrace 2,3 \rbrace$, is this means $\tau$ is not topology?
No, it is not a topology because there is no n in N with X = A$_n$.
Exercise. Show it is a base for a topology for X.