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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?

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  • $\begingroup$ You should use \emptyset for the empty set, not \phi. $\endgroup$ – parsiad Feb 18 '19 at 2:51
  • $\begingroup$ $A_2 \cup A_3 = \{2, 3\}$ and $A_6 = \{2, 3\}$. Your last claim is not correct. $\endgroup$ – parsiad Feb 18 '19 at 2:54
  • $\begingroup$ @parsiad Thanks, I edited it. I think $A_6 =\lbrace 2,3,6 \rbrace$, is not correct? $\endgroup$ – Dima Feb 18 '19 at 2:56
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    $\begingroup$ Sorry, all of your claims were correct :-) Any set in $\tau$ that contains $2$ and $3$ must also contain $6$. $\tau$ is not a topology. $\endgroup$ – parsiad Feb 18 '19 at 3:07
  • $\begingroup$ @parsiad Thank you so much. $\endgroup$ – Dima Feb 18 '19 at 3:11
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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.

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