# Let $X=\Bbb{N} \setminus \lbrace 1 \rbrace$, $A_n=\lbrace d\in X : d|n \rbrace$, for $n\in \Bbb{N}$. Is $\tau=\lbrace A_n : n\in \Bbb{N}$ topology?

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?

• You should use \emptyset for the empty set, not \phi. – parsiad Feb 18 '19 at 2:51
• $A_2 \cup A_3 = \{2, 3\}$ and $A_6 = \{2, 3\}$. Your last claim is not correct. – parsiad Feb 18 '19 at 2:54
• @parsiad Thanks, I edited it. I think $A_6 =\lbrace 2,3,6 \rbrace$, is not correct? – Dima Feb 18 '19 at 2:56
• 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. – parsiad Feb 18 '19 at 3:07
• @parsiad Thank you so much. – Dima Feb 18 '19 at 3:11

No, it is not a topology because there is no n in N with X = A$$_n$$.