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please why $\tau=\{\emptyset\}\cup\{n\mathbb{Z}, n\in \mathbb{N}\}$ is not a topology on $\mathbb{Z}$?

$\emptyset \in \mathbb{Z}$ and $\mathbb{Z}=1\mathbb{Z}\in \tau$

let $A=n_1\mathbb{Z}$ and $B=n_2\mathbb{Z}$ then $A\cap B= \max\{n_1,n_2\}\mathbb{Z}\in \tau$

then $\tau$ is stable by finite intersection .

let $(A_i)_{i\in I}$ a family of sets from $\tau$ : $A_i=n_i\mathbb{Z}$

then $\bigcup A_i=\min_{I\in I}n_i \mathbb{Z}$

thank you

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$2\mathbb{Z}\cup 3\mathbb{Z}\ne n\mathbb{Z}$ for any $n$.

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  • $\begingroup$ why ? I don't understand ? and the intersection is right? ? $\endgroup$ – Poline Sandra Mar 11 at 20:52
  • $\begingroup$ @PolineSandra your family is a base for a topology but not a topology. As to this non-equality: $2$ is in the left hand side and can only be in $n\mathbb{Z}$ for $n=1,2$ but the first is too large (as $5$ is not in the left hand side), the second too small as $3$ is in the left hand side and not in $2\mathbb{Z}$. $\endgroup$ – Henno Brandsma Mar 11 at 22:40
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It is not true that $\cup _{i \in I}A_i = \min_{i \in I}n_i \mathbb{Z}$. Can you come up with a counter-example where $I$ has size $2$?

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As you said, $\emptyset \in \tau$ and $\mathbb{Z}=1\mathbb{Z} \in \tau$.

Also, let $A=a\mathbb{Z}$ and $B=b\mathbb{Z}$. We have that $A \cap B = \textrm{lcm}(a,b)\mathbb{Z} \in \tau$.

But this set is not a topology, because it doesn't satisfy the union property. Just take any $A=a\mathbb{Z}$ and $B=b\mathbb{Z}$ such that $\textrm{gcd}(a,b)=1$. For example: $3\mathbb{Z} \cup 5\mathbb{Z} \neq n\mathbb{Z}, \ \forall \ n \in \mathbb{Z}$. Note that this union contains all integers that are multiple to $3$ and all integers that are multiple to $5$.

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