# Why is $\{\emptyset\}\cup\{n\mathbb{Z}, n\in \mathbb{N}\}$ not a topology on $\mathbb{Z}$?

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

$$2\mathbb{Z}\cup 3\mathbb{Z}\ne n\mathbb{Z}$$ for any $$n$$.
• @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 $2\mathbb{Z}$. – Henno Brandsma Mar 11 at 22:40
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 $$2$$?
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 $$5$$.