# Prove if the following sets are topologies in $\mathbb{Z}$

I've been solving the following problem from the start of my general topology course, and I'd like to check if my answers are correct. The problem is:

Study if the following sets of subsets of $$\mathbb{Z}$$ are topologies or not:

1. $$\mathcal{A}_1=\{\emptyset,\{1,2\},\{1,2,3\},\{2,3,-4\},\{1,2,3,-4\},\mathbb{Z}\}$$
2. $$\mathcal{A}_2=\{\emptyset\}\cup\{n\mathbb{Z}:n\in\mathbb{N}\}$$
3. $$\mathcal{A}_3=\{A\subset\mathbb{Z}: 0 \in A\}$$
4. $$\mathcal{A}_4=\{A\subset Z: A \text{ is infinite }\}\cup\{\emptyset\}$$

My solutions (I concluded no one is a topology in $$\mathbb{Z}$$, all using the definition of topology):

1. NOT, because $$\{1,2\}\cap\{2,3,-4\}=\{2\}\notin \mathcal{A}_1.$$
2. NOT, because $$\nexists\phantom{,} n\in\mathbb{N}: n\mathbb{Z}=3\mathbb{Z}\cup 7\mathbb{Z}$$ (this is because $$\text{gcd}(3,7)=1$$, but $$1\mathbb{Z}=\mathbb{Z}\neq3\mathbb{Z}\cup 7\mathbb{Z}$$).
3. NOT, because $$\emptyset\notin\mathcal{A}_3.$$
4. NOT, because if we consider the sets $$A=\{z\in\mathbb{Z} : z \text{ is even}\}$$ $$B=\{z\in\mathbb{Z} : z \text{ is odd}\}\cup \{2\},$$ it's clear that both $$A,B\in\mathcal{A}_4$$ because both are infinite sets, but $$A\cap B=\{2\}$$, hence we conclude that $$A\cap B\notin\mathcal{A}_4$$ (because $$\{2\}$$ is a finite set), so we conclude $$\mathcal{A}_4$$ is NOT a topology in $$\mathbb{Z}$$.

Are my solutions correct? Thanks in advance.

• Yes, this fine. – Brian M. Scott Feb 22 at 23:28
• @BrianM.Scott Great, thanks! – Alejandro Bergasa Alonso Feb 22 at 23:29
• Perfecto!!!!!!! – Riemann'sPointyNose Feb 22 at 23:31
• Maybe I am wrong, but I am not quite sure about the third one. I would consider the empty set to be a subset of every set. – Octavius Feb 22 at 23:34
• @Octavius But it's not the same. $\mathcal{A}_3$ does not need to verify that $\emptyset\subset\mathcal{A}_3$, but rather that $\emptyset\in\mathcal{A}_3$ (notice the difference here between "$\subset$" and "$\in$"), and it's true that for every set $X$ then $\emptyset\subset X$, but in general $\emptyset\notin X$. Thanks for your comment anyway! – Alejandro Bergasa Alonso Feb 23 at 7:21