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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.

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    $\begingroup$ Yes, this fine. $\endgroup$ – Brian M. Scott Feb 22 at 23:28
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    $\begingroup$ @BrianM.Scott Great, thanks! $\endgroup$ – Alejandro Bergasa Alonso Feb 22 at 23:29
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    $\begingroup$ Perfecto!!!!!!! $\endgroup$ – Riemann'sPointyNose Feb 22 at 23:31
  • $\begingroup$ 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. $\endgroup$ – Octavius Feb 22 at 23:34
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    $\begingroup$ @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! $\endgroup$ – Alejandro Bergasa Alonso Feb 23 at 7:21

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