How to prove that it is a topology on this set?

Question

How to prove that this $$D_n=\{p\in E, p|n\}$$ can define a topology on the set $$E=\{n\in\mathbb{N}, n\geq2\}$$

Answer 1

@Osama Ghani asked, this question really boils down to whether you are looking for the basis of a topology or the topology itself. If you are looking for the topology itself, then no, the set of all such $D_n$'s is not a topology.

To prove it, let $\mathcal{T} = \{D_n, n\in \mathbb{N} \}$. Then $D_2 = \{2\}$ and $D_3 = \{3\}$, but $D_2 \cup D_3 = \{2,3\} \notin \mathcal{T}$ since any number divisible by 2 and 3 is at least 6, and so would also be divisible by 6. (We can check 2,3,4,5 and show that none of them are divisible by both 2 and 3, and any number larger than 5 that is divisible by 2 and 3 is also divisible by 6).

But, if you were to define $\mathcal{B} = \{D_n, n\in \mathbb{N} \}$ and take all intersections and finite unions, you can easily show you would have a topology. But, that is how you can extend any subset of the powerset of a set into a topology on that set, so there is nothing special about sets of divisors.