For each prime $p$ let $S_p=\{n\in \mathbb N: \text{n is a multiple of } p\}$. $\mathscr S=\{1\}\cup \{S_p|\text {p is prime}\}$.

Question

For each prime $p$ let $S_p=\{n\in \mathbb N: \text{n is a multiple of } p\}$. $\mathscr S=\{1\}\cup \{S_p|\text {p is prime}\}$. Prove that $\mathscr S$ is a sub basis for a topology on $\mathbb N$. Describe the topology generated by $\mathscr S$.

Proof:

$\mathscr S$ is a sub basis for a topology on $\mathbb N$

$n\in \mathbb N$ either $n$ is prime or composite number(it has prime factors). so $n \in S_p$ for some $p$ prime.

$S_2=2\mathbb N$

$S_3=\{3,6,9,...\}$

$S_5=\{5,10,15...\}$

...

$S_p=\{p,2p,3p,...,\}$

Basis for the topology $\mathscr B=\{B\in \mathbb{N}:$ B is the intersection of finite number of members of $\mathscr S\}$. I am not able to see, How would the topology look like. Please help me.

Answer 1

All it takes for a collection C of subsets of S, to
be a subbase for some topology for S is $\cup$C = S.

A base for the topology given by that subbase is
{ $S_n$ : n positive square free integer }.

The topology is coarser than the set of upper sets of P(N) determined by the order a <= b when a divides b.