# Is it the discreate topology

• Aug 28th 2011, 02:10 AM
rqeeb
Is it the discreate topology
Consider the Natural numbers $\displaystyle N$. and $\displaystyle t$ the collection of all subsets $\displaystyle G$ which satisfy the condition: if $\displaystyle {n\in\mathbb{G}}$ and $\displaystyle m|n$, then $\displaystyle {m\in\mathbb{G}}$. Show that $\displaystyle t$ is a topology on $\displaystyle N$. Is it the discrete topology.

I can show all the conditions for $\displaystyle t$ eccept that $\displaystyle {N\in\mathbb}$$\displaystyle t and \displaystyle {(the empty set)\in\mathbb}$$\displaystyle t$

For $\displaystyle G=\emptyset$, the statement $\displaystyle n\in G$ is always false and it implies everything, in particular $\displaystyle m\in G$ if $\displaystyle m\mid n$. If $\displaystyle n\in \mathbb N$ and $\displaystyle m\mid n$, then $\displaystyle m$ should be an integer.
Hint for the question: if $\displaystyle n=p_1p_2$ where $\displaystyle p_1$ and $\displaystyle p_2$ are prime numbers then a set of the topology which contains n should contain $\displaystyle p_1$ and $\displaystyle p_2$.