Is it the discreate topology

**rqeeb** · August 28th 2011, 02:10 AM

Consider the Natural numbers $N$ . and $t$ the collection of all subsets $G$ which satisfy the condition: if ${n\in\mathbb{G}}$ and $m|n$ , then ${m\in\mathbb{G}}$ . Show that $t$ is a topology on $N$ . Is it the discrete topology.

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

Is it the discrete topology?? I have no idea about this.

help please

**girdav** · August 28th 2011, 02:19 AM

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

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