Re: topology with metric and discrete.

[William Elliot]

On Thu, 14 Sep 2006, mina_world wrote:

> > > > On N, the metric d(n,m) = |1/n - 1/m| induce
> > > > the discrete topology.

> oh, sorry. possible.
> let e = 1/n - 1/(n+1).
>
Indeed, you simplified your min(thing).

> if x in N and x =/= n,
> then |1/n - 1/x| >= e.
> because,
> if x > n, |1/n - 1/x| = 1/n - 1/x >= 1/n - 1/(n+1) = e.

Ok.

> if x < n, |1/n - 1/x| = 1/x - 1/n >= 1/(n-1) - 1/n
> = 1/{n(n-1)} > 1/{n(n+1)} = 1/n - 1/(n+1) = e.
>
What if n = 1?