Prove that $\mathbb{N}$, along with the metric $d(m,n)=\lvert m^{-1}-n^{-1}\rvert$, is a discrete metric space.
I am stuck with this one, I don't know how to proceed ?
Any help will be appreciated.
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$\begingroup$ As before, can you show that singletons are open here as well? $\endgroup$ – Prahlad Vaidyanathan Oct 4 '13 at 13:57
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$\begingroup$ @PrahladVaidyanathan How do i do that here ? the distance between any two different natural numbers is always going to be greater than zero but less than 1. So, what should be the radius of the ball that i choose ? $\endgroup$ – johny Oct 4 '13 at 14:07
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Fix a point $m\in \mathbb{N}$, and you want to show that $\{m\}$ is an open set. In other words, you want to find $r > 0$ such that $$ B(m,r)\cap \mathbb{N} = \{m\} $$ So for any $n\in \mathbb{N}$, consider $$ \left | \frac{1}{m} - \frac{1}{n} \right | $$ If $n\to \infty$, this distance approaches $|1/m| > 0$. In particular, there is some $n_0\in \mathbb{N}$ such that $$ d(m,n) > \frac{1}{2|m|} \quad\forall n \geq n_0, n\neq m $$ Now consider the set $$ S = \{d(m,n) : n < n_0, n\neq m\} $$ This is a finite set, and every number inside is positive, hence its minimum is a positive number. Take $$ r = \frac{1}{2} \min\{ \min(S), |1/m|\} > 0 $$ Then, for any $n \in \mathbb{N}, n\neq m$, it follows that $$ d(m,n) > r $$ Hence, $$ B(m,r)\cap \mathbb{N} = \{m\} $$ Does this help?
answered Oct 4 '13 at 14:16
