I was trying to prove this result.
I started out by taking some arbitrary subset, S of N,and finding its boundary points.
Boundary points of S is the set of all points x of N whose distance from S and S complement is 0. But because the metric space is the set of natural numbers, therefore the minimum distance between any two points is 1.
So,
dist(x,S) = inf{d(x,a)|a is in S} = 1
dist(x,S complement) = inf{d(x,b)|b is in S complement} =1
These two distances are the same,but are not equal to zero, which would imply that boundary of S is empty.
And hence the intersection of S and its boundary is empty.
So, S is open.
Also because S was arbitrary, we will have that all the subsets of N under the usual metric inherited form R are open.
Thus, N is a discrete metric space.
Is this argument correct ?
Best Answer
If you want to show that all subsets are open, the easiest way to do so is to show that singletons are open (and hence all subsets would be - why?) : So just take a singleton $\{n\} \subset \mathbb{N}$, then for $r = 1/2$, then open ball $$ B(n,r)\cap \mathbb{N} = \{n\} $$ Hence, $\{n\}$ is open, and you're done!
Edit: Your argument is almost correct. You choose a point $x\in \mathbb{N}$ and you want to check where $d(x,S) = d(x,S^c) = 0$. There are two cases, either $x\in S$, or $x\in S^c$. In the first case, $$ d(x,S) = 0, \text{ but } d(x,S^c) \geq 1 $$ as you have observed. And in the second case, the reverse is true.
Your statement that $d(x,S) = d(x,S^c) = 1$ is not true for any $x\in \mathbb{N}$!