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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 ?

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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 $x\in \mathbb{N}$!

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  • $\begingroup$ Your argument is much better than what i did, but could you still please tell me whether my attempt was correct or not ? $\endgroup$ – johny Oct 4 '13 at 13:41
  • $\begingroup$ @johny : I have edited my post to include a comment on your argument as well. $\endgroup$ – Prahlad Vaidyanathan Oct 4 '13 at 13:57
  • $\begingroup$ Alright, got it. Thank you ! $\endgroup$ – johny Oct 4 '13 at 13:58

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