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Prove that the set $ \displaystyle{\mathbb{N} =\{1,2,3, \cdots \} }$ is nonwhere dense in metric space $ \displaystyle{ \left( \mathbb{R} ,|\cdot| \right)}$ .

I have found a solution in two steps:

  1. I prove that $\displaystyle{\bar{\mathbb{N}}=\mathbb{N}}$
  2. $\displaystyle{ \text{int}\mathbb{N} = \emptyset }$

From these two we get that $ \displaystyle{ \text{int}\left(\bar{\mathbb{N}} \right) = \emptyset}$ so we are done.

I wonder if I we can found a solution proving that every interval $[a,b]$ has a sub-interval whose intersection with $\mathbb{N}$ is the empty set.

Any help?

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1 Answer 1

up vote 4 down vote accepted

Suppose that $a<b$. If $(a,b)\cap\Bbb N=\varnothing$, then $[a,b]$ certainly has a non-empty open subinterval disjoint from $\Bbb N$. Otherwise, let $n\in(a,b)\cap\Bbb N$. If $b\le n+1$, then $(n,b)$ is a non-empty open subinterval of $[a,b]$ disjoint from $\Bbb N$; if $b>n+1$, $(n,n+1)$ is such a subinterval. Thus, in all cases such a subinterval does indeed exist.

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Thank you for your time! –  passenger Mar 28 '12 at 22:21

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