Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
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.

share|cite|improve this answer
Thank you for your time! – passenger Mar 28 '12 at 22:21

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.