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 $(\mathbb{N},d_2)$ is a complete metric space.

Attempt: So I need to show that every Cauchy sequence in this metric space converges. Presumably all of these convergent Cauchy sequences would be eventually constant -- otherwise they wouldn't converge in $(\mathbb{N},d_2)$.

share|cite|improve this question
Just a side-question, I'm not familiar with this: what is $d_2$ here? – InvisiblePanda Feb 29 '12 at 9:45
the usual metric. – Emir Feb 29 '12 at 9:47
Hello @Emir, I was pretty sure that it meant a metric, regarding the context, but not so sure of why there was an index 2. I was asking out of pure curiosity and hoping you could give me a short definition of the 'usual metric'. Is it just $d_2(m,n):=|m-n|$? If so, your attempt is the right way. Nevermind, Brian M. Scott already formalized the argument a bit ;) – InvisiblePanda Feb 29 '12 at 10:06
I have downvoted this because you assume everyone knows what $d_2$ means in your post. You should include definitions the next time. In the body of the question, not in later comments. – Asaf Karagila Feb 29 '12 at 10:21
@Rand,@Asaf: $d_2$ is a fairly common notation for the Euclidean metric. It’s probably worth asking for confirmation, but I can understand why Emir wouldn’t have thought it necessary to mention. – Brian M. Scott Feb 29 '12 at 10:29

HINT: Suppose that $\langle n_k:k\in\mathbb{N}\rangle$ is a Cauchy sequence in $\langle\mathbb{N},d_2\rangle$. This means that for each $\epsilon>0$ there is a $k_\epsilon\in\mathbb{N}$ such that $|n_i-n_j|=d_2(n_i,n_j)<\epsilon$ whenever $i,j\ge k_\epsilon$. What happens when you look at $\epsilon=1$ (or any smaller positive value)? What can you say about integers $a$ and $b$ if $|a-b|<1$?

share|cite|improve this answer
For integers, we have $|a-b|<1\implies a=b$. How do I know that eventually $\epsilon\leq 1$, though? – Emir Feb 29 '12 at 10:12
@Emir: ‘Eventually’ makes no sense there: the definition of Cauchy sequence is that for every positive $\epsilon$ such a $k_\epsilon$ exists. In particular, there must be a $k_1$ such that $|n_i-n_j|<1$ whenever $i,j\ge k_1$. – Brian M. Scott Feb 29 '12 at 10:23
@Emir: $(n_k)_{k\in\mathbb{N}}$ is Cauchy, and as Brian M. Scott wrote, for each $\epsilon>0$, there will be such a $k_\epsilon$. So if you choose any $\epsilon>0$ - in particular you may choose $\epsilon$ to be 1 - there will be an index s.t. the desired inequality holds for all elements of the sequence with bigger indices. – InvisiblePanda Feb 29 '12 at 10:25
so I just need to say $\forall\epsilon>0, \exists k_{\epsilon}:\forall i,j>k_{\epsilon},d(n_i,n_j)<\epsilon\implies\exists k_1$ s.t. $d(n_i,n_j)<1$ whenever $i,ij\geq k_1$ which converges in $\mathbb{N}$, hence $\mathbb{N}$ is complete? – Emir Feb 29 '12 at 11:14
@Emir: I’d say a little more at the end, but yes, that’s the heart of it. I’d way that since $d(n_i,n_j)<1$ for $i,j\ge k_1$, and the $n_i$ are integers, we must have $d(n_i,n_j)=0$ for $i,j\ge k_1$ and hence $n_i=n_{k_1}$ for all $i\ge k_1$. Therefore the sequence converges to $n_{k_i}$, and $\langle\mathbb{N},d\rangle$ is complete. – Brian M. Scott Feb 29 '12 at 12:52

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.