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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 $(\mathbb{N},d_2)$.

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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 $|a-b|<1$?

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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

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