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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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    @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.2012-02-29

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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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    @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.2012-02-29