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$(\mathbb{Z}, \rho)$ is a metric space with the metric: $\rho(m, n) = |e^{in} - e^{im}|,\ i^2 = -1,\ m, n \in \mathbb{Z}$
So I need construct completion set of integers. How can I do this?

My thoughts:
$f(n) = e^{in}$ monotonically increased function therefore for $\rho$ the conditions of metric are fulfilled. So if I could construct function $f: X \to Y$ (where $Y$ is set with fixed interval or semi-interval) then perhaps I could define a metric on $Y$ (let it be called $d$) and construct an isomorphism between $(X, \rho)$ and $(Y, d).$

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    $\begingroup$ Unfortunately, the $i$ ruins the monotonicity by making $f(n)\notin\Bbb R$ for all $n\neq 0.$ You won't get an interval. It is still injective, though, so $\rho$ is a metric on $\Bbb Z.$ $\endgroup$ – Cameron Buie Oct 12 '17 at 12:44
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One way to determine the completion of a space is to make the correct wild guess about what the completion is. Although maybe the guess does not need to be so wild, maybe there's a clue.

For this problem, you can get a clue by looking at the formula $$\rho(m,n) = |e^{in} - e^{im}| $$ Break this formula into parts: define a function $$f : \mathbb{Z} \to \mathbb{C} $$ by the formula $$f(n) = e^{in} $$ Denote the usual Euclidean metric on $\mathbb{C}$ as $$d(w,z) = |w-z| $$ The formula for $\rho$ can then be rewritten as $$\rho(m,n) = d(f(n),f(m)) $$ Furthermore, $f$ is a one-to-one function, because $2\pi$ is irrational: if $f(m)=f(n)$ then $e^{in}=e^{im}$ so $e^{i(n-m)}=1$ so $n-m$ is a multiple of $2 \pi$, so $n=m$.

It follows that $f$ is an isometry from $Z$ to its image $f(Z) \in \mathbb{C}$.

Now there's a general theorem: for any complete metric space $M$ and any subset $A \subset M$ endowed with the restricted metric, the completion of $A$ is the closure of $A$. And the Euclidean metric on the plane $\mathbb{C}$ is a complete metric space.

So, the completion of $Z$ in the metric $\rho$ is isometric to the closure of $f(Z)$ in $\mathbb{C}$.

Can you determine the closure of $f(Z)$ in $\mathbb{C}$?

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  • $\begingroup$ Thank you for your answer! So, I think the closure is the circle: $S^1 = \{ z \in \mathbb{C} : |z| = 1\}$? $\endgroup$ – Pennywise Oct 12 '17 at 19:44
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    $\begingroup$ That's correct. $\endgroup$ – Lee Mosher Oct 13 '17 at 1:02

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