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I'm supposed to show that the set of real numbers, $R$, is an incomplete metric space with metric $d(x,y)=|\text{tan}^{-1}(x)-\text{tan}^{-1}(y)|$.

My issue: What we usually do in such problems is that we find a cauchy sequence which doesn't converge in same metric space. Now since all of the terms of the sequence $(x_n)_{n\geq1}$ are supposed to come from real numbers, shouldn't the limit itself be in real numbers too? Also, we were taught a theorem in class "Every Cauchy sequence of real numbers converges" which means that the limit will exist too. Then why would it be incomplete? Thanks!

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marked as duplicate by José Carlos Santos real-analysis Jan 28 at 9:52

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Consider $u_n=n$, $u_n$ is a Cauchy sequence since $lim_ntan^{-1}(n)=\pi/2$ and does not have a limit since there does not exists a number $x$ with $tan^{-1}(x)=\pi/2$.

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  • $\begingroup$ Oh got it! So the convergence is determined by the metric right? Like "Every Cauchy sequence of real numbers converges" is probably true only for the usual metric. Am I correct? $\endgroup$ – Ankit Kumar Jan 28 at 9:54

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