Real numbers equipped with the metric d(x,y)=|arctan(x)−arctan(y)| d (x,y) = | \arctan(x) – \arctan(y)| is an incomplete metric space

I have to show that the real numbers equipped with the metric
d(x,y)=|arctan(x)arctan(y)| is an incomplete metric space.

Certainly, I have to search for a Cauchy sequence of real numbers with respect to given metric that must not be convergent. But I am unable to figure out that.
Can anybody help me with this.

Thanks for helping me.

Answer

Sorry for reviving such an old problem…

Anyways, what is important here is that arctan is a bijection from R to (π/2,π/2), and it is an isometry if we give (π/2,π/2) the metric it carries as a subspace of R with the usual metric. If f:XY is a surjective isometry then the Cauchy sequences in Y are the images of Cauchy sequences under f, and the convergent sequences in Y are the images of convergent sequences under f, so Y is complete if and only if X is. Thus (R,d) is complete if and only if (π/2,π/2) with the metric dist(x,y)=|xy| is complete. But (π/2,π/2) is not complete since {π/21/n}n=1 is Cauchy and does not converge in (π/2,π/2).

Attribution
Source : Link , Question Author : Srijan , Answer Author : Adam

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