Prove that (X,d) is complete Let $X=\mathbb{R}$ and $d(x,y)=\min(1, |x-y|)$
I am trying to prove that this metric space is compete. I know that means that every Cauchy sequence converges. So when |x-y| < 1, that would just be the standard metric on R which I know is complete. I'm a bit confused for when d(x,y)=1 though. Would this mean that the Cauchy sequences would both be equal and constant sequences?

The idea here is that completeness of a metric space is only a local property. By taking the minimum you are just throwing out information about points that are far away from one another, effectively saying all I know is that these points are just far away. But the Cauchy sequences (the sequences you care about) always get close enough so that this new metric does not matter. So the first part of your idea is basically right. 
