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?
 A: First, some intuition: Convergence is a property that only care about "really big" values of $N$. If you change the behavior of the beginning of a sequence, that's never going to change the convergence behavior. Similarly, it only cares about "really small" values of $\epsilon$. Your metric only differs from the usual metric for "largish" (specifically, not arbitrarily small) values of $\epsilon$, so it shouldn't have any effect. Exactly the same sequences will converge, and they'll converge to exactly the same values as in the usual metric.

Now to prove this fact: Take some Cauchy sequence, $a_k$. We wish to prove that this sequence converges.
It follows from the definition of $d$ that $d(x,y)\leq |x-y|$ holds for all $x,y$. Considering $a_k$ in the metric space with the usual notion of distance, we know it has some limit, $L$, since $\mathbb{R}$ is complete. In order to show that $a_k\to L$ under the new metric, we need to show that $$\forall\epsilon>0,\exists N\text{ such that }n>N\Rightarrow d(a_n,L)<\epsilon$$
We already know that $$\forall\epsilon>0,\exists N\text{ such that }n>N\Rightarrow |a_n-L|<\epsilon$$ and so we can just evoke the fact that $d(x,y)\leq |x-y|$ to find out that $$\forall\epsilon>0,\exists N\text{ such that }n>N\Rightarrow d(a_n,L)\leq|a_n-L|<\epsilon$$
A: 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. 
