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?

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 $$\forall\epsilon>0,\exists N\text{ such that }n>N\Rightarrow d(a_n,L)\leq|a_n-L|<\epsilon$$
