Show that a set in a metric space is complete Let X be the closed unit ball in $(\mathbb R ^2,||.||)$ where $||.||$ is the euclidean norm. Let :
$$d(x,y)= \left \{
   \begin{array}{r c l}
      ||x-y||  & \text{if x,y and 0 are on the same line}\\
      ||x||+||y||    & \text{otherwise}
   \end{array}
   \right .$$
I want to show that $(X,d)$ is complete.
I have tried different approaches :

*

*Using the bound $d(x,y)\geq|\space ||x|| - ||y|| \space | $ then using that $(X,||.||)$ is complete (beacause it is compact) I can prove that if $(x_n)$ is a Cauchy sequence in $(X,d)$ then $||x_n||$ converges in $(X,||.||)$. But I cant see how to go from there.

*Let $(x_n)$ a Cauchy sequence in $(X,d)$. If there is only a finite number of terms such that $x_n,x_m,0$ are not on the same line then I can prove that the sequence is convergent in $(X,d)$. In the same way if there is only a finite number of terms such that $x_n,x_m,0$ are on the same line then I can prove that the sequence converge to 0 in $(X,d)$. But I don't know what to do there is only an infinite number of terms such that $x_n,x_m,0$ are on the same line and an infinite number of terms such that $x_n,x_m,0$ are not on the same line.

 A: $(X,d)$ certainly isn't compact ($S^1$ is closed and discrete). So 1. is a dead end. Just show that $(\Bbb R^2,d)$ is complete and then so will all closed subsets be, including the closed unit ball.
If $(x_n)_n$ is a Cauchy sequence in $(\Bbb R^2,d)$, then the definition makes it clear that for infinitely many indices, either $x_n$ lie on a common line through the origin, and then the metric is just the usual (of $\Bbb R$) on that line, and we can apply completeness to that subsequence, or the $x_n$ lie within a small closed ball around the radius (in the usual metric) and then we can apply completeness of a closed ball to show convergence of that subsequence too. Finally we then apply that if a Cauchy sequence has a convergent subsequence it converges as a whole to that same limit.
E.g. if infinitely $x_n$ (say $n \in M \subseteq \Bbb N$) lie outside  $\{x\in \Bbb R^2\mid \|x\| \le r\}$ then if $x_n,x_m$ for $x \notin M$ don't lie on the same line, their distance would be at least $2r$ and so couldn't be part of a Cauchy sequence.
