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Consider the metric space $(\mathbb{R},d)$ with $d(x,y)=|x-y|$ for $x,y \in \mathbb{R}$ with $x \neq 0 \neq y$ and $d(0,x)=1+|x|$ for $x \in \mathbb{R}_{0}$ and $d(0,0)=0$.

I figure this metric space is not complete, since we can take the Cauchy sequence $(x_{n})_{n}$ with $x_{n}= \frac{1}{n}$. Then for $\epsilon >0$ we can find a $n_{0} \in \mathbb{N}$ so that for $n,m \geq n_{0}$ it holds that $d(x_{n},x_{m})=|\frac{1}{n} - \frac{1}{m}| < \epsilon$ since $\frac{1}{n} \neq 0 \neq \frac{1}{m}$. But $\lim_{n \rightarrow \infty}d(x_{n},x_{m})=d(0,x_{m})=1+ \frac{1}{m}$ and this converges to $1$, while $\frac{1}{m} \rightarrow 0$ and $d(0,0)=0$. Thus the Cauchy sequence has no limit.

Is this correct? And if so, how should I find a completion of this metric space?

simp
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    To prove that it isn't complete you should show that $\frac{1}{n}$ cannot converge to a nonzero real, and cannot converge to $0$. Have you checked that $d$ was a distance ? – Maxime Ramzi Jun 18 '17 at 11:07
  • The way you defined your metric would give $d(0,0) = 1 \neq 0$, contradicting the definition of a metric. So I think there is something wrong with the statement of the problem at the moment. – Andre Jun 18 '17 at 11:09
  • I was given this metric so I think it is well-defined. :) Also, $d(0,0)=0$ here because $d(0,x)=1+|x|$ for $x \in \mathbb{R}_{0}=\mathbb{R} \backslash { { 0 }}$, I'll adjust it in my question. – simp Jun 18 '17 at 11:13
  • I have checked now that $d$ is a distance. – simp Jun 18 '17 at 11:19

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