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Given that $(X,d)$ is a metric space, with $d$ the French railway metric. Show that $(X,d)$ is a complete metric space.

Suppose $\{x_n\}$ is a Cauchy sequence in $(X,d)$. Choose $N$ such that for all $m,n$ $\geq$ N: d($x_{n},x_{m}$) < $\epsilon$. (Note that: d($x_{n},x_{m}$) = d($x_{n},p)+d(p,x_{m}$) , $p \in X$). Then for all $n \geq N: d(x_{n},x) \leq$ .....

Thats where im stuck. Help would be much appreciated.

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    How would you define the French railway metric? – 2017-02-27
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    The definition given: Let (F,d) be a metric space and p in F. Then for all x,y in F: x $\not=$ y implies d(x,y) = d(x,p) + d(p,y). – 2017-02-27
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    This is all that is given in the problem. The metric (X,d) is a general metric. – 2017-02-27
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    @pjs36: That _is_ the definition of that particular kind of metric. It's also known as a _post office metric_. – 2017-02-27
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    @HenningMakholm Thanks, I do see it now, I just hadn't read closely enough earlier. – 2017-02-27

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If $x_n$ is a non-constant Cauchy sequence in a post office metric with center $p$, then we will show that $x_n\to p$. Pick $\varepsilon>0$. Then, there exists $N\in \mathbb N$ so that $d(x_n, x_m)<\varepsilon$ if $n,m\geq N$. Now:

$$d(x_n,p)\leq d(x_n,p)+d(p,x_m)=d(x_n,x_m)<\varepsilon$$

so, for an arbitrary $\varepsilon$, there exists $N$ so that $d(x_n, p)<\varepsilon$ if $n\geq N$.

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    A Cauchy sequence that converges towards the center doesn't need to be eventually constant. – 2017-02-27
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    It's not convergence towards the center. From that inequality, you get $d(x_p, x_q)<\varepsilon$ for all $p,q\in\mathbb N$, and $\varepsilon$ was arbitrary. – 2017-02-27
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    I dont get how you derived this. Is $x_{p}$ the so called centre of the metric? – 2017-02-27
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    @fly: Suppose our space is $\mathbb R$ with $d(x,y)=|x|+|y|$, which is a post office metric with center $0$. Then the sequence $(\frac1n)_n$ is Cauchy and is _not_ eventually constant. – 2017-02-27
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    @HenningMakholm that distance does not verify $d(x,y)=d(x,z)+(z,y)$, as the OP posted in their question. – 2017-02-27
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    @fly: Yes it does, when $z$ is the center point: $d(x,y)=|x|+|y|=|x|+|0|+|0|=|y| =d(x,0)+d(0,y)$. – 2017-02-27
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    Oh, I think I misread the definition, then. I was supposing $d(x,y)=d(x,z)+d(z,y)$ for every $z$. – 2017-02-27
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    you now have t en z to be the center point. Only one point can be the center point. – 2017-02-27
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    Yes, I understand it now. I'll try to provide a useful answer now. – 2017-02-27
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    @flytothesurface Thanks :) – 2017-02-27
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    However, there's still the possibility of an eventually constant sequence converging to a non-central point. Even in the post-office metric we still have $d(x,x)=0$ for every $x$, as an explicit exception to the rule about going through the center. – 2017-02-27
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    @HenningMakholm yes, that's true. Thanks. – 2017-02-27