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Definition. Function $ d:X\times X\rightarrow \mathbb{R} $ is called metric, if d satisfes following axioms:

  1. $\quad d(x,y)=0\Longleftrightarrow x=y$
  2. $\quad d(x,y)=d(y,x)$
  3. $d(x,y)\leq d(x,z)+d(z,y)$

$\forall x,y,z\in X$. Metric space is denoted as $(X,d)$

Let $ (\mathbb{R},d) $ be metric space. Is it true, that for any metric $ d $, from $ d(x_{n},x)\longrightarrow0 $ follows $ d(x_{n}-x,0)\longrightarrow C $, when $ n\longrightarrow\infty $, where $C\in \mathbb{R}$ is constant? Actually, it is not true, but it is terribly difficult to find counterexample

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    You forgot $d(x,y) \geq 0$ for all $x, y \in X$.2012-02-13

2 Answers 2

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As I said in a general metric space you don't have subtraction and zero. However you can imagine a metric on $\mathbb R$ which is the euclidean metric for all points except for zero: $d(0,0)=0,\ d(0,x)=\max(|x|,1)\ \forall x\neq 0, d(x,y)=|x-y| \ \forall x,y\neq0\ \ \ $ it is a good counterexample. (Of course you have to assume $x_n \neq x$.)

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    @JonasMeyer: Nice observation. I'd like to add that it is *isometric* with $\mathbb{R} \setminus \{0\} \cup \{ (0,1) \} \subseteq (\mathbb{R}^2, d_{\max})$, where $d_\max$ is the square norm on $\mathbb{R}^2$.2012-01-01
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Let $f\colon \mathbb{R} \to \mathbb{R}^2$ be $ f(x) = \begin{cases} 0 & x = 0 \\ x^{-2} & x \not = 0 \end{cases} $ Let $d$ be the metric on $\mathbb{R}$ such that $d(x,y)$ is the distance in $\mathbb{R}^2$ from $(x,f(x))$ to $(y,f(y))$. This is a metric because the metric on $\mathbb{R}^2$ is a metric.

Now let $a_n = 1+1/n$. Then $(a_n)$ converges to $1$ in this metric (basically because $f$ is continuous in the usual sense at $x=1$). But the distances from $1-a_n$ to $0$ go to infinity.

This illustrates a different way to make metric on $\mathbb{R}$ with interesting convergence properties.