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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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    In a general metric space you don't have things like subtraction and zero.2011-12-16
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    Could you please try to make your title a bit more descriptive and say what exactly the assumptions are, that you put on $(R,d)$ and the $-$ operation?2011-12-16
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    It does not seem to be terribly difficult to find an example. In general, "difficulty" is a very subjective thing.2011-12-17
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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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    This space is homeomorphic to the space you get by taking the unit circle and removing the north and south poles, and placing a point at its center, with the Euclidean metric in the plane. $0\leftrightarrow\text{center}$, $(0,\infty)\leftrightarrow\text{right half of circle}$, $(-\infty,0)\leftrightarrow\text{left half of circle}$.2011-12-16
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    But the substraction operation is not continuous in this case...2011-12-17
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    @savicko1 Sorry, your answer is incorrect. If $x\neq 0$, then $d(x,0)$ is not defined2011-12-17
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    @cibulis: Under savick01's definition, if $x\neq 0$, $d(x,0)=\max(|x|,1)$. This was not stated explicitly, but to be a metric $d$ must be symmetric, so it was left to be inferred.2011-12-17
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    @Jonas Meyer: If I choose savicko1 metric, $d(x_{n},x)=|x_{n}-x|\longrightarrow 0$ means, that $d(x_{n}-x,0)=max(|x_{n}-x|,1)\longrightarrow 1$, so it means that $d(x_{n}-x,0)$ converges, therefore Savicko1 metric is not useful2011-12-17
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    Just to point out one way to see this is a correct example. The sequence $1, 1.1, 1, 1.01, 1, 1.001, \ldots$ converges to $1$ in this metric but if we subtract $1$ we get $0, .1, 0, .01, \ldots$, and the corresponding sequence of distances from $0$ is $0, 1, 0, 1, \ldots$, which does not converge.2011-12-17
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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.