# Metric space. Real difficult exercise

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. –  savick01 Dec 16 '11 at 19:54
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? –  t.b. Dec 16 '11 at 19:58
It does not seem to be terribly difficult to find an example. In general, "difficulty" is a very subjective thing. –  Carl Mummert Dec 17 '11 at 18:17
You forgot $d(x,y) \geq 0$ for all $x, y \in X$. –  Stefan Smith Feb 13 '12 at 0:33

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. $0\leftrightarrow\text{center}$, $(0,\infty)\leftrightarrow\text{right half of circle}$, $(-\infty,0)\leftrightarrow\text{left half of circle}$. –  Jonas Meyer Dec 16 '11 at 20:27
But the substraction operation is not continuous in this case... –  user17786 Dec 17 '11 at 1:05
@savicko1 Sorry, your answer is incorrect. If $x\neq 0$, then $d(x,0)$ is not defined –  Minkow Dec 17 '11 at 10:20
@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. –  Jonas Meyer Dec 17 '11 at 10:52
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. –  Carl Mummert Dec 17 '11 at 18:17

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.

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