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Let $d : \mathbb{R}^n \times \mathbb{R}^n \to [0, \infty]$ be defined by $$ d(x,y) = \left\{ \begin{array}{ll} 0 & : ~ x = y \\ ||x|| + ||y|| & : ~ x \ne y \end{array} \right. $$ where $||\cdot ||$ denotes the usual norm of $\mathbb{R}^n$.

  1. Show that $d$ is a metric.

  2. Draw the $\varepsilon$-Spheres $B_{\varepsilon}(x_0) := \{ x \in \mathbb{R}^2 ~|~ d(x,x_0) < \varepsilon \}$ for $x_0 = (0,0)$ and $x_0 = (1,1)$ and $\varepsilon = \frac{1}{2}, 1, \frac{3}{2}$.

  3. Characterize the open, closed and compact sets with respect to this metric.

  4. Is $(\mathbb{R}^n, d)$ complete?

Number 1) is simple, for 2) I got:

If $x_0 = (0,0)$ then $$ d(x,x_0)= \left\{ \begin{array}{ll} 0 & \textrm{ for } x = (0,0) \\ \sqrt{x^2 + y^2} & \textrm{ otherwise } \end{array} \right. $$ and if $x_0 = (1,1)$ then $$ d(x,x_0) = \left\{ \begin{array}{ll} 0 & \textrm{ for } x = (1,1) \\ \sqrt{2} + \sqrt{x^2 + y^2} & \textrm{ otherwise } \end{array} \right. $$ and the pictures are simple spheres with the point $x_0$ in the sphere ($x_0 = (0,0)$) or isolated outside ($x_0 = (1,1)$).

But with 3) I have my problems, i conjecture that $$ B_{\varepsilon}(x) \quad \textrm{ is open iff } \quad ||x|| - \varepsilon > 0 $$ and going on I know that finite intersections of open sets are open, but then had I got all open sets by this construction? And what about the other properties, how can I characterize them, do you have any hints?

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Note that you are asked to provide the topology induced by $d$, not what balls for $d$ are open in the usual topology on $\Bbb R^n$. – Lord_Farin Oct 15 '12 at 17:04
isn't it the case that a topolgy induced by a metric d consists of all open balls? – Stefan Oct 15 '12 at 17:07
Yes, these balls are defined by $B_\epsilon^d(x) := \{y \in \Bbb R^n: d(x,y) < \epsilon\}$. This is different from $B_\epsilon (x) = \{y \in \Bbb R^n: \|x-y\|< \epsilon\}$ (the latter is for the Euclidean metric $d(x,y) = \|x-y\|$). Reading again, I think we may have confused one another. Which of my two mentioned types of balls did you intend with $B_\epsilon(x)$? – Lord_Farin Oct 15 '12 at 17:09
By the way, this is sometimes called the British Rail metric, or something similar with regard to another highly centralised transport system. – Chris Eagle Oct 15 '12 at 17:17
i mean the balls defined by $B_{\varepsilon}(x_0) := \{ x \in \mathbb{R}^2 ~|~ d(x,x_0) < \varepsilon \}$ – Stefan Oct 15 '12 at 21:37
up vote 3 down vote accepted

For every $x \ne 0$ we can find $0 < \epsilon < \|x\|$ so that $B_\epsilon(x) = \{x\}$. Therefore, for every $x \ne 0$, $\{x\}$ is open. Also, evidently $B^d_\epsilon (0) = B_\epsilon(0)$; i.e. all Euclidean balls around $0$ are open in the topology induced by $d$ (denote it with $\tau_d$ from now on).

Thus, every open in $\tau_d$ is of the form:

$$B_\epsilon(0) \cup S$$

with $\epsilon \ge 0, B_\epsilon(0)$ an Euclidean ball around $0$ (remark the empty case $\epsilon = 0$) and $S \subseteq X \setminus \{0\}$.

The closed sets are then characterised as $\complement(B_\epsilon(0)) \cap \complement(S)$ with $B_\epsilon(0)$ and $S$ as above. Now $\complement (S)$ contains $0$ but is otherwise an arbitrary subset of $X$. Since $\complement(B_\epsilon(0))$ does not contain $0$ if $\epsilon > 0$, the only requirement remaining for $C \subseteq S$ to be closed in this case, is that it be contained in the complement of some $B_\epsilon(0)$. If $\epsilon = 0$, then since $\complement(B_\epsilon(0)) = X$, the only remaining requirement is that $0 \in C$.

This is summarized by saying that $C \subseteq X$ is closed iff:

$$0 \in C \quad \text{or}\quad \inf \{\|c\|: c \in C\} =: \operatorname{dist}(C, 0) > 0$$

The compact sets are easily shown to be all finite $S \subseteq X$.

Finally $(\Bbb R^n, d)$ is complete since a Cauchy sequence $(x_n)_n$ either takes finitely many values (and trivially converges) or has for $\epsilon >0$ that $d(x_n,x_m) < \epsilon \implies d(x_n,0) <\epsilon$ (since $x_n \ne x_m$ infinitely often), and we conclude $\displaystyle \lim_{n \to \infty} x_n = 0$ in this case.

EDIT: Grateful thanks to fgp for pointing out clumsy errors in my analysis of the problem.

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In the last paragraph, you cannot technically conclude from $d(x_n,x_m) < \epsilon$ that $d(x_n,0) < \epsilon$ - it could also be that all the $x_i$ are equal after some point. The convergent sequences are thus those which either converge to zero in the traditional sense, or all trivially convergent because they take only finitely many values. The overall conclusion that the metric is complete is correct, though. – fgp Oct 15 '12 at 17:48
Agreed, I jumped to a conclusion there; thanks for pointing it out. – Lord_Farin Oct 15 '12 at 17:53
Your characterization of the closed sets isn't right - the set $\{0\}$ is closed, since $X \setminus \{0\}$ is open (it's the union of all ${x}$ for $x\neq0$, which are all open sets), yet it doesn't fit your description. – fgp Oct 15 '12 at 17:54
In fact, all sets which contain $0$ are closed, since all sets which do not contain $0$ are open.... – fgp Oct 15 '12 at 17:57
Yes, I figured it out; at first, it seemed almost like all sets were closed, but this would be ridiculous, rendering all sets open as well. After finding the closed sets if $\epsilon > 0$ I failed to reconsider $\epsilon = 0$. I hope everything is correct now; thanks again. – Lord_Farin Oct 15 '12 at 17:59

Regarding (2), the $B_e(x_0)$ are not all spheres, I believe - some of them contain just a single point. You have $d(x,y) \geq ||x||$, so $B_\epsilon(x_0)$ is $\{x_0\}$ if $||x_0|| \geq \epsilon$, since you then have $d(x_0,y) \geq \epsilon$ for all $y \neq x_0$. Thus $$ B_\epsilon(x_0) = \{x_0\} \bigcup \{y : ||y|| < \epsilon - ||x_0||\} $$ The second set is empty if $\epsilon \leq ||x_0||$.

Regarding (3) - You need to characterize the open sets of the topology induced by $d$. Thus, all $B_\epsilon(x)$ are open by definition. And so are all (finite or infinite) unions of sets $B_{\epsilon_i}(x_i)$. I guess what they want you to do is to find a more direct way to characterize the open sets than that. Usually, the set of all open sets is not easily described without resorting to at least infinit unions, but here you can do much better. Look at the balls $B_{||x||}(x)$ for arbitrary $x$, in particular...

One interpretation of this metric, btw, is that to go from $a$ to $b$, you always have to go from $a$ to $0$ first, and then from $0$ to $b$. Thus, for every point $a$, the closest "neighbour" of $a$ that isn't $a$ itself is always $0$, and its distance is $||a||$. The matric is therefore sometimes called french railroad metric, alluding to the (surely exaggerated!) claim that to get from one city to another by railroad in france, you always have to go via paris.

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$B_\epsilon (x_0)$ always contains $x_0$. – Chris Eagle Oct 15 '12 at 17:18
@ChrisEagle Yeah, noticed my error only after I posted unfortunately. Fixed now, though. – fgp Oct 15 '12 at 17:23

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