Metric space - complete - discrete

a)Let $M\neq \emptyset$ be a set. Show that,

$d(x,y):= \begin{cases}0&\text{if x=y}\\1&\text{otherwise}\end{cases}$

is a metric on $M$. This is a discrete metric. Formulate and prove an easy criterion for the convergence of a sequence $(x_n)_{n\in N}$ in (M,d). Is M complete?

b)Let V be a normed linear space with norm $\|\cdot\|$. Show that:

(i) $d(x,y):= \|x-y\|$ defiens a metric on V.

(ii) $|\|x\|-\|y\||\leq \|x-y\|$ for all $x,y\in V$. (d is named the induced metric of $\|\cdot\|$)

c) Show that the discrete metric on $R^2$ isn't induced by any norm.

We started talking about the concept of a metric space last week in class, which is why I'm not yet fully sure how to deal with exercises concerning this type of topic.

Anyway, here are my thoughts:

a): In class we said that if $x=y$ then for a metric space it follows that $d(x,y)=0$, right? But here comes the problem: I'm not sure how to show that with the second condition it's still a metric. I thought about just proving the general conditions for a metric space. Since $d(x,y)=0$ for $x=y$ I followed with $1+1=d(x,y)+d(y,x)\geq d(x,x)=0$. Is that enough to prove that M is a metric space?

b) (i) To be honest I'm not sure how to show that. In class the prof mentioned that basically any norm can define a metric. And I'm not yet fully familiar with the basic norms used in mathemtics. Just out of memory I can remember 4 norms: $\|\cdot\|~;~\|\cdot\|_1~;~\|\cdot\|_2~;~\|\cdot\|_\infty$, but not sure how each of them is defined.

(ii) I can imagine that this is true, but I'm not sure how to show that mathemtically.

c) The prof didn't talk about discrete metrics, so I read about it on Wiki. It's basically the metric that is defined in a), right? Also not sure what an induced metric is. The wiki entry says that it's some sort of tensor. So basically I'm also lost here.

Could anyone give me some hints to approach these?

@Konstantin

a) You need to read several times the axioms to be verified by a metric in order to be fluent with them! In particular for the triangle inequality, you need to verify that for all $x,y,z \in M$ you have $$d(x,z) \le d(x,y) + d(y,z)$$ You can't start with $d(x,x)$

b)(i) Look at definitions for norm and metric precisely. The proof will come straight afterwards.

b)(ii) Start with $$\Vert x \Vert = \Vert (x-y) + y \Vert \le \Vert x-y \Vert +\Vert y \Vert$$

c) If a norm would induce the discrete metric, you would have for all $x,y$: $d(x,y)=\Vert x-y \Vert$. What do you think then for $x \neq y$ of $d(2x,2y)$?

• About a: I somehow managed to show that all the conditions are fulfilled, but how can I show that it's complete? About c: I got $d(2x,2y)=\|2x-2y\|=2\|x-y\|$ and because $\|x-y\|=1$ for $x\neq y$ and $d(2x,2y)=2\|x-y\|=2\cdot 1=2\neq 1$ is a contradtiction and it's not discrete. Is that argumentation correct? – Konstantin Jun 16 '15 at 12:08
• Yes it is correct. – mathcounterexamples.net Jun 16 '15 at 12:13
• I'm kind of struggling with the criterion for a sequence in M. I was thinking of using the cauchy convergence criterion since it's the only criterion that came to mind, but I don't think it can be applied here. Any ideas? – Konstantin Jun 16 '15 at 14:39