I was reading Metric space and saw that in one book they defined $R$ as a discrete metric space. But I have read early in the same book that R is normed linear space with usual metric $d (x,y)$$=$$||x-y||$. And I read another remark that any metric defined in the normed linear space will be associated with this metric. But discrete metric does not satisfy the homogeneity condition of a norm. I.e

$||ax||$$=$$|a||x|$. For any scalar $a $$>$$0$.

But $d (x,y)$$\neq $$d (ax,ay) $ for all scalar $a>0$. Then how R can be a discreete metric space.?? Any help would be appreciated. Thanks

Edit: another thing that should be kept In mind. Is $R$ is a discreete metric space. Then it will be bounded. So can $R $ be discreete metric space.

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    $\begingroup$ They either defined two metrics in $\Bbb R$ or you misread. $\endgroup$ – Will M. Mar 20 '17 at 4:03
  • $\begingroup$ You can define many different metrics in $\Bbb R$. The standard metric in $\Bbb R$ is defined by the euclidean norm $|{\cdot}|$. But the discrete metric is not induced from a norm. A norm can induce a metric but this doesnt mean that any metric is induced from a norm. $\endgroup$ – Masacroso Mar 20 '17 at 4:08
  • $\begingroup$ Yes. But R is discreete metric space then R will be bounded $\endgroup$ – user426700 Mar 20 '17 at 4:17

What you should be aware of is that strictly speaking $\mathbb R$ is not a metric space. $\mathbb R$ is just a set (of real numbers), in order to be a metric space one need to have a metric too.

Normally one assumes that we have the metric $d(x,y) = |x-y|$, which makes $(\mathbb R, d)$ a metric space. Normally one sloppily just writes $\mathbb R$.

You could define another metric, for example the discrete metric $e(x,y)$ (that is $0$ if $x=y$ and $1$ if $x\ne y$). This makes $(\mathbb R, e)$ a metric space too.

There's no requirement of a metric to have the homogenity property. In fact there's no requirement for a metric space to be linear to start with, that is scaling the element does not need to make sense. All the other properties of a metric is met by the discrete metric so there's actually nothing that hinders it (ie $(\mathbb R, e)$) from being a metric space.

Note that it's the same with normed spaces. They aren't just a set, it's a set together with a norm - so $\mathbb R$ is strictly speaking not a normed space, but $(\mathbb R, |\cdot|)$ is. What we can say about a norm on $\mathbb R$ need to be of a specific form, for we have $||x|| = ||x\cdot x|| = |x|\cdot||1||$, that is every norm is a scaled absolute value.

Note also that to be really strict we shouldn't call $\mathbb R$ a linear space. We should then include the operations of scaling and vector addition. However this often becomes impractical - instead the convention is that unless explicitly stated one assumes certain operations to be used (we assume that addition is the normal addition, multiplication is the normal multiplication, scaling too, absolute value is the norm and so on).

  • $\begingroup$ So if we can define any function on a set that follows conventional rules for a function to be metric. But changes some fundamental property of the set. We can still call the function a metric??? $\endgroup$ – user426700 Mar 20 '17 at 7:46
  • $\begingroup$ @user426700 How can a function define the fundamental property of the set? The fundamental properties of a set lies only in the elements that are in the set. However the requirements on a metric is only those axioms for metrics - otherwise anything goes (however keep in mind that things that are true for one metric need not be true for another). $\endgroup$ – skyking Mar 20 '17 at 7:53

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