# What's the need of defining notion of distance using norm function in a metric space?

I have started studying normed spaces. I wonder what's the need of defining notion of distance using norm function. For example , we know that $\mathbb{R}$ is a metric space with respect to usual metric defined by $d(x,y) = \mid x-y \mid$.

Now, I am studying $\mathbb{R}$ is a metric space with respect to metric induced by norm defined by $d(x,y) = \parallel x - y\parallel$.

Edit 1: I mean can't we simply study metric spaces using distance function which doesn't involve norms? Why we have introduced concept of norms?

I have no problem in understanding things related with norms. But this question is troubling me which might sound trivial.

Thanks for helping me.

-
I'm not sure I understand your question. Are you asking "Do there exists metrics which are not norms?" – matt Jun 1 '12 at 6:18
More structure means stronger results. – copper.hat Jun 1 '12 at 6:26
@srijan Many metric spaces have distance functions which are not norms. For example the discrete metric en.wikipedia.org/wiki/Discrete_metric. Every norm is a metric, but not every metric is a norm. – matt Jun 1 '12 at 6:33
Normed spaces have an underlying vector space. The corresponding distance has nice properties such as translation invariance, homogeneity, etc. Much of our Euclidean intuition palsy well in normed spaces. – copper.hat Jun 1 '12 at 6:33
The phrase "norm defined by $d(x,y) = \lVert x-y \rVert$" in the question isn't quite right. The norm is denoted by $\lVert x \rVert$, while the metric induced by the norm is $d(x,y) = \lVert x-y \rVert$. – Rahul Jun 1 '12 at 6:41

Norms only make sense on vector spaces, whereas metrics can be defined on arbitrary sets.

Supposing we are in a vector space, the principles that distinguish a norm from a metric are 1) translation invariance, and 2) homogeniety. Metrics induced by a norm are always translation invariant and homogeneous, and given a translation invariant homogeneous metric you can build a norm via, $$||x||:=d(x,0).$$

Thus the core question is: what is the intuition behind translation invariance and homogeniety, and why are they interesting?

1) Translation invariance means that normed spaces look the same everywhere, in some sense. Any property that depends on pairwise distances between points will be the same if you translate all the points over. $$d(x+h,y+h)=||x-y|| \text{, independent of h.}$$

2) Homogeniety means that it is meaningful to put "units of measurement" on your space. If you measure two vectors in meters then take the norm distance between them, you get the same result as if you measured them in inches, then took the norm distance between them, then converted that distance from inches to meters. $$d(ax,ay) = |a|d(x,y).$$

-
I think this answer highlights the key special features of distances described by norms. Another aspect to consider is that most metric spaces that appear in analysis are normed vector spaces, and there are important ideas analysts use with norms that really don't work in general metric spaces. For an analogy, just because vector spaces over a field are particular examples of modules over a commutative ring doesn't mean we should only think about vector spaces within the setting of modules; linear algebra over fields has its own distinctive features. – KCd Jun 1 '12 at 15:42
@Nick Thank you very much for providing me useful information. – srijan Jun 1 '12 at 16:24
Glad to help - this is something that used to bother me as well. Norms are used in so many places, it becomes second nature so lot of mathematicians don't bother to really think about what they mean. – Nick Alger Jun 3 '12 at 1:18

As copper.hat wrote, more structure means stronger results. As an example take the theorem

Any two norms on $\mathbb{R}^n$ are equivalent.

In particular, any norm on $\mathbb{R}^n$ generates the same topology as the "usual" Euclidean norm (i.e., you get the same notions of convergence, open sets and so on as for the Euclidean metric.) Moreover, equivalence of norms also implies same notions of completeness.

On the other hand, as you learned in this question about the arctan metric, this is not true for different metrics on $\mathbb{R}^n$. The arctan metric on $\mathbb{R}$ generates the "usual" topology coming from the Euclidean metric, but $\mathbb{R}$ with that metric is not complete.

-
Thanks for replying. I am confused with notion of Completeness will also remain same? Since completeness is not a topological property. Point out me if i am wrong. – srijan Jun 1 '12 at 10:50
@srijan: How right you are! Edited ... – Hendrik Vogt Jun 1 '12 at 10:56