What is the intuition and motivation behind a norm on a space? I have a question regarding norms. I am looking for an intuitive understanding of what norms do. From what I know so far, in any arbitrary space, if I have a norm, then it appears that it allows me to define distances. That is, if I have two points in the space, I know where they are in relation to one another. Is the motivation then that I can sort of "classify" and discriminate where things are once I have a metric or norm on the space? Also, if a space doesnt have a metric or norm, what would that mean? Thank you
 A: Considering $\mathbb R^n$ with its standard linear structure and inner norm given by $x\cdot y=\sum _k x_ky_k$ the Cauchy-Schwarz ($|x\cdot y|\le \|x\| \|y\|$) inequality allows one to define angles between vectors and relate the angle to the norms. Here the norm $\|x\|$ is defined to be $\sqrt {x\cdot x}$, i.e., it is derived from the inner product. 
Now the whole idea of defining the inner product the way it is is coming from the law of cosines, so it is entirely geometric. So, one can say that a vector space with an inner product is an axiomatization of angles and lengths. 
The idea behind a normed space is to axiomatize lengths, but not angles. This is important for two reasons. The first is that it allows one to ignore angles when these are irrelevant. Secondly, and somewhat more importantly, quite often a given vector space of interest, such as the $\ell_p$ spaces for $p\ne 2$, does not have a notion of angles, but does have a perfectly good norm defined on it. 
A norm is sufficient to define a metric function and thus obtain a topology on the space, which is why having a norm is important.
A very healthy way to think of a normed space is as a fusion between a vector space and a metric space, as follows. A metric on a vector space $V$ is translation invariant when $d(x+z,y+z)=d(x,y)$ for all vectors $x,y,z$. The metric is homogenous when $d(\alpha x, \alpha y)=|\alpha |d(x,y)$ for all vectors $x,y$ and scalars $\alpha$. Now it is trivial to verify that the metric induced by a normed space is translation invariant and homogenous. The interesting fact is that any translation invariant and homogenous metric defines a norm on the vector space. There is thus a bijective correspondence between norms on a vector space $V$ and metrics on $V$ which are translation invariant and homogenous. 
A: Consider the $l_{\infty}$ metric on $\mathbb{R}^2$ which arises from a norm:
$$d:\mathbb{R}^2 \times \mathbb{R}^2 \rightarrow \mathbb{R}$$
$$((a_1,a_2),(b_1,b_2))\mapsto max\{|a_1-b_1|,|a_2-b_2|\}$$
This metric might be applied to a lathe/CNC machine for instance. 
Suppose you have an automated manufacturing machine which would move a drill around over the top of a flat sheet of metal. The machine can move the drill in two dimensions over the sheet metal before dipping and drilling into the metal. Suppose also  that this movement is controlled by two motors in the $x$ and $y$ direction which each have the same maximum speed.
If you have a particular design you want this machine to drill and you're trying to minimize the time taken to manufacture it, then you want to know how long the drill will take to get from hole to hole. In this case, if both of the movement motors can move at the same time, then the limiting factor that slows down the automated drill is not the euclidean distance from one hole to another, but the greater of the $x$ and $y$ distances-- whichever motor takes longer to move the drill in its direction. 
So in this case, to weigh up in which order you should have the machine drill the holes, you would not worry about the euclidean distance but instead measure the distance from one hole to the next using the $l_\infty$ metric.
This is one application of a metric. 

A metric which arises from a norm also has the two properties:
$$d(x+c,y+c)=d(x,y)$$
$$d(\lambda x, \lambda y)=|\lambda|\cdot d(x,y)$$
The first property, translational invariance, means that no matter where the drill currently is, its distance to a hole which is, say, $2$cm forward and $1$cm right will always be the same. It wouldn't make sense in this case if the distance to a hole which is $2$cm forward and $1$cm right depended on where the drill head currently was!
The second property means that a hole which is $4$cm forward and $2$cm right would be exactly twice as far away.

I hope this gives you some intuition about the properties and applications of metrics and norms.
