Vector space, norm and metric This is a follow up question to my previous question.
It is perfectly clear to me what a metric space is. Basically, one can see that notions like convergence and continuity only depend on some sort of closeness or distance, so we can define a distance and then talk about these things.
Now it might happen that we have a vector space and we would like to talk about these concepts. Now in some cases we want the metric to respect the vector space operations because otherwise we can't really make use of them.
This means that the distance should have some additional properties:
$$(i) d(w,v)=d(w+u,v+u)$$
$$(ii) d(\alpha u,\alpha v)=\lvert\alpha\rvert d(u,v)$$
In most cases we introduce a norm which in turn induces a metric with these properties. The norm itself can be seen as a generalization of length, size or magnitude, so it is also an inuitive concept.
Now I've got 2 questions:
1)I am wondering if the only reason to introduce a norm is to be able to talk about convergence and continuity or are there other reasons why one would want a generalization of length for arbitrary vector spaces. Are there other things we can do once we have a norm? If yes, please provide an example. If not why don't we just define a norm as a metric with the additional properties (i) and (ii). I am guessing it´s because the properties of the norm are sufficient and we want our definitions to be short.
2) What is the intuition for the triangle inequality for an arbitrary norm in any vector space? In $\mathbb{R^{n}}$ it makes sense to require that the length of a sum of two vectors is smaller than the sum of the length since the directions might be different. For example, why should this be true if I have some norm for a function space? For a distance this is intuitive since it basically says that the shortest path between 2 points is a line between the points.
Thanks very much!
Edit:
Regarding 2) it might make sense to view the norm as a generalization of the absolute value rather than the length in $\mathbb{R^{n}}$. Intuitively the absolute value gives the magnitude of a number. The triangle inequality can be seen as the property of subadditivity. In higher dimensions it then makes sense to say the magnitude of a vector is its length. Of course it satisfies the properties of a norm.
 A: (1) Well, apart from convergence and continuity, it allows us to talk about completeness. Complete normed spaces are called Banach spaces and are vastly important. But we can talk about completeness in the general setting of a metric space, so this doesn't answer your question. My guess for defining norms as we do, instead of requiring structure-preserving properties out of a metric, is that norms themselves can be induced out of inner products in a natural way: in some way, norms are at the "structural midpoint" between a purely topological notion (a metric) and a purely linear one (an inner product).
(2) One reason for requiring that metrics (and thus norms) satisfy the triangle inequality is that, otherwise, you can't really deal with convergence in the way we are used to, and this would make the whole edifice of analysis come crumbling down. Convergent sequences wouldn't be Cauchy, so completeness would be harder to define. Convergent sequences wouldn't be bounded. Differentiable functions wouldn't be continuous. The Banach fixed point theorem would not hold, so you wouldn't be able to prove ODE solution uniqueness. From a theoretical point of view, it would be a mess.
This is ultimately because, out of all the axioms in the definition of a metric, the triangle inequality is the one encoding the notion of "nearness" required for the basic idea of convergence.
