Why don't we study 'metric vector spaces' on their own? I recently took a mastercourse on functional analysis and I was wondering why we 'skip' the metric structure on vector spaces. I have seen that
$$\{\text{vector spaces}\}\supsetneq\{\text{topological vector spaces}\}\supsetneq\{\text{locally convex vector spaces}\}\supsetneq\{\text{normed vector spaces}\}\supsetneq\{\text{innerproduct spaces}\}.$$
Isn't it natural to include vector spaces endowed with a metric (that may not be induced by a norm) in this sequence? I can't seem to find any literature about it.
 A: In general topology, a metric space has much more structure than a topological space, and metrization theorems form a rich area of research. In topological vector spaces, things are rather different: a tvs is metrizable if and only if it is (Hausdorff and) first countable. If this is the case, then there also is a translation-invariant metric. (Both of these results can be found, for instance, in [Köt83, §15.11.(1)]. Further properties of metrizable tvs can be found in some of the more advanced functional analysis textbooks; for instance, see [Köt83, §15.11 and §18.2] or [Jar81, §2.8].) Topological vector spaces are already uniform spaces, which perhaps explains why being metrizable is not as important as it is for general topological spaces.
While it is true that the metrizable theory is not nearly as rich as the locally convex theory, metrization properties do occasionally come up in functional analysis. A few examples:

*

*As pointed out by others, the open mapping theorem holds for arbitrary (not necessarily locally convex) F-spaces. More generally, the Baire category theorem is a very powerful topological tool. It requires a complete metric (a complete uniform structure is not enough), but local convexity is not needed.


*If an ordered tvs $E$ is completely metrizable and has a closed and generating positive cone, then every positive linear functional on $E$ is automatically continuous. (See [AT07, Corollary 2.34].)


*If $E$ is a separable metrizable lcs (in particular, if $E$ is a separable normed space), then $E'$ is weak-$*$ separable (cf. [Köt83, §21.3.(5)]). In the proof, it is crucial that equicontinuous subsets of $E'$ are weak-$*$ metrizable (under the present assumptions), so here the metrizability of a subset of a topological vector space matters too.


*The spaces $L^p[0,1]$ for $0 < p < 1$ form an important example of a topological vector space $E$ whose dual $E'$ does not separate points on $E$. (In fact, these spaces have no continuous linear functionals, apart from $0$.) Nevertheless, these spaces are completely metrizable.
In a master's course, you can only do so much, so it is reasonable to focus on the most important concepts and theorems. Metrizable spaces do not play a crucial role in the theory — when metrizable tvs have special properties, it is usually because they belong to a larger class (every metrizable lcs is bornological, every F-space is a Baire space, etc.).
On the other hand, locally convex spaces are central to the theory, and this is why your masters course or textbooks treat locally convex spaces in great detail but do not say much about metric vector spaces.
References.
[AT07]: Charalambos D. Aliprantis, Rabee Tourky, Cones and Duality (2007), Graduate Studies in Mathematics 84, American Mathematical Society.
[Jar81]: Hans Jarchow, Locally Convex Spaces (1981), Mathematische Leitfäden, Teubner.
[Köt83]: Gottfried Köthe, Topological Vector Spaces I, Second revised printing (1983), Grundlehren der mathematischen Wissenschaften 159, Springer.
A: If the metric is not related with the vector space structure, there is not much to talk about. 
As you say, we could require that the metric is translation invariant. And there is another operation on a vector space, which is multiplication by scalars: does it make any sense to say that $2x$ is not at twice the distance from the origin than $x$ is? So you want to assume that the metric scales with scalar multiplication. That is, the metric satisfies 


*

*$d(x,y)=d(x+z,y+z)$

*$d(\lambda x,\lambda y)=|\lambda|\,d(x,y)$
With those two assumptions, $\|x\|=d(x,0)$ is a norm that induces the metric $d$. 
So, there is very little room for endowing a vector space with a meaningful metric that does not come from a norm but still somehow interacts naturally with the vector space structure. And if the metric does not match with the vector space structure, then you have no reason to pay attention to the topology and the vector space structure at the same time. 
