TITLE: Is there such thing as an unnormed vector space?
QUESTION [18 upvotes]: I learned about Banach spaces a few weeks ago. A Banach space is a complete normed vector space. This of course made me wonder: are there unnormed vector spaces? If there are, can anyone please provide any examples? 
Some thoughts:
A complete space is where all Cauchy sequences converge.
A normed vector space is a vector space (say, over $\mathbb{R})$ on some norm $N$ (which is a function that maps $N\to\mathbb{R}$), where the norm obeys the triangle inequality, the norm of a vector is non-negative, and if you have a scalar being multiplied by a vector, you can factor the scalar out, but it'll have absolute value braces. 
I'm not really sure what is needed in order to have an unnormed vector space (perhaps the vector space necessarily needs to be infinite dimensional?). Perhaps something really weird like the zero space?
Thanks for any insight.

REPLY [3 votes]: While you might be able to define a norm, there are real world circumstances where you might not want to. Let's take an example motivated more from science/engineering than mathematics. Consider the vector space defined by the tuple x = (a mm, b kg) where a and b are real numbers and mm and kg mean "millimeter" and "kilogram." To a scientist/engineer, the notion of a naive L1 metric might seem ludicrous, because you can't add 1mm + 1kg. Yes, the metric $\lVert x \rVert = |a| + |b|$, which scales the first element by 1/mm and the second by 1/kg, is a metric, but it's a completely arbitrary one: why is 1mm in some sense "equivalent" to 1kg? You might have a hard time arguing with me if I instead declare that you should scale the second component by 1/g rather than 1/kg, but of course that will give us very different notions of distance. (There are exceptions: e.g., in relativity theory, the speed of light makes 1s equivalent to 3e8m, but in other contexts one might not agree with that equivalence.)
From the standpoint of a mathematician, I suppose the question is what is the right topology for such a vector space. If you decide to allow open sets of the form $I_a \otimes I_b$, where $I_a$ and $I_b$ are open intervals with respect to the first and second tuple element and $\otimes$ indicates the Cartesian product, then the entire space is second-countable and satisfies the Urysohn metrization theorem. However, if instead you say that the topology is defined through open sets only of the form $I_a \otimes b$ and $a \otimes I_b$ (i.e., have zero "width" along one of the two coordinates), and if the numerical coefficients can be any real number, then I think the space is rigorously non-metrizable.