TITLE: What is the difference between metric spaces and vector spaces?
QUESTION [32 upvotes]: Does a metric space have an origin? That is, does it have $(0,0)$.
Does a vector space have an origin?
It seems whatever you can do in a metric space can also be done in a vector space. Is this true?

REPLY [3 votes]: This is an ancient question; it also has been correctly answered. But maybe I can provide a perspective from Applied Mathematics.
A vector space with a finite number of dimensions $n$ must have a global coordinate structure (actually, many such structures) specified by $n$ vectors. For example, $\mathbb R^3$ has a global coordinate structure whose basis vectors are
$$ X = (1, 0, 0) $$
$$ Y = (0, 1, 0) $$
$$ Z = (0, 0, 1) $$
What I mean by "coordinate structure" is that points of $\mathbb R^3$ can be specified as a linear combination of these three vectors:
$$ v = (a X + b Y + c Z, d X + e Y + f Z, g X + h Y + i Z) $$
For the canonical basis above this is easy to see; there are more complicated choices of basis, but at any rate, this works for every element of $\mathbb R^3$.
Now take something as planet Earth. Obviously Earth isn't a mathematical object, and admits multiple mathematical models -- cosmologists see it as three-dimensional and who knows what contemporary physicists are up to. But we don't see Earth in three dimensions -- it's the literal 2D ground on which we walk. Still, it's notorious that a proper two-dimensional coordinate system for the Earth cannot be found.
One way to put this is that this model of the Earth is a two-dimensional manifold embedded in $\mathbb R^3$. But technically manifolds don't need to be embedded in anything; they just need to have some local structure that's out of scope for us. But the point is that we can easily define distances on Earth: two possible choices are geodesic (shortest flyover) and shortest-possible walking distance. We don't care how high the Everest is in a putative third dimension, we care how far it is on foot.
So that's a metric space (please check the axioms to convince yourself this is true); but not a vector space.
My current research has to do with Nearest Neighbors methods for data science. A conceptual point that I stress is that NN methods are useful when the data can be given a distance structure, but not a (credible) vector space structure. Take something like underwear: we can clearly say that U1 is closer to U2 than to U3, but it's much harder to imagine that there are basis vectors such that all underwear is just combinations of it. I.e. often the domain problems that arrive at the data scientist's desk have definite similarity structure but not a clear global coordinate structure. 
(Of course, the Nearest Neighbor algorithm ultimately does calculations with computer arrays representing vectors; the entire challenge is to find "metric learning" algorithms that figure out the appropriate distance function near each point and map the data onto something on which the standard vector distances represent the abstract, ground-truth distances between objects. But this too is besides the point.)