Why are convex sets assumed / defined to be subsets of vector spaces (and not of more general spaces)? Roughly speaking, my question is: Why are convex sets assumed to be subsets of vector spaces?
Well, one might answer that convex sets are defined to be subsets of vector spaces, but this is not the concern of my question.
Are there spaces with a "weaker" structure (in the sense that any vector space has or induces this structure) such that one can give a meaningful definition of the property of a subset that any convex combination of elements of that subset is contained in the subset?
I wonder for the following reason (unimportant for the answer of the above question; omit if uninterested): 
There are attempts in theoretical physics to construct a mathematical framework, called "generalized probabilistic theories", in which (essential parts of) quantum theory can be formulated as a "special theory", i.e. as a special case of the framework. One starts by defining a set of states (call it $S$). This set is assumed to be convex (in the sense that combinations of the form $\sum\limits_i \alpha_i \omega_i$ with $\sum\limits_i \alpha_i = 1$ and $\omega_i \in S \ \forall i$ lie in $S$). Then, without any further argument, one deals with vector spaces for which $S$ is a subset. I wonder if it is necessary to deal with vector spaces or if it is possible to deal with weaker structures.
 A: If you have a meaningful way to interpret $\sum_i\alpha_i\omega_i$ for $\alpha$s that sum to $1$, then you almost have a vector space already. More precisely, the set of all formal linear combinations of elements of $S$, modulo multiples of $\alpha_1\omega_1+\cdots+\alpha_n\omega_n-\sum_i\alpha_i\omega_i$ will constitute a vector space. Thus, assuming that $S$ is a subset of a vector space does not lose generality.
Because vector spaces are more well-behaved and well-studied than other weaker structures you could work with, it is convenient to make this assumption.
A: http://en.wikipedia.org/wiki/Convex_metric_space
However, this sounds like it is too weak for your purposes, as the notion of "linear combination" is lost altogether.
A: Editing Note: cleaned up link rot
If you are wedded to the idea of using real numbers with convexity you might be interested in [CONVEX SPACES 1: DEFINITIONS AND EXAMPLES by Tobia Fritz.https://arxiv.org/pdf/0903.5522.pdf] If you want to see if real numbers are really necessary you might wish to read Theory of Convex Structures by Marcel van de Vel. 
Let's define two different structures. A closure system is a set $X$ together with a collection $\mathcal{C}$ of subsets of $X$ that satisfies:


*

*We have $X \in \mathcal{C}$;

*For all $\mathcal{A} \subseteq \mathcal{C}$ we have $\cap \mathcal{A} \in \mathcal{C}$.


For all $A \subseteq X$ define $\mathsf{cl}\,A = \cap \{ C \in \mathcal{C} \colon A \subseteq C \} $.
A convex structure is a closure system that satisfies for all $A \subseteq X$ we have $\mathsf{cl}\, A = \cup \{ \mathsf{cl} \, F \colon F \text{ is a finite subset of } A \} $.
The trick to using these systems is to figure out how to describe items that are useful in real vector 'spaces without using any numbers. Lets give an example. Suppose that $V$ is a real vector space. For all $E \subseteq A \subseteq V$ we will say that $E$ is an extreme subset of $A$ if and only if for all $x_{0}, x_{1} \in A$ if $\frac{1}{2}x_{0} + \frac{1}{2}x_{1} \in E$ then $x_{0}, x_{1} \in E$. Most references also require that $E$ is not empty. We can eliminate numbers and algebraic operations as follows: Suppose that $\langle X, \mathcal{C} \rangle$ is a closure system. For all $E \subseteq A \subseteq X$ we will say that $E$ is an extreme subset of $A$ if and only if for all $D \subseteq A$ we have $E \cap \mathsf{cl}\, D = E \cap \mathsf{cl}( E \cap D)$. In a real vector space this is equivalent to the usual definition. We will say that $E$ is a minimal extreme subset of $A$ if and only if for all extreme subsets $E^*$ of $A$ if $E^* \subseteq E$ then $E^* = \varnothing$ or $E^* = E$.
Here is an example. Let $X = [0, 1] \times S^{1}$. A set $C \subset X$ will be said to be convex if and only if for all $c_{0}, c_{1} \in C$ every geodesic segment from one of the points to the other is contained in $C$. In this situation $X$ has two minimal extreme subsets, namely $\{ 0 \} \times S^{1}$ and $\{ 1 \} \times S^{1}$. Even in an example this nice we have lost the notion that minimal extreme subsets are singletons.
Here is another example. Suppose that $V$ is a real vector space. Often we use continuous linear functionals to separate points. Suppose that $f \colon V \rightarrow \mathbb{R}$ is such a function. If $r \in \mathbb{R}$ then $f^{-1}(r)$ is an extreme subset of both $f^{-1}(-\infty , r]$ and $f^{-1}[r, \infty)$.
We can partition $V$ into three pairwise disjoint pieces: $O_{0} = f^{-1}(-\infty , r)$, $E = f^{-1}(r)$, and $O_{1} = f^{-1}[r, \infty)$. Then $E$ is an extreme subset of both $O_{0} \cup E$ and $O_{1} \cup E$. By breaking up a closure system in this way we can use the $O_{i}$ to form a subbase for a topology for a closure system.
A: As convexity by its very definition implies linear combination of elements of the space, and given the fact that vector space seems to be the minimal structure supporting this algebraic property between element of space, I would be surprised if such generalization could exist. 
Best regards
A: If you have a Module $M$ over a ordered ring (such as $\mathbb{Z}$ or $\mathbb{Q}$) $R$, you can use the exact same definition, without changing absolutely anything of "convex set in vector spaces" and it will of course make sense, since modules are exactly vector spaces, only over a ring instead of a field, but that is not used at all in the definition of convex! Just for completion, this definition is given of "convex subset" in the wiki article linked:

A set $C$ in $S$ is said to be convex if, for all $x$ and $y$ in $C$
  and all $t$ in the interval $[0, 1]$, the point $(1 − t)x + ty$ also
  belongs to $C$.

Note that we NEED the ring to be ordered, because if not, wich ring elements are in $[0, 1]$ is impossible to say.
