Base of a cone in a vector space: can one always choose a convex base? Let $C$ be a pointed convex cone in a vector space $V$. This means that $C$ satisfies the three following axioms: 


*

*$C + C \subset C$, 

*$\mathbb{R}_+ \cdot C \subset C$, and 

*$C \cap (-C) = \{ 0 \}$. 


Say that $K \subset C$ is a base of $C$ if, for every $x \in C \setminus \{ 0 \}$, there is a unique $\lambda > 0$ such that $\lambda \cdot x \in K$. Usually one demands also that $K$ be convex, a hypothesis that I do not make here. Now if $C$ has at least one base, does a convex base always exist (possibly under additional topological assumptions)? 
One approach I have tried is the following: if $K$ and $K'$ are two bases of $C$, say that $K \leqslant K'$ if 


*

*$\operatorname{co}(K) \supset \operatorname{co}(K')$ and 

*for every $x \in K$ there exists some $0 < \lambda \leqslant 1$ such that $\lambda \cdot x \in K'$. 


where $\operatorname{co}$ denotes the convex hull operator. Then $\leqslant$ is a partial order, and one may wish to apply Zorn's lemma and get a maximal element to find the desired convex base. However I cannot manage to show that the partial order is inductive. 
Would you have any suggestion or any other idea? 
 A: This is an extended version of my observations in the comments. The upshot is that there exist pointed convex cones without a convex base, but every cone has a base. Hence what the OP is trying to do is bound not to work.
(1) There are pointed convex cones that do not have a convex base. To see this, 
take $V=\mathbb{R}^2$ as a simple example, with $C$ given by all those $(x,y)\in\mathbb{R}^2$ for which $x>0$, or $x=0$ and $y>0$. One can visualize the non-existence of a convex base by intersecting $C$ with a line not passing through the origin, and then rotating this line all the way around, noticing that it never hits all the rays of $C$.
More formally, suppose that $K$ is a convex base. This means that there is exactly one point of the form $(0,y_0)\in K$ with $y_0\in K$; up to rescaling, we may assume $y_0=1$. Similarly, we can assume that $(1,0)\in K$ is the only element in $K$ on the $x$-axis.
But then also the ray generated by $(1,-1)$ must intersect $K$ somewhere, say at $(t,-t)$ for $t>0$. By convexity and $(0,1)\in K$, this would imply that the point
$$
\frac{t}{t+1}(0,1) + \frac{1}{t+1}(t,-t) = \left(\frac{t}{t+1},0\right)
$$
is in $K$ as well, in contradiction to the assumption that $(1,0)$ is the only point in $K$ on the $x$-axis.
(2) Every convex cone has a base (not necessarily convex). This is a simple consequence of the axiom of choice: we choose one representative of each ray $(C\setminus\{0\})/\mathbb{R}_{>0}$.
Arguably, the non-existence of a convex base in (1) is due to the cone not being closed. In fact, in finite dimensions, one can show that every pointed closed convex cone has a base, using the Hahn-Banach theorem as suggested by Willie Wong in the comments. However, constructing a convex base like this requires a functional which is strictly positive on the whole cone, which one can obtain from the Hahn-Banach theorem only in certain situations, such as in finite dimensions. In fact, even a pointed closed cone in an infinite-dimensional locally convex space does not necessarily have a convex base: the following example is my own rephrasing of Exercise 1.7.6 in Aliprantis/Tourky, "Cones and Duality", following Willie Wong's request in the comments.
Let $A$ a set and $\mathcal{B}(A,\mathbb{R})$ the vector space of bounded real-valued functions on $A$. This has an obvious pointed convex cone given by the set of all functions $f$ that are pointwise nonnegative, i.e. for which $f(x)\geq 0$ for all $x\in A$. It is also a normed space via the supremum norm, and the cone is closed.
Now suppose that the cone has a convex base $K$. For nonzero $x\geq 0$, I will say that $x$ lies "above" $K$ if $x$ needs to be scaled down in order to hit $K$. Then for every $x\in A$, consider the indicator function $\chi_{\{x\}}$, and take $A'$ to be the set of all $x$ for which $\chi_{\{x\}}$ is above $K$. I claim that $A'$ is finite: if $n$ of those indicator functions are above $K$, then so is their average, and hence their sum is above $nK$; but this would mean that $\chi_{A'}$, which is greater than all those finite sums in the ordering, is above all $nK$, which is absurd. Hence $\chi_{\{x\}}$ is above $K$ for only finitely many $x$. Similarly, for every $m\in\mathbb{N}$, the function $m\chi_{\{x\}}$ can be above $K$ for only finitely many $x$. Hence the total number of $x$'s, i.e. the cardinality of $A$, is at most countable.
In other words: if $A$ is uncountable, then no convex base exists, although the pointed convex cone is about as nice as it gets!
A: @TobiasFritz's answer is very excellent. Here is another nice example of a Banach lattice with order unit. Consider a measurable set $(S,\Sigma)$ and a $\sigma$-ring $N\subseteq \Sigma$, elements of which we call null. Now consider $L_\infty(\Sigma|N)$ of all equivalence classes of bounded measurable functionals such that $f\sim g$ if they differ on a null set in $N$.   The space $L_\infty(\Sigma|N)$ is an ordered Banach lattice (with usual norm), the positive cone has non-empty interior. the positive cone has a convex base if and only if there is a (finitely additive) probability measure on $\Sigma$ that gives positive probability to every non-null set. The characterisation of the existence of such a probability is given by Kelly (1956). 
