I recently tried to see in how far polyhedral geometry can be reduced to the study of convex sets with finitely many faces. In other words, I tried to replace "finitely generated" by "finite face lattice". One advantage is that this includes more objects, such as subspaces or halfspaces of an infinite dimensional ambient space.
I quickly noticed that this was too naive and too many unwanted objects fell under this new definition. For example, open cones have only the trivial faces, hence finitely many. I do not want to assume a topology, so I have limited tools to exclude open cones. My idea was to use algebraic duality as a closure operator (see below), and I believe this should suffice to capture what I have in mind. I did however not succeed in proving this ... which brings me to my question below. For simplicity, I ask only about cones.
Let $V$ be a vector space over some complete ordered field, so essentially over $\Bbb R$. I do not assume a topology on $V$, nor that $V$ is of finite dimension. Let $C\subseteq V$ be a convex cone (i.e. closed under addition and scalar multiplication with non-negative scalars) with three properties:
$C$ has finitely many faces. Here a face is any convex sub-cone $F\subseteq C$ so that for all $x,y\in C$ holds $x + y \in F\Rightarrow x,y\in F$.
$C$ is dual closed i.e. identical to its algebraic double dual. Equivalently, there are linear forms $\phi_i\in V^*,i\in I$ over some potentially infinite index set $I$ so that $C=\{x\in V \mid \phi_i(x)\ge 0 \text{ for all } i\in I\}$.
$C$ is pointed, i.e. $x\in C\Rightarrow -x\notin C$ for all non-zero $x$.
Question: Is $C$ finitely generated?
Let me briefly explain why the assumptions are necessary. We need the field to be complete because $\langle (0,1),(\sqrt 2,1)\rangle\cap \Bbb Q^2$ is a cone over $\Bbb Q$ with all the above properties, but is not finitely generated. Dual closed takes care of the previously problematic open cones and is, I believe, the best we can do in absence of a topology. Finally, we need pointed: if $C$ is finitely generated and $S$ is an infinite dimensional subspace, then $C+S$ is dual closed with finitely many faces, but not finitely generated.
