A lattice $N$ is a free $\mathbb{Z}$-module of finite rank. Let $V$ be the real vector space $N\otimes_\mathbb{Z} \mathbb{R}.$ A cone is a set $\sigma = \{ r_1 v_1 + \ldots + r_k v_k \in V : r_i\geq 0 \}$ where the $v_i\in V.$ The dual space of $V$ is $V'$ and the dual cone of $\sigma$ is $$\sigma^{\vee}:=\{ f\in V' : f(x)\geq 0 \ \text{for all} \ x\in \sigma \}.$$
I want to prove that if $x_o\notin \sigma$ then there is some $f\in \sigma^{\vee}$ such that $f(x_0) < 0.$ I think we can prove this by first noticing that $V$ is a finite dimensional real vector space. As such, we can endow some locally convex Hausdorff topology on it (for example the one induced by an inner product defined by $ \langle e_i, e_j \rangle = \delta_{ij}$ for some basis $\{ e_i \})$. Then one can construct the desired $f$ as t.b. has done at the end of his answer here.
The result I want to prove is purely algebraic, there are no topologies a priori on any of the objects. As such the above proof feels unsatisfactory. Can someone find a proof that does not put a topology on $V$ first?