Arbitrage-Free Cash Flows (Part of Proof, Geometric Problem) (I've seen following notation, used in this question, on YouTube videos and in several undergradute text books, so I will not explain it. It is pretty standard!)
Fix a set of times $0=t_0<\cdots<t_n$ and consider a market consisting of $m$ cash flows: $$\{(c_{k,0},t_0),(c_{k,1},t_1),\ldots,(c_{k,n},t_n)\}$$ for $k=1,\ldots,m$. Since times are fixed represent the cash flows more compactly as $m$ elements $\mathbf c_1,\ldots,\mathbf c_m\in\mathbb R^{n+1}$. Let $\mathbb C$ be a linear subspace of $\mathbb R^{n+1}$ and let $\mathbf c$ be any non-zero linear combination of $\mathbf c_1,\ldots,\mathbf c_m$ such that $\mathbf c\in\mathbb C$.
(i) $\nexists\ \mathbf c\in\mathbb C$ such that $\mathbf c\geq 0$.
(ii) $\exists\mathbf d\in\mathbb R^{n+1}$ with $\mathbf d>0$ satisying $\mathbf c^T\mathbf d=0$ for all $\mathbf c\in\mathbb C$.
(i) implies (ii).
To show (ii) implies (i) is easy. If $\mathbf d>0$ and $\mathbf c^T\mathbf d=0$ then some $\mathbf c\in\mathbb C$ has negative elements.
Problem: How to show (i) implies (ii)? I interpret (i) as there exists no linear combination $\mathbf c$ such that there is arbitrage. But why does $\mathbf c$ need to be restricted to $\mathbb C$? And what is the meaning of (ii)?
 A: Let's first start by answering a few of the interpretation questions. $\mathbf c_1,\ldots,\mathbf c_m\in\mathbb R^{n+1}$ are cash flows, thus any linear combination of those cash flows represents a certain portfolio. Hence, you are "restricted" to $\mathbb{C}$ because that's the only cash flows we seem to know about, and thus linear combinations of those are the only portfolios we can make.
Next, a claim is made about the nature of $\mathbb{C}$: if $\mathbb{C}$ satisfies condition (i), then it should also satisfy condition (ii) and visa versa. 
This theorem is a variant on Farkas' lemma. The usual approach to prove this lemma is to rely on a hyperplane separation theorem. I will not prove the theorem, you can look up a proof for it, but I think it is intuitive enough. It essentially states that two disjoint convex sets can be separated by a hyperplane.
A more precise formulation we can use in the present context:
Let $\mathbb{V}$ and $\mathbb{W}$ be two disjoint non-empty convex subsets of $\mathbb{R}^{n+1}$, then there exists an $\mathbf a \in \mathbb{R}^{n+1}$ with $\mathbf a \neq \mathbf 0$ and a $b \in \mathbb{R}$ such that $\mathbf a^T\mathbf x \leq b , \; \forall \mathbf x \in \mathbb{V}$ and $\mathbf a^T\mathbf x \geq b , \; \forall \mathbf x \in \mathbb{W}$. 
To apply this theorem to prove that (i) implies (ii), we'll set $\mathbb{V}=\mathbb{C}$ and $\mathbb{W}=\{\mathbf x \in \mathbb{R}^{n+1}\setminus\{\mathbf 0\}| \mathbf x \geq 0\}$, the cone of all vectors with positive components, null vector excluded. It is clear that those sets satisfy the required condition since by (i) there is no $\mathbf c$ in $\mathbb{C}$ that can also be an element of $\mathbb{W}$. Also a linear subspace is convex and non-empty and it is not hard to show that $\mathbb{W}$ is convex and non-empty as well. 
Therefore $\exists \mathbf a \in \mathbb{R}^{n+1}$ with $\mathbf a \neq \mathbf 0$ and a $b \in \mathbb{R}$ such that $\mathbf a^T\mathbf x \leq b , \; \forall \mathbf x \in \mathbb{C}$ and $\mathbf a^T\mathbf x \geq b , \; \forall \mathbf x \in \mathbb{W}$. 
Observe that the null vector $\mathbf 0$ is in $\mathbb{C}$ and therefore $\mathbf a^T\mathbf 0 = 0 \leq b$. Thus, we also have for the elements $\mathbf x \in \mathbb{W}$ that $ \mathbf a^T\mathbf x \geq b \geq 0$. In particular, $\mathbb{W}$ contains the standardbasis $\{\mathbb{e}_k\}_{k \in \{0,\ldots,n\}}$. Plugging those in the last inequality shows that all the components of $\mathbf a$ are positive or $\mathbf a \geq 0$.
But this now implies that $b=0$. Indeed, if $b>0$ then the equation $\mathbf a^T\mathbf x = b$ represents a hyperplane cutting off the coordinate axes in positive values, but this would contradict the theorem that the hyperplane separates $\mathbb{W}$ from $\mathbb{C}$. 
Therefore we boiled it down to $\exists \mathbf a \in \mathbb{R}^{n+1}$ with $\mathbf a \neq \mathbf 0$ and $\mathbf a \geq 0$ such that $\mathbf a^T\mathbf x \leq 0 , \; \forall \mathbf x \in \mathbb{C}$ and $\mathbf a^T\mathbf x \geq 0 , \; \forall \mathbf x \in \mathbb{W}$. 
Finally, suppose $\exists \mathbf c \in \mathbb{C}$ such that $\mathbf a^T\mathbf c < 0$. But since $\mathbb{C}$ is a linear subspace, $-\mathbf c$ is also an element of $\mathbb{C}$ and $\mathbf a^T(-\mathbf c) \leq 0$. But simply multiplying the equation for $\mathbf c$ by $(-1)$ leads to a contradiction. Therefore $\forall \mathbf x \in \mathbb{C}: \mathbf a^T\mathbf x = 0$. 
So, we are done with the proof, except for one detail: we need $\mathbf a >0$. But we have $\mathbf a \geq 0$. I'll leave that for you to handle.
So, we see that (i) implies that the linear subspace $\mathbb{C}$ is a subspace of a hyperplane that does not intersect with the cone $\mathbb{W}$. That's the geometrical interpretation for (ii).
