Proof of the No Arbitrage Theorem

The no-arbitrage theorem is a theorem of the alternative. Let $\Delta$ denote the set of $y\in R^{S+1}_+$ such that $\sum_ny_n =1$ . If $Mz>0$ has a non-negative solution, then it has one in $\Delta$.

Lemma. If $A,B\subset\R^m$ are convex, $A$ closed and $B$ compact, then there is a $\pi$ that separates them; that is, $\sup_{a\in A}\pi a<\inf_{b\in B}\pi b$.

(Sufficiency) If $\tilde\pi M=0$, then $\pi Mz=0$ for all $z$. Since $\pi\gg0$, $Mz\not>0$.

(Necessity) Suppose that $\langle M\rangle\cap\R^{S+1}_+=\{0\}$. Let $D=\{y:y=Mz, z\in\R^J\}$. The no-arbitrage condition implies that $\Delta\cap D=\emptyset$. The lemma will say there is a $\pi$ that separates them.

Suppose $\pi_s\leq 0$. Then the infimum over $\delta\leq 0$. But $0\in D$ contradicting the conclusion of the lemma. Thus $\pi\gg0$.

If $\pi d\neq 0$ for some $d\in D$ then there is a $z\in\R^J$ such that $\pi Mz\neq 0$. Expanding positively or negatively in the $z$ direction as needed, we see that $\sup_{d\in D}\pi d=+\infty$, contradicting the conclusion of the lemma. ∎