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For holdings $h\in {\mathbb{R}}^n$, let $V_0=h\cdot P_0$ and $V_1=h\cdot P_1$. Under the investment-consumption arbitrage convention used here, an arbitrage satisfiesThe initial consumption is $-V_0$, and the terminal consumption is $V_1$: there is a possible gain with no external funding and no negative consumption. A pure-investment arbitrage additionally requires $V_0=0$. This convention is important because the last part distinguishes initial consumption from pure investment.

Write $(a,b,c)$ for holdings of cash, the stock and the European put option, and let $\gamma =-V_0$ be initial consumption. At the given price, $a=-10b-c-\gamma$, and the three terminal payoffs, in descending order of the stock price, areNonnegative terminal payoffs require $c+\gamma \le 2b$ and $c-\gamma \ge b$. Together with $\gamma \ge 0$, these imply $b\ge 2\gamma \ge 0$. If $b=0$ they force $\gamma =c=a=0$, which is not an arbitrage. If $b>0$, the highest-state payoff is at least $3b>0$, so every such choice is an arbitrage. Consequently the complete set isThe pure-investment arbitrages are exactly the choices $\gamma =0$:For example, $(a,b,c)=(-11,1,1)$ costs zero and pays $(4,1,0)$.