# Showing a basic market admits no arbitrage

I'm learning the fundamentals of financial mathematics and came across the following problem I cannot solve

### Setting

We work in $$\left(\Omega, \mathcal{F},\left(\mathcal{F}_t\right)_{t=0}^1, \mathbb{P}\right)$$. Let $$d=1, T=1$$ and assume the discounted price equals the non-discounted price.

Take $$S_0^1 \in \mathbb{R}_{+}$$, and $$S_1^1 \in\left\{\alpha S_0^1, \beta S_0^1\right\}$$ each with positive probability s.t. $$0<\alpha<\beta$$.

I want to show that $$\alpha<1<\beta$$ iff there is no arbitrage. Additionally I'd like to find an example which shows that if $$\mathcal{F}_0$$ is not the trivial $$\sigma$$-algebra, then there exists an arbitrage.

### Attempt

I know that for there to be an arbitrage I'd need to find $$H_1$$ s.t. $$\mathbb{P}\left(H_1 \cdot \left(S_1^1-S_0^1\right) \geq 0\right)=1 \text { and } \mathbb{P}\left(H_1 \cdot \left(S_1^1-S_0^1\right)>0\right)>0 .$$ but I don't know what to base the proof on besides that. I would be grateful for any help!

Suppose $$0 < \alpha < \beta \leq 1$$ and consider the strategy of shorting one share for $$S_0^1$$ at $$t = 0$$. At $$t = 1$$, we can buy back the share for a price $$S_1^1$$. Observe our profit is non-negative with probability one and strictly positive with non-negative probability. A similar argument can be used to construct an arbitrage if $$1 \leq \alpha < \beta$$.
This proves if there is no arbitrage, then $$\alpha < 1 < \beta$$. It technically proves the converse.