Consider a financial market with $d+1$ assets in a one-period model. The $0-$th asset is considered the risk-free asset, the others are risky. The vector $\overline{\pi} \in \mathbb R^{d+1}$ denotes the initial price vector at $t=0$, and the random vector $\overline{S} \in \mathbb R^{d+1}$ denotes prices at $t=1$. Further we have $\pi^{0}=1,\; S^{0}=1+r$, as the $0-$th asset is risk-free.

Here a strategy $\overline{\xi}\in \mathbb R^{d+1}$ is called an arbitrage opportunity if $\overline{\xi}⋅\overline{\pi}≤0$ but $\overline{\xi}⋅\overline{S}\geq 0$ and $\mathbb P(\overline{\xi}⋅\overline{S}> 0)>0$. Further the market model is called non-redundant if
$\overline{\xi}\cdot \overline{S}=0 \; \;\mathbb P\text{-a.s.}\implies \overline{\xi}=0$.

By the Law of One price we know that in an arbitrage-free model, for any $\overline{\rho},\; \overline{\xi}$ such that $\overline{\xi}\cdot\overline{S}=\overline{\rho}\cdot\overline{S}$ a.s. it must follow that $\overline{\xi}\cdot\overline{\pi}=\overline{\rho}\cdot\overline{\pi}$.

Now as in <https://quant.stackexchange.com/q/17839/42184> we need to show that for $w>0$,

$$\mathcal{R}\_{w}:=\{\overline{\xi}\in \mathbb R^{d+1}:\overline{\xi}\cdot \overline{\pi}=w,\;\overline{\xi}\cdot\overline{S}\geq 0\; \text{a.s.}\}\;\; \text{ is compact} $$

As we are in finite-dimensions, I assume that we need to use $\text{compact}\iff \text{closed and bounded}$.

Closedness is no problem at all, using convergence arguments. However, I am struggling to show boundedness. As I have not used the no arbitrage arguments or the non-redundancy condition, I assume that I will need to use them in my argument. Any ideas?