# Show that in an arbitrage-free and non-redundant market a certain set is compact

Some notation: We consider a financial market with $d+1$ assets, the $0$-th asset is considered the risk-free asset, the others are the risky ones. The vector $\overline \pi \in \mathbb R^{d+1}$ denotes the pricing vectors, and the random vector $\overline S = (S^0,S^1,\ldots, S^d)$ denotes the vector of the random variables corresponding to the assets. Further we have $\pi^0 = 1, S^0 \equiv 1 + r$, as these are risk-free.

Here a portfolio $\overline \xi \in \mathbb R^{d+1}$ is called an arbitrage opportunity if $\overline \xi \cdot \overline \pi \le 0$ but $\overline \xi \cdot \overline S \ge 0$ P-a.s. and $P(\overline \xi \cdot \overline S > 0) > 0$. And the market model is called non-redundant if $$\overline \xi \cdot \overline S = 0 ~\mbox{P-a.s.} \Rightarrow \overline \xi = 0$$ i.e. non of the assets $S^i$ could be synthezied by investing in the others.

Show that in a non-redundant and arbitrage-free market model the set $$\{ \overline \xi \in \mathbb R^{d+1} : | \overline \pi \cdot \overline \xi = w \mbox{ and } \overline \xi \cdot \overline S \ge 0 ~ \mbox{P-a.s.} \}$$ is compact for any $w > 0$.

First, as this is a subset of the euclidean space $\mathbb R^{d+1}$, compactness is equivalent with bounded and closed. But already showing boundedness gives me a hard time, I do not see where to start?

Also, I guess a non-redundant market model also implies that it is arbitrage-free, so the presupposition of just arbitrage-free'ness should be enough.

• Is there a reference book, or some other background, for this question? – Gordon May 21 '15 at 17:56
• Is this continuous or discrete? If it is discrete, then how many states you can have? – Gordon May 21 '15 at 21:07
• It is an exercise form H. Föllmer, Stochastic Finance, 3rd edition, 1. Chapter. The number of assets is finite (and thereby discrete, see my preliminary remarks, their number is $d$), but everything else is continuous, as for example the measure space on which the $S^i$ are defined could be arbitrary, and of course the values the $S^i$ could attain are not restricted too, does this answer your question? – StefanH May 21 '15 at 21:21
• I think if we assume the set is compact, then we have no arbitrage. I guess you could try to find a proof for that in some text and see if the converse is proved in the same text. ? or ? – BCLC Dec 16 '15 at 17:53