Show that $R_{w}$ is bounded/compact using no arbitrage or non-redundancy arguments 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?
 A: Here is a contradiction idea (inspired from proof of Theorem 1.32 from Follmer's Stochastic Finance which is the source material of this question): Take a sequence in your set $R_w$ such that $\limsup_{n \rightarrow \infty} |\overline{\xi_n}| = \infty$, and consider looking at $\eta_n = \frac{\overline{\xi_n}}{|\overline{\xi_n}|}$. Note that $\eta_n$ is a bounded sequence in $\mathbb R^n$, so a subsequence of it (by Bolzano-Weierstrass) will converge to an $\eta$ with $|\eta|=1$ (norm is continuous). However, this is a problem, since $\eta$ now satisfies $\eta \cdot\overline{\pi} = 0$, and $\eta\cdot\overline{S}\geq 0 \, \,\mathbb P$almost surely . Now, note that by nonredundancy, it must be the case that $\mathbb P(\eta ⋅\overline{S}> 0)>0$, because otherwise $\eta = 0$, which contradicts $|\eta| = 1$. Thus, $\eta$ is an arbitrage, which should not happen in an arbitrage-free market.
A: This is just a re-statement of the following geometric fact about finite dimensional Hilbert spaces:
Proposition Let $V$ be a finite dimensional Hilbert space, $\bar{\pi} \in V$, and $\mathcal{S} = \{ \bar{S}_{\omega}\}_{\omega \in \Omega} \subset V$ such that:

*

*$\bar{\pi}$ lies in the interior of the positive cone generated by $\mathcal{S}$, i.e. if $\langle \bar{\xi}, \bar{\pi} \rangle \leq 0$, then either $\langle \bar{\xi}, \bar{S}_{\omega} \rangle < 0$ for some $\omega$ or $\langle \bar{\xi}, \bar{S}_{\omega} \rangle = 0$ for all $\omega$.


*$\mathcal{S}$ is separating in the sense that, if $\bar{\xi} \neq 0$, then $\langle \bar{\xi},  \bar{S}_{\omega} \rangle \neq 0$ for some $\omega$.
Then , for all $w > 0$,
$$ 
\mathcal{R}_w =  \{\bar{\xi}:  \langle \bar{\xi}, \bar{\pi} \rangle = w, \; \langle \bar{\xi}, \bar{S}_{\omega} \rangle \geq 0  \} 
$$
is compact in $V$.
In your context, $V = \mathbb{R}^{1+d}$, $\Omega$ is the underlying probability space, Condition 1 is no-arbitrage, and Condition 2 is non-redundancy.
(Replace $\Omega$ by a set with probability $1$ if necessary. No-arbitrage and non-redundancy are really $\omega$-by-$\omega$ conditions.)
In the finite dimensional case (finitely many assets/commodities), the geometry of the statement is clear---e.g. take a basis $\bar{S}_{1}, \cdots, \bar{S}_{1+d}$ in
$\mathbb{R}^{1+d}$, and a $\bar{\pi}$ that lies in the interior of the positive cone they generate.
(Condition 2 implies that $|\Omega| \geq 1+d$, i.e. non-redundancy means there cannot be more assets than states. This means the linear span---equivalently, its closure---of $\mathcal{S}$ is $V$.)
As you said, closedness and convexity of $\mathcal{R}_w$ is clear.
$\mathcal{R}_w$ is an intersection of closed and convex sets,
$$
\mathcal{R}_w =  \bar{\pi}^{-1}( w ) \bigcap \left( \bigcap_{\bar{S} \in \mathcal{S}} \bar{S}^{-1}[0, \infty) \right)
$$
where, as a slight abuse of notation, $\bar{\pi}(\cdot)$ also denotes the linear functional/dual element $\bar{\xi} \mapsto \langle \bar{\xi},  \bar{\pi} \rangle$, same for $\bar{S}(\cdot)$.
Boundedness can be shown via the very nice argument given by @E-A. (When $\Omega$ is finite, you can also argue directly using
$$
\bar{\xi} \in \mathcal{R}_w \; \Rightarrow \; 0 \leq \sum_{k=1}^{K} q_k \langle \bar{\xi},  \bar{S}_{\omega_k} \rangle = (1+r) w,
$$
where $q_k$'s form an equivalent martingale measure.
)
Compactness of $\mathcal{R}_w$ does not hold in general for infinite dimensional Hilbert spaces. $\mathcal{R}_w$ remains closed but is no longer bounded in general.
For example, take $\mathcal{S} = \{ e_k, k \geq 1 \}$ to be an orthonormal basis of a separable Hilbert space $V$, and $\bar{\pi} = \sum\limits_{k \geq 1} \frac{1}{k^2} e_k$. Then both Conditions 1 and 2
hold. However, $\mathcal{R}_w$ is not bounded---e.g. consider
$$
\bar{\xi}_n = (w - \frac{1}{n}) e_1 + n e_n, \;\; n \geq 2.
$$
in $\mathcal{R}_w$.
Where the boundedness argument breaks down is that the unit sphere in $V$ is only weakly compact when $V$ is finite dimensional.
A sequence of unit vectors $\frac{\bar{\xi}_n}{\|\bar{\xi}_n\|}$ need not have a non-zero weak limit---which would then be an arbitrage portfolio if $\limsup\limits_n \|\bar{\xi}_n\| = \infty$.
