Severi-Bouligand tangents, Frenet frames and Riesz spaces

Abstract: It was recently proved that a compact set $X\subseteq \mathbb R^2$ has an outgoing Severi-Bouligand tangent vector $u\not=0$ at $x\in X$ iff some principal ideal of the Riesz space $\mathcal R(X)$ of piecewise linear functions on $X$ is not an intersection of maximal ideals. "Outgoing" means $X\cap [x,x+u]={x}$. Suppose now $X\subseteq \mathbb{R}^n$ and some principal ideal of $\mathcal R(X)$ is not an intersection of maximal ideals. We prove that this is equivalent to saying that $X$ contains a sequence ${x_i}$ whose Frenet $k$-frame $(u_1,\ldots,u_k)$ is an outgoing Severi-Bouligand tangent of $X$. When the ${x_i}$ are taken as sample points of a smooth curve $\gamma,$ the Frenet $k$-frames of ${x_i}$ and of $\gamma$ coincide. The computation of Frenet frames via sample sequences does not require the knowledge of any higher-order derivative of $\gamma$.