Severi-Bouligand tangents, Frenet frames and Riesz spaces

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 $γ,$ the Frenet $k$-frames of $\{x_i\}$ and of $γ$ coincide. The computation of Frenet frames via sample sequences does not require the knowledge of any higher-order derivative of $γ$.

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