  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 $\gamma$.
