*Hopefully this answer to my question is correct despite my minimal knowledge of semi-algebraic things.*

**$X$ is not in general algebraic:** Consider the case of a square, i.e. $f=(4,4)$ (using the notation from my post, this should actually be $(4,4,1)$, but it seems that the number of $n$-dimensional faces of an n-dimensional polytope is usually left out since it’s always one). Then $X$ is exactly the set of 4-tuples $(x_1,\ldots,x_4)$ of points in $\Bbb R^2$ such that $x_4$ is not in the convex hull of $\{x_1,\ldots,x_3\}$. Suppose $X$ is algebraic, and fix $x_1,\ldots,x_3$. Then so is the set of all $x_4$ not in the convex hull of $\{x_1,\ldots,x_3\}$. But this set is $\Bbb R^2$ minus a (solid) triangle, and such a set is not algebraic.

**$X$ is semi-algebraic:** By the definition of “face”, $X$ is exactly the set of tuples $(x_1,\ldots,x_{f_0})$ of points in $\Bbb R^n$ satisfying the following (lines are numbered and indented for clarity):

- For all $k\in\{0,\ldots,n-1\}$,
- $\quad$there exist exactly $f_k$ subsets $I\subseteq \{1,\ldots, f_0\}$ such that
- $\quad\quad$there exists an affine linear functional $g\colon \Bbb R^n\to \Bbb R$ such that
- $\quad\quad\quad g(x_i)\geq0$ for all $i\in \{1,\ldots, f_0\}$
- $\quad\quad\quad$and $g(x_i)=0$ for all $i\in I$.

Lines 3-5 describe a semi-algebraic set when $I$ and $k$ are fixed. Since there are only finitely many possible $I$ and $k$, we get that $X$ is indeed semi-algebraic.