Let $\mathcal{F}_n$ be the $\mathbb{R}$-vector space of functions $f : \mathbb{R}^n \rightarrow \mathbb{R}$. For all $X \subset \mathbb{R}^n$, we denote $\mathbb{1}_X$ the indicator function of $X$. Let $\mathcal{U}_n$ be the vector subspace of $\mathcal{F}_n$ generated by the functions $\mathbb{1}_P$ where $P$ is a polytope of $\mathbb{R}^n$. We assume that $n > 1$. Let $f \in \mathcal{U}_n$. For $z \in \mathbb{R}$, we define the function $f_z : \mathbb{R}^{n-1} \rightarrow \mathbb{R}$ by $f_z(x_1, \ldots, x_{n-1}) = f(x_1, \ldots, x_{n-1}, z)$ for all $(x_1, \ldots, x_{n-1}) \in \mathbb{R}^{n-1}$. Prove that $f_z \in \mathcal{U}_{n-1}$.
Let $\mathcal{F}_n$ be the $\mathbb{R}$-vector space of functions $f : \mathbb{R}^n \rightarrow \mathbb{R}$. For all $X \subset \mathbb{R}^n$, we denote $\mathbb{1}_X$ the indicator function of $X$. Let $\mathcal{U}_n$ be the vector subspace of $\mathcal{F}_n$ generated by the functions $\mathbb{1}_P$ where $P$ is a polytope of $\mathbb{R}^n$.
We assume that $n > 1$. Let $f \in \mathcal{U}_n$. For $z \in \mathbb{R}$, we define the function $f_z : \mathbb{R}^{n-1} \rightarrow \mathbb{R}$ by $f_z(x_1, \ldots, x_{n-1}) = f(x_1, \ldots, x_{n-1}, z)$ for all $(x_1, \ldots, x_{n-1}) \in \mathbb{R}^{n-1}$. Prove that $f_z \in \mathcal{U}_{n-1}$.