I'm trying to understand the following definition:
Let $x \in A \subset \mathbb R^n$. We say that $A$ is polyhedral at $x$ iff there is a neighborhood $U$ of $x$ and a polyhedron $B$ such that $A \cap U = A \cap B$.
Let $A = \{x \in \mathbb R^n: \sum_{i=1}^n x_i = 1, \ x_i \geq 0\}$.
Is $A$ polyhedral at each of its points?
Intuitively, it seems to me like the answer should be yes: $A$ is a convex polytope, which is just a special kind of polyhedron, and polyhedra should be polyhedral at each of their points (right?). I'm not really sure how to formally prove this from the definition though.
Edited (seems the in the equality of the question was actually $A\cap U=A\cap B$, although that seems strange to me).
Your set $A$ is a polyhedron, because it is the intersection of $\sum_{i=1}^n x_i \geq 1$, $\sum_{i=1}^n x_i \leq 1$, and $x_i\geq0$.
Thus, for every point $x$ of $A$ and nhood $U$ of $x$, you can pick $U=\mathbb{R}^n$ and $B=A$ and you have $A\cap U=B\cap A$, that is, $A$ is polyhedral at $x$.