 Solution

Use the finite-dimensional vector-space topology on $V^{\ast }$: relative to the dual basis, it is the ordinary Euclidean topology on ${\mathbb{R}}^{|I|}$. Each coordinate map $f\mapsto f(e_i)$ is continuous, sois closed, while $A_i$, $A_i^{-}$, and the finite intersection $C={\bigcap }_i{A}_i$ are open. Their closures are

Every ${\sigma }^{\ast }(w)$ is an invertible linear map with inverse ${\sigma }^{\ast }(w^{-1})$. Linear maps between finite-dimensional topological vector spaces are continuous, so both it and its inverse are continuous. Hence every ${\sigma }^{\ast }(w)$ is a homeomorphism.

Solved by gpt-5.6-sol high.