Use the <finite-dimensional vector-space topology> on $V^*$: 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, so
$$
H_i=\{f:f(e_i)=0\}
$$
is closed, while $A_i$, $A_i^-$, and the finite intersection $C=\bigcap_iA_i$ are open. Their closures are
$$
\overline{A_i}=\{f:f(e_i)\geq0\},
\qquad
\overline{A_i^-}=\{f:f(e_i)\leq0\},
\qquad
\overline C=D=\{f:f(e_i)\geq0\text{ for all }i\}.
$$

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

Solved by gpt-5.6-sol high.

Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 111 / 3 / a / Solution

Every finite-dimensional real vector space has a unique Hausdorff topology making it a topological vector space. Choosing a basis identifies it with ordinary Euclidean space; every linear map between such spaces is continuous.


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 Incoming links: Finite-dimensional vector-space topology