 Solution

The seminorms in $\mathcal{R}$ have inverse images of intervals that constrain one coordinate at a time. Finite intersections of these sets are exactly the standard basic neighbourhoods for the product topology, so $(X\times Y,\mathcal{R})$ is the product of locally convex spaces. If $F\in (X\times Y{)}^{\ast }$, thenbelong to $X^{\ast }$ and $Y^{\ast }$ and $F(x,y)=f(x)+g(y)$. Conversely every such pair defines a continuous functional, so $(X\times Y{)}^{\ast }=X^{\ast }\oplus Y^{\ast }$.

For the open convex sets $K_1,\dots ,K_n$, considerThis set is open and convex, and $0\notin C$ because the $K_i$ have empty intersection. Separate $0$ from $C$ by a continuous linear functional on the product. By the dual description just proved, it has the formfor $f_j\in X^{\ast }$, and $L$ is nonzero with one strict sign on $C$. DefineIf some vector belonged to every $T(K_i)$, choose $x_i\in K_i$ with the same image. Then every $f_j(x_j-x_n)=0$, making $L(x_1-x_n,\dots ,x_{n-1}-x_n)=0$, a contradiction. Hence ${\bigcap }_{i}T(K_i)=\varnothing$, proving finite-dimensional separation of open convex sets.

Solved by gpt-5.6-sol high.