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M. Winter
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The comment of Robert brought me onto the right track. Say $I=\{1,...,m\}$, then

\begin{align} \Bbb R^n\setminus C^\circ &= \{\,y\in\Bbb R^n\mid \exists x\in C\colon\langle x,y\rangle >1\,\} \\&= \{\,y\in\Bbb R^n\mid \exists x\in\Bbb R^n\colon p_1(x)\le 0 \land\cdots\land p_m(x)\le0 \land \langle x,y\rangle>1\,\} \\&=\pi\, \{(x,y)\in\Bbb R^n\times\Bbb R^n\mid p_1(x)\le 0 \land\cdots\land p_m(x)\le0 \land \langle x,y\rangle>1\,\}, \end{align}

where $\pi$ is the projection $(x,y)\mapsto y$.

So $\Bbb R^n\setminus C^\circ$ is the projection of a semi-algebraic set. By Tarski-Seidenberg $\Bbb R^n\setminus C^\circ$ is a semi-algebraic set, and so is its complement $C^\circ$.

What I don't yet see is why it is necessarily of the form $(*)$, i.e. using only intersections.

M. Winter
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