I'd like to know if the reasoning below is correct or not.
I have this problem: $\underset{x}{\operatorname{min}}\begin{Bmatrix} \operatorname{c}^T\operatorname{x} : A(\zeta)\operatorname{x} \leq \operatorname{b},\forall \zeta \in U \end{Bmatrix}$ with $\operatorname{c}\in \mathbb{R}^n, \operatorname{b}\in \mathbb{R}^m, A\in \mathbb{R}^{m \times n}$ and $\zeta \in \mathbb{R}^L$. This is a LP problem under uncertainty, where the uncertainty is now detectable only at the level of eligible region with particular reference to the matrix of coefficients $A$, for $\zeta \equiv \zeta(\omega),\omega \in \Omega$ the single realization of the uncertain parameters of $A$. Generally I know that a LP problem without uncertainty is a convex problem, since the constraints system describes a convex set intended as the intersection of a finite number of half-spaces. And if we consider the same problem under uncertainty, I know that it's possible to represent the uncertainty set through intervals, i.e. to say $U=\begin{Bmatrix} A: A^- \leq A \leq A^+ \end{Bmatrix}$, with $A^-$ and $A^+$ the estimates of lower and upper bounds of the interval of $A$. Obviously, in spite of this assumption, remain infinite values that the parameters of $A$ could assume, which is why I think it's out of question the convexity of $U$: number of potential half-spaces is now uncountable. However we can consider $U$ like $\operatorname{conv}(U)$, i.e. the smallest convex set that includes $U$, because testing the feasibility of a solution on $U$ for a maximization problem means evaluating the objective function respect to $A^+$ which will coincide (no escaping) with the value of objective evaluating on his convex hull. So a LP problem under uncertainty that should be unconvex can always be considered in the same way of a convex problem.
Is it right? Thanks in advance for any clarification.