This is a follow-up of this.
Suppose I have a union set, $\bigcup\limits_{\{\mathbf{q}_k\}_{k=1}^n:\mathbf{q}_k \in\mathbf{P},\forall k\in\{1,2,...,n\}}A(\{\mathbf{q}_k\}_{k=1}^{n})$, where $A(\{\mathbf{q}_k\}_{k=1}^{n})$ is a $n$-dimensional hypercube defined as
$$A(\{\mathbf{q}_k\}_{k=1}^{n})=\left\{ \mathbf{x}\in\mathbb{R}^n \mid 0\leq x_j\leq \min_{k\in\{1,2,...,n\}\setminus\{j\}} \langle( \mathbf{y}(j),\mathbf{q}_k\rangle, \forall\,j\in\{1,2,\dots,n\} \right\}$$
where $\mathbf{y}(j)\in\mathbb{R}^m$ is a non-negative constant vector and indexed by $j,\forall\,j\in\{1,2,...,n\}$, which means the value of each $\mathbf{y}(j)$ might be different among $j$ and each of element in $\mathbf{y}(j)$ is non-negative. $\mathbf{q}_k\in\mathbb{R}^m$ is a probabilistic vector for all $k\in\{1,2,...,n\}$, that is $||\mathbf{q}_k||_1=1$ and each element of $\mathbf{q}_k$ is non-negative. $\mathbf{P}$ is a standard probability simplex with $m$ dimensions.
Moreover, I have shown that $\bigcup\limits_{\{\mathbf{q}_k\}_{k=1}^n:\mathbf{q}_k \in\mathbf{P},\forall k\in\{1,2,...,n\}}A(\{\mathbf{q}_k\}_{k=1}^{n})$ is closed, bounded, and convex.
The question is, is the $\bigcup\limits_{\{\mathbf{q}_k\}_{k=1}^n:\mathbf{q}_k \in\mathbf{P},\forall k\in\{1,2,...,n\}}A(\{\mathbf{q}_k\}_{k=1}^{n})$ still a convex polytope?
The difficulty of this problem is that the upper bound of each $x_j$ is a non-linear function of $\{\mathbf{q}_k\}_{k=1}^{n}$. Hence, I think it is hard to find an equivalent representation of a convex polytope (if it is a convex polytope) with some standard basis of $\mathbb{R}^m$ to decompose $\mathbf{q}_k$ as the aforementioned post.