Suppose I have a union set, $\bigcup\limits_{\mathbf{q} \in\mathbf{P}}A(\mathbf{q})$, where $A(\mathbf{q})$ is a $n$-dimensional hypercube defined by
$$A(\mathbf{q}) = \left\{ \mathbf{x}\in\mathbb{R}^n \mid 0\leq x_j\leq \langle \mathbf{y}(j),\mathbf{q}\rangle, \forall\,j\in\{1,2,\dots,n\} \right\}$$
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. $\mathbf{q}\in\mathbb{R}^m$ is a probabilistic vector, that is $||\mathbf{q}||_1=1$ and each element of $\mathbf{q}$ is non-negative. $\mathbf{P}$ is a standard probability simplex with $m$ dimensions.
Moreover, I have shown that $\cup_{\mathbf{q}\in\mathbf{P}}A(\mathbf{q})$ is closed, bounded, and convex.
The question is, is the $\cup_{\mathbf{q}\in\mathbf{P}}A(\mathbf{q})$ still a convex polytope?
Based on its closed, bounded, and convex conditions of the union of the sets, I think it is a convex polytope, not a ball, but I got stuck while trying to show that there are finite numbers of half-spaces that can form $\cup_{\mathbf{q}\in\mathbf{P}}A(\mathbf{q})$, which is a definition of a polytope. I think I got stuck because there may not exist a dominant $\mathbf{q}$, that is, different $\mathbf{q}$ may cause different upper bounds of the same $x_j$, due to $\mathbf{y}(j)$ may be different among $j$. I guess there exists some tradeoffs.
I can find $n$ vertices of $\cup_{\mathbf{q}}A(\mathbf{q})$, i.e., $\langle \mathbf{y}(j),\mathbf{q}^*(j)\rangle, \forall\,j\in\{1,2,…,n\}$, where $\mathbf{q}^*(j)=\arg \max_{\mathbf{q}\in\mathbf{P}}\langle \mathbf{y}(j),\mathbf{q}\rangle, \forall\,j\in\{1,2,…,n\}$.
Edit: I think the problem can be firstly reduced into $\mathbb{R}^2$ space, that is, for an infinite union of sets of rectangles, if it is bounded, closed, and convex, whether the infinite union is a circle or a convex polygon. However, in this case, I still have no idea.
Edit: The reason why I state the union is infinite union is because I thought the number of unions on the standard simplex is infinite. Sorry for any misunderstanding.
Best Answer
Answer to the Revised Question. Indeed, the union $P = \cup_{\mathbf{q}} A(\mathbf{q})$ is a polytope. To show this, let
$$ \tilde{P} = \texttt{ConvexHull}(\cup_{i=1}^{m} A(\mathbf{e}_i))$$
where $\mathbf{e}_1, \ldots, \mathbf{e}_m$ are the standard basis vectors of $\mathbb{R}^m$. Since $\tilde{P}$ is the convex hull of a finite union of polytopes, $\tilde{P}$ itself is also a polytope (which is spanned by the union of vertices of $A(\mathbf{e}_i)$'s.)
Proof. Since $P$ is a convex set containing each of $A(\mathbf{e}_i)$, it is clear that $\tilde{P} \subseteq P$. So it suffices to prove that $P \subseteq \tilde{P}$.
To this end, let $\mathbf{x} \in P$ be arbitrary. Then there exists a probability vector $\mathbf{q}$ such that $\mathbf{x} \in A(\mathbf{q})$. To show that $\mathbf{x} \in \tilde{P}$, it suffices to show that there exist $\mathbf{x}^{(i)} \in A(\mathbf{e}_i)$, $i = 1, \ldots, m$, such that
$$ \mathbf{x} = q_1 \mathbf{x}^{(1)} + \cdots + q_m \mathbf{x}^{(m)}. \tag{*} $$
To this end, we note:
\begin{align*} \mathbf{x} \in A(\mathbf{q}) &\iff x_j = c_{j} \langle \mathbf{y}(j), \mathbf{q} \rangle \quad \text{for some $c_1, \ldots, c_n \in [0, 1]$} \\ &\implies x_j = \sum_{i=1}^{m} q_i c_{j} \langle \mathbf{y}(j), \mathbf{e}_i \rangle \quad \text{for some $c_1, \ldots, c_n \in [0, 1]$}. \end{align*}
So, if we set $\mathbf{x}^{(i)} = \sum_{j=1}^{n} c_{j} \langle \mathbf{y}(j), \mathbf{e}_i \rangle \mathbf{e}_j$, then $\mathbf{x}^{(i)} \in A(\mathbf{e}_i)$ for each $i$ and $\text{(*)}$ holds, concluding the desired claim. $\square$
Answer to the Original Question. If some of the hypercubes are allowed to be not "parallel" to one another, then there is a counter-example.
Let me construct a counter-example in 2D case. We begin by noting the following observation:
Since its proof is not so fun to write down rigorously, let me provide an animated illustration of how the proof works:
In light of this lemma, we may weaken the condition on $A(i)$'s by allowing them to be polygons with no acute interior angles. Now we are ready to construct a counter-example:
Set $P_1 = (0, 1)$, $P_2 = (-1, 0)$, $P_3 = (0, -1)$, and $Q_k = (\sin(\pi/2k), \cos(\pi/2k))$ for $k = 1, 2, 3, \ldots$
For each $i$, let $A(i)$ be the polygon with vertices $P_1, P_2, P_3, Q_1, \ldots, Q_i$.
Then $\cup_{i=1}^{\infty} A(i)$ is the compact convex hull spanned by $P_1, P_2, P_3, Q_1, Q_2, \ldots$, which is clearly not a polygon: