Countable union of cartesian product. Let $I^1_j$ $1$-dimensional intervals( of the form $[a,b)$) and $I^n_j$ be $n$-dimensional interval(of the form $[a_1,b_1)\times \ldots\times [a_n,b_n)$) such that $I_j\times I^n_j$ are pairwise disjoint, suppose that $\displaystyle\bigcup_{j \in \Bbb N}I^1_j\times I^n_j=I$ (union is disjoint)is an $(n+1)$-dimensional interval. 
Can we say $$\bigcup_{j \in \Bbb N}I^1_j\times I^n_j=I=\bigcup_{j\in \Bbb N}I^1_j \times \bigcup_{j\in \Bbb N}I^n_j \text{  ?}$$
The only thing I figured out is that the RHS unions are not necessarily disjoint. 
 A: In the specific situation in the question, we have the equality
$$\bigcup_{j\in\mathbb{N}} (I_j^1\times I_j^n) = I = \left(\bigcup_{j\in\mathbb{N}} I_j^1\right)\times \left(\bigcup_{j\in\mathbb{N}} I_j^n\right)$$
provided that none of the $I_j^1$ or $I_j^n$ are empty. If some $I_j^1$ or $I_j^n$ are empty, the product on the right could be a proper superset of $I$.
That these are intervals is irrelevant, as is the countability of the union or the fact that the union on the left is disjoint.
All that matters is that we have two families $\{ X_\alpha : \alpha \in A\}$ and $\{ Y_\alpha : \alpha \in A\}$ of nonempty sets, indexed by the same index set, and
$$\bigcup_{\alpha\in A} (X_\alpha \times Y_\alpha) = \mathcal{X}\times \mathcal{Y}.$$
Then we also have
$$\bigcup_{\alpha \in A} X_\alpha = \mathcal{X}\tag{x}$$
and
$$\bigcup_{\alpha\in A} Y_\alpha = \mathcal{Y},\tag{y}$$
and therefore
$$\left(\bigcup_{\alpha\in A} X_\alpha\right) \times \left(\bigcup_{\alpha\in A} Y_\alpha\right) = \mathcal{X}\times \mathcal{Y}.\tag{p}$$
Since we assumed all $Y_\alpha \neq \varnothing$, we have $\pi_1(X_\alpha\times Y_\alpha) = X_\alpha$ for all $\alpha$, and hence $(\text{x})$:
$$\mathcal{X} = \pi_1(\mathcal{X}\times\mathcal{Y}) = \pi_1 \left(\bigcup_{\alpha\in A} (X_\alpha\times Y_\alpha)\right) = \bigcup_{\alpha\in A} \pi_1(X_\alpha\times Y_\alpha) = \bigcup_{\alpha\in A} X_\alpha.$$
The assumption $X_\alpha\neq\varnothing$ for all $\alpha$ yields $(\text{y})$ in the same way, using the projection $\pi_2$ to the second factor.
