Does the binomial theorem hold by default for the Cartesian Product of intervals? I was wondering if the binomial theorem holds for Cartesian products of intervals?
For example, does this inequality hold:
$$[(a,b) + (c,d)] ^2=(a,b) \times (a,b)+2\cdot(a,b) \times (c,d)+(c,d) \times (c,d)\ \text{?}$$
I somehow feel that this is a dumb question but still hope that I get a clarifying answer. Thank you in advance... 
 A: If $I_{1}$ and $I_{2}$ are sets (such as intervals of real numbers), define $I_{1} + I_{2}$ to be their disjoint union and $I_{1} \times I_{2}$ to be their Cartesian product.
Since the Cartesian product distributes over disjoint unions,
$$
(I_{1} + I_{2}) \times (I_{1} + I_{2})
  = (I_{1} \times I_{1}) + (I_{1} \times I_{2}) + (I_{2} \times I_{1}) + (I_{2} \times I_{2}).
$$
An analogous identity holds for an $n$-fold Cartesian product.
Moreover, $I_{2} \times I_{1}$ is naturally identified with $I_{1} \times I_{2}$, and the "cross term"
$$
2(I_{1} \times I_{2})
$$
may be interpreted as "two copies of $I_{1} \times I_{2}$". In this sense (up to re-ordering factors in an $n$-fold Cartesian product) an $n$-fold product of a disjoint union satisfies a formal binomial theorem
$$
(I_{1} + I_{2})^{n} = \sum_{k=0}^{n} \binom{n}{k} I_{1}^{n-k} \times I_{2}^{k}.
$$
This can be visualized for $n = 2$ or $n = 3$ (below) using the types of diagram often used to prove the product rule for derivatives. (The colors signify "factors", and the heavy lines show one of the "summands".)

