Are Unions/Intersections of Cartesian Products of Indexed Sets "Distributive"?

Question

In a situation like $\bigcup\limits_{x \in [0,1]} [x,1] \times [0,x^2]$, could one correctly assume that $\bigcup\limits_{x \in [0,1]} [x,1] \times \bigcup\limits_{x \in [0,1]}[0,x^2]$ also holds true? That is to say, is it "distributive" in some sense of the word?

In general, if $\bigcup\limits_{i \in I} A_i \times B_i$ or $\bigcap\limits_{i \in I} A_i \times B_i$ for some indexed sets $A_i, B_i$, does it hold that:

$$\bigcup\limits_{i \in I} A_i \times B_i = \bigcup\limits_{i \in I} A_i \times \bigcup\limits_{i \in I} B_i$$ $$\bigcap\limits_{i \in I} A_i \times B_i = \bigcap\limits_{i \in I} A_i \times \bigcap\limits_{i \in I} B_i$$

in general?

Answer 1

It is false for unions. To see why, consider the following Battenberg cake (image from Wikipedia), whose cross-section is identified with the subset $[0,1]^2 \subseteq \mathbb{R}^2$ of the plane:

Let $A_1 = B_1 = [0,\frac{1}{2}]$ and $A_2 = B_2 = [\frac{1}{2},1]$. Then

• $\bigcup_{i \in \{ 1,2 \}} (A_i \times B_i) = [0,\frac{1}{2}]^2 \cup [\frac{1}{2},1]^2$ consists only of the white squares in the cake; but
• $\left( \bigcup_{i \in \{ 1,2 \}} A_i \right) \times \left( \bigcup_{i \in \{1,2\}} B_i \right) = [0,1]^2$ is the entire cake.

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It is true for intersections. To see why, just chase the definitions. Indeed, $\bigcap_{i \in I} (A_i \times B_i)$ consists of those things which are elements of all the sets $A_i \times B_i$. Now:

• For each $i \in I$, an element of $A_i \times B_i$ is a pair $(a,b)$;
• Such a pair is an element of all the sets $A_i \times B_i$ if and only if $a \in A_i$ for each $i \in I$ and $b \in B_i$ for each $i \in I$;
• But this is equivalent to saying that $(a,b) \in \left( \bigcap_{i \in I} A_i \right) \times \left( \bigcap_{i \in I} B_i \right)$.