# Is it always true that $(A_1 \cup A_2) \times (B_1 \cup B_2)=(A_1\times B_1) \cup (A_2 \times B_2)$

Is it always true that

$(A_1 \cup A_2) \times (B_1 \cup B_2)=(A_1\times B_1) \cup (A_2 \times B_2)$?

I don't believe this is true. I have tried to draw pictures to help me get on the right path, but I think that the union makes this untrue. for example, if $a \in A_1$ and $b \in B_2$, then $(a,b)$ would not be in $(A_1\times B_1) \cup (A_2 \times B_2)$. Is this a correct assumption?

• You are correct. Assuming $a\not\in A_2$ Commented Jan 21, 2013 at 21:32
• Is it true that one is subset of another? Commented Oct 24, 2013 at 15:41
• Related (at a higher level): math.stackexchange.com/questions/2878687/… Commented Mar 22, 2021 at 23:33

No, they behave like $+$ and $\cdot$: $$(A_1\cup A_2)\times(B_1\cup B_2)=(A_1\times B_1)\cup (A_1\times B_2)\cup (A_2\times B_1)\cup (A_2\times B_2)$$

• How and why could one divine or presage that $\bigcup$ behaves like $+$ here and $\text{the Cartesian Product}$ like scalar multiplication?
– user53259
Commented Sep 2, 2013 at 15:35
• @Berci: Is $A_1 \times B_1 \cup A_1 \times B_2 = A_1 \times (B_1 \cup B_2)$? Commented Apr 3, 2019 at 17:33

This is not true. Think of the sets $A$ and $B$ as singletons.

If $A_1 = \{0\}$, $A_2 = \{1\}$, $B_1 = \{0\}$ and $B_2 = \{1\}$ then $(A_1 \cup A_2) \times (B_1 \cup B_2)$ is like a 2-by-2 grid, but $(A_1\times B_1)$ is the bottom-left point and $(A_2\times B_2)$ is the top-right.

ADDITION What is true, however, is that

$$(A_1 \cup A_2) \times (B_1 \cup B_2)=(A_1\times B_1) \cup (A_2 \times B_2) \cup (A_1 \times B_2) \cup (A_2 \times B_1)$$

which is essentially the distributed property.

• Could you please elucidate what goaded or motivated you to "think of the sets $A$ and $B$ as singletons"?
– user53259
Commented Sep 2, 2013 at 15:40

Let $A_1=\varnothing =B_2$. The LHS will be $A_2\times B_1$ while the RHS will be $\varnothing \cup \varnothing=\varnothing$.

Also, a more general result is this:

$$\prod_{i\in I}\biggl( \bigcup_{j\in J} A_{i, j} \biggr) = \bigcup_{f\colon I\to J}\biggl( \prod_{i\in I} A_{i, j} \biggr)$$

Note that this uses axiom of choice for LHS's inclusion in RHS.