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

elementary-set-theory

Solution 1:

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

Solution 2:

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.

Solution 3:

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

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