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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?

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You are correct. Assuming $a\not\in A_2$ – Thomas Andrews Jan 21 '13 at 21:32
Is it true that one is subset of another? – Idonknow Oct 24 '13 at 15:41

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

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How and why could one divine or presage that $\bigcup$ behaves like $+$ here and $\text{the Cartesian Product}$ like scalar multiplication? – Timere Sep 2 '13 at 15:35

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.

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Could you please elucidate what goaded or motivated you to "think of the sets $A$ and $B$ as singletons"? – Timere Sep 2 '13 at 15:40

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

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