Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
2  
You are correct. Assuming $a\not\in A_2$ –  Thomas Andrews Jan 21 '13 at 21:32
    
homework should not be used as a standalone tag; see tag-wiki and meta. –  Martin Sleziak Feb 1 '13 at 9:35
    
Is it true that one is subset of another? –  Idonknow Oct 24 '13 at 15:41
add comment

3 Answers

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

share|improve this answer
    
How and why could one divine or presage that $\bigcup$ behaves like $+$ here and $\text{the Cartesian Product}$ like scalar multiplication? –  LePressentiment Sep 2 '13 at 15:35
add comment

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.

share|improve this answer
    
Could you please elucidate what goaded or motivated you to "think of the sets $A$ and $B$ as singletons"? –  LePressentiment Sep 2 '13 at 15:40
add comment

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

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.