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Let $A \times \emptyset = \{(x,y)| x\in A, y \in \emptyset \}$. We know there is no element in $\emptyset$. But how does it follow that $A \times \emptyset = \emptyset $?

Daniel
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    Suppose by contradiction that you're able to pick $(x,y) \in A \times \emptyset$ $\ldots$ – Dominique Feb 16 '13 at 21:10
  • @Dominique Damn you are right on the bullseye :) Thanks – Daniel Feb 16 '13 at 21:12
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    Also: http://math.stackexchange.com/questions/177190/is-the-product-of-two-sets-well-defined-if-one-is-empty http://math.stackexchange.com/questions/51401/how-do-the-sets-emptyset-times-b-a-times-emptyset-emptyset-times-em – Asaf Karagila Feb 16 '13 at 22:05

1 Answers1

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Claim: $A\times B=\emptyset$ iff $A=\emptyset$ or $B=\emptyset$

Proof: If $A=\emptyset$ or $B=\emptyset$, then there is no $(a,b)$ such that $a\in A$ and $b\in B$. Therefore $A\times B$, which is the set of these pairs, is empty.

If $A\neq\emptyset$ and $B\neq\emptyset$, there exists $a\in A$ and $b\in B$, thus $(a,b)\in A\times B$. Therefore $A\times B\neq\emptyset$.

Ryan
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Salech Alhasov
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