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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 $?

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marked as duplicate by Rahul, Micah, Asaf Karagila, Henry T. Horton, Alexander Gruber Mar 5 '13 at 1:39

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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
up vote 13 down vote accepted

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 $b\in B$. Therefor $A\times B$, which is a set of these pairs is empty.

If $A\neq\emptyset$ and $B\neq\emptyset$, exist $a\in A$ and $b\in B$, thus $(a,b)\in A\times B$. Therefor $A\times B\neq\emptyset$.

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