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.

This question already has an answer here:

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 $?

share|improve this question

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

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

7  
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
1  

1 Answer 1

up vote 5 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 $A\times B\neq\emptyset$.

share|improve this answer

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