# Prove that $X\times Y=\emptyset$ $\iff$ $X=\emptyset$ or $Y=\emptyset$.

• Prove that $$X\times Y=\emptyset$$ $$\iff$$ $$X=\emptyset$$ or $$Y=\emptyset$$.

My proof. I will do contrapositive.

LHS. Assume $$X\neq\emptyset$$ and $$Y\neq\emptyset$$, so there is $$a\in X$$ and $$b\in Y$$ such that $$(a,b)\in X\times Y$$. Thus, $$X\times Y\neq\emptyset.$$

RHS. Assume $$X\times Y\neq\emptyset.$$ Then, there is a $$(a,b)\in X\times Y$$ such that $$a\in X$$ and $$b\in Y$$, hence $$X\neq \emptyset$$ and $$Y\neq\emptyset$$.

Can you check my proof? Thanks..

• looks good to me – Tortuga Mar 5 at 18:34
• The proof is correct. It is straight from the definitions, no tricks. You should feel more confident in solving such exercises. – Mark Mar 5 at 18:34
• Your proof is correct. – White Crow Mar 5 at 18:35
• Thanks for comments... – PozcuKushimotoStreet Mar 5 at 18:36
• You can do it directly. If $A =\emptyset$ then there can't be any $(a,b)\in A\times B$ as there are no $a \in A$. And if $A\times B=\emptyset$ then for any $a \in A$ and $b \in B$ we cant have $(a,b) \in A\times B$ so for any $a \in A$ there can be no $b\in B$ and for any $b\in B$ there can be no $a\in A$ so either $A$ or $B$ is empty. – fleablood Mar 6 at 1:03