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  • 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..

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    $\begingroup$ looks good to me $\endgroup$ – Tortuga Mar 5 at 18:34
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    $\begingroup$ The proof is correct. It is straight from the definitions, no tricks. You should feel more confident in solving such exercises. $\endgroup$ – Mark Mar 5 at 18:34
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    $\begingroup$ Your proof is correct. $\endgroup$ – White Crow Mar 5 at 18:35
  • $\begingroup$ Thanks for comments... $\endgroup$ – PozcuKushimotoStreet Mar 5 at 18:36
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    $\begingroup$ 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. $\endgroup$ – fleablood Mar 6 at 1:03

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