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Let $X$ be a set. What is $X\times \emptyset$ supposed to mean? Is it just the empty set?

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And more can be said: a cartesian product is empty if and only if one of the two factors is empty.

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    $\begingroup$ My formal proof: dcproof.com/Siminone.htm $\endgroup$ – Dan Christensen Jul 31 '12 at 19:25
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    $\begingroup$ A side note: This proof also works fine for the Cartesian product of finitely many sets. For arbitrary products, this statement is precisely the axiom of choice. $\endgroup$ – Nate Eldredge Jul 31 '12 at 22:44
  • $\begingroup$ Any of two factors. Your theorem is uncertain about two factors are empty ;) $\endgroup$ – Little Alien Oct 26 '16 at 10:18
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$X \times \emptyset = \emptyset$. In fact, if not we have $x \in X $ and $y \in \emptyset$ such that $(x,y) \in X \times \emptyset$. But $y \in \emptyset$ is impossible.

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Recall the definition of $A\times B$: $z\in A\times B$ if and only if $z=\langle a,b\rangle$ where $a\in A$ and $b\in B$. That is $A\times B$ is the set of all ordered pairs whose first coordinate is in $A$ and second in $B$.

If $B$ is empty then there are no ordered pairs $\langle a,b\rangle$ such that $b\in\varnothing$, therefore $A\times\varnothing=\varnothing$. Similarly $\varnothing\times B=\varnothing$.

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