Let $X$ be a set. What is $X\times \emptyset$ supposed to mean? Is it just the empty set?
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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.
David Mitra
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user29999
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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.
Siminore
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1My formal proof: http://www.dcproof.com/Siminone.htm – Dan Christensen Jul 31 '12 at 19:25
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1A 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. – Nate Eldredge Jul 31 '12 at 22:44
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Any of two factors. Your theorem is uncertain about two factors are empty ;) – Little Alien Oct 26 '16 at 10:18
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Dan's website is not useful for actual mathematical work, and moreover has some bogus articles. – user21820 Jun 07 '22 at 17:21
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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 $\varnothing\times B=\varnothing$.
Asaf Karagila
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