# Why is the Cartesian product of a set $A$ and empty set an empty set? [duplicate]

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 $A \times \emptyset = \emptyset$?

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

• Suppose by contradiction that you're able to pick $(x,y) \in A \times \emptyset$ $\ldots$ – Dominique Feb 16 '13 at 21:10
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 $$b\in B$$. Therefore $$A\times B$$, which is the set of these pairs, is empty.
If $$A\neq\emptyset$$ and $$B\neq\emptyset$$, there exists $$a\in A$$ and $$b\in B$$, thus $$(a,b)\in A\times B$$. Therefore $$A\times B\neq\emptyset$$.