Set theory proof with empty set Prove that if $A \times B = A \times (C \setminus B) $ $then: A \times (B \bigcup C) = \emptyset$ I get that $B=(C\setminus B)$ so that means either C is an empty set ~~or C and B have no element in common.~~ (Striked text is probably wrong) What can I do from here on ? Thanks.

We do not necessarily have $B=C\setminus B$, for example we might have $A=\emptyset$ and $B,C$ arbitrary!

Instead, just try a proof of the contraposition:

Assume $A\times (B\cup C)\
e \emptyset$. Then there exists an element $z\in A\times (B\cup C)\
e \emptyset$ and this is of the form $z=(x,y)$ with $x\in A$ and $y\in B\cup C$.

* If $y\in B$ then $y\
otin C\setminus B$, so we have $(x,y)\in A\times B$ and $(x,y)\
otin C\setminus B$.
* If $y\
otin B$, then from $y\in B\cup C$ we have $y\in C\setminus B$, so $(x,y)\
otin A\times B$, but $(x,y)\in A\times(C\setminus B)$.



In both cases $(x,y)$ witnesses that $A\times B\
e A\times (C\setminus B)$.

In summary, $A\times (B\cup C)\
e \emptyset$ implies $A\times B\
e A\times (C\setminus B)$. In other words, $A\times B= A\times (C\setminus B)$ implies $A\times (B\cup C)= \emptyset$.