Use contradiction to show that if $A$ × $B$ and $C$× $D$ are disjoint, then either $A\cap C\:=\:\varnothing$ or $B\cap D\:=\:\varnothing$

My thinking:

Assume, for contradiction, $A\cap B\ne \varnothing $ and $A\cap B\ne \varnothing $.

Let $x\in A\cap C$ and $y\in B\cap D$

If $x\in A$, the by defintion of cartesian product, $x\in$ $A$ × $B$

If $y\in B$, the by defintion of cartesian product, $y\in$ $B$ × $D$

Since $A × B$ and $C × D$ are disjoint, it must be that $x \notin A\cap C$ and $x\notin B\cap D$, hence a contradiction.

Is this proof valid? I'm not too sure...

Solution 1:

If $x \in A$ then $x$ cannot be in $A \times B$. Your proof is wrong.

Let $x \in A \cap C$ and $y \in B \cap D$. Then $x \in A, x \in C, y \in B$ and $y \in D$. Hence, $(x,y) \in (A \times B) \cap (C\times D)$.