Suppose $A , B , C$ are arbitrary sets and we know that $( A times B ) cap ( C times D ) = emptyset$ What conclusion can we draw?

$(Atimes B)cap(Ctimes D)=emptysetimplies$

$forall (a,b) in (Atimes B), (a,b)notin (Ctimes D) implies$

$neg[(ain A land ain C) land (bin B land bin D)]implies$

$neg (ain A land ain C) lor neg (bin B land bin D)implies$

$(anotin A lor anotin C)lor(bnotin Blor bnotin D)$

Since $ain A land bin B$, $(Atimes B)cap(Ctimes D)implies anotin C lor bnotin D$

$anotin C lor bnotin Dimplies neg(ain C land bin D)implies$

$(Acap C = emptyset) lor (Bcap D=emptyset)implies$

$(Acap C)times (Bcap D) = emptyset$

It is not necessarily true that both $Acap C = emptyset$ and $Bcap D=emptyset$. Only one or the other must be. From this it follows that $(Acap C)times (Bcap D) = emptyset$ is always true.

For case of $Acap D = emptyset$, consider $(a,b) in (Atimes B)$, such that $ (a,b)notin (Ctimes D)$. This does not imply that $anotin D$. It just means $anotin C$ or $bnotin D$.

The first answer is correct, because in fact, $(A times B) cap (C times D)$ $=(A cap C) times (B cap D)$ for any four sets $A,B,C$ and $D$.

Only $(1)$ is true.

Start proving by contradiction. If possible let $(Acap C)times (Bcap D)neq emptyset$ and $(x,y)in(Acap C)times (Bcap D)$. Then show that $(x,y)in (Atimes B)cap(Ctimes D)$.