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

Mathematics Asked by Anonymous Molecule on November 29, 2021

  1. $( A cap C ) times ( B cap D ) = emptyset$
  2. $B cap D = emptyset$
  3. $A cap C = emptyset$
  4. $A cap D = emptyset$

I feel like answers $2,3$ and $4$ could be correct. For example by substituting $A=C=emptyset$, or substituting $B=D=emptyset$, or substituting $A=D=emptyset$.

Please help!

3 Answers

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

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

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

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

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

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 $bnotin D$.

Answered by RJM on November 29, 2021

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 $D$.

Answered by Geoffrey Trang on November 29, 2021

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 $(x,y)in (Atimes B)cap(Ctimes D)$.

Answered by user598858 on November 29, 2021

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