Set theory cartesian products and difference












0














First of all hello to everyone!



I have been stuck on the same exercise and I'm looking for proper guidance in order to solve this task.
I have $Atimes(Bsetminus C)=(Atimes B) setminus (Atimes C)$ ( sorry for not using the proper math format on this page, I'm new so I don't know how :D), and the problem is not proving the right side but the left.($Atimes (Bsetminus C)$).



I get to the point where ($xin A$ and $yin B$) and ($xin A$ and $ynotin C$) and don't know what to do?



Can someone please explain ?






elementary-set-theory







share|cite|improve this question




















  • 1




    what do you mean by proving the left side? You are supposed to prove an equality of sets so you have to show left is a subset of right and right is a subset of left.
    – Anurag A
    Nov 24 at 19:27
















0














First of all hello to everyone!



I have been stuck on the same exercise and I'm looking for proper guidance in order to solve this task.
I have $Atimes(Bsetminus C)=(Atimes B) setminus (Atimes C)$ ( sorry for not using the proper math format on this page, I'm new so I don't know how :D), and the problem is not proving the right side but the left.($Atimes (Bsetminus C)$).



I get to the point where ($xin A$ and $yin B$) and ($xin A$ and $ynotin C$) and don't know what to do?



Can someone please explain ?






elementary-set-theory







share|cite|improve this question




















  • 1




    what do you mean by proving the left side? You are supposed to prove an equality of sets so you have to show left is a subset of right and right is a subset of left.
    – Anurag A
    Nov 24 at 19:27














0












0








0







First of all hello to everyone!



I have been stuck on the same exercise and I'm looking for proper guidance in order to solve this task.
I have $Atimes(Bsetminus C)=(Atimes B) setminus (Atimes C)$ ( sorry for not using the proper math format on this page, I'm new so I don't know how :D), and the problem is not proving the right side but the left.($Atimes (Bsetminus C)$).



I get to the point where ($xin A$ and $yin B$) and ($xin A$ and $ynotin C$) and don't know what to do?



Can someone please explain ?






elementary-set-theory







share|cite|improve this question















First of all hello to everyone!



I have been stuck on the same exercise and I'm looking for proper guidance in order to solve this task.
I have $Atimes(Bsetminus C)=(Atimes B) setminus (Atimes C)$ ( sorry for not using the proper math format on this page, I'm new so I don't know how :D), and the problem is not proving the right side but the left.($Atimes (Bsetminus C)$).



I get to the point where ($xin A$ and $yin B$) and ($xin A$ and $ynotin C$) and don't know what to do?



Can someone please explain ?






elementary-set-theory




elementary-set-theory






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 24 at 19:24









MPW

29.8k11956




29.8k11956










asked Nov 24 at 19:20









PotentialMemory

1




1








  • 1




    what do you mean by proving the left side? You are supposed to prove an equality of sets so you have to show left is a subset of right and right is a subset of left.
    – Anurag A
    Nov 24 at 19:27














  • 1




    what do you mean by proving the left side? You are supposed to prove an equality of sets so you have to show left is a subset of right and right is a subset of left.
    – Anurag A
    Nov 24 at 19:27








1




1




what do you mean by proving the left side? You are supposed to prove an equality of sets so you have to show left is a subset of right and right is a subset of left.
– Anurag A
Nov 24 at 19:27




what do you mean by proving the left side? You are supposed to prove an equality of sets so you have to show left is a subset of right and right is a subset of left.
– Anurag A
Nov 24 at 19:27










1 Answer
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If I understand, you are trying to prove membership in RHS implies membership in LHS.



You should then start by assuming $(x,y)in Atimes B$ and $(x,y)notin Atimes C$, since that's what it means to have $(x,y)in (Atimes B)setminus (Atimes C)$.



Your goal is to prove that $xin A$ and $y in Bsetminus C$, since that's what it means to have $(x,y)in Atimes (Bsetminus C)$.



This implies that $xin A$, that $yin B$, and that $ynotin C$. Why? Because if $(x,y)notin Atimes C$, then either $xnotin A$ or $ynotin C$ -- but you already know that $xin A$, so it must be the case that $ynotin C$.



You should be able to proceed from here.




For the other direction, start by assuming that $xin A$ and $y in Bsetminus C$, since that's what it means to have $(x,y)in Atimes (Bsetminus C)$. Then $yin B$ and $ynotin C$, so surely "$xin A$ and $yin B$" is true, so $(x,y)in Atimes B$. You should be able to finish the proof from here. Can you see that it is also true that $(x,y)notin Atimes C$?






share|cite|improve this answer























  • I have actually done that I'm trying to prove the LHS and that's where I'm having troubles.
    – PotentialMemory
    Nov 24 at 19:34










  • Okay, then I misunderstood. But the proof is completely reversible. Try it!
    – MPW
    Nov 24 at 19:38











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If I understand, you are trying to prove membership in RHS implies membership in LHS.



You should then start by assuming $(x,y)in Atimes B$ and $(x,y)notin Atimes C$, since that's what it means to have $(x,y)in (Atimes B)setminus (Atimes C)$.



Your goal is to prove that $xin A$ and $y in Bsetminus C$, since that's what it means to have $(x,y)in Atimes (Bsetminus C)$.



This implies that $xin A$, that $yin B$, and that $ynotin C$. Why? Because if $(x,y)notin Atimes C$, then either $xnotin A$ or $ynotin C$ -- but you already know that $xin A$, so it must be the case that $ynotin C$.



You should be able to proceed from here.




For the other direction, start by assuming that $xin A$ and $y in Bsetminus C$, since that's what it means to have $(x,y)in Atimes (Bsetminus C)$. Then $yin B$ and $ynotin C$, so surely "$xin A$ and $yin B$" is true, so $(x,y)in Atimes B$. You should be able to finish the proof from here. Can you see that it is also true that $(x,y)notin Atimes C$?






share|cite|improve this answer























  • I have actually done that I'm trying to prove the LHS and that's where I'm having troubles.
    – PotentialMemory
    Nov 24 at 19:34










  • Okay, then I misunderstood. But the proof is completely reversible. Try it!
    – MPW
    Nov 24 at 19:38
















0














If I understand, you are trying to prove membership in RHS implies membership in LHS.



You should then start by assuming $(x,y)in Atimes B$ and $(x,y)notin Atimes C$, since that's what it means to have $(x,y)in (Atimes B)setminus (Atimes C)$.



Your goal is to prove that $xin A$ and $y in Bsetminus C$, since that's what it means to have $(x,y)in Atimes (Bsetminus C)$.



This implies that $xin A$, that $yin B$, and that $ynotin C$. Why? Because if $(x,y)notin Atimes C$, then either $xnotin A$ or $ynotin C$ -- but you already know that $xin A$, so it must be the case that $ynotin C$.



You should be able to proceed from here.




For the other direction, start by assuming that $xin A$ and $y in Bsetminus C$, since that's what it means to have $(x,y)in Atimes (Bsetminus C)$. Then $yin B$ and $ynotin C$, so surely "$xin A$ and $yin B$" is true, so $(x,y)in Atimes B$. You should be able to finish the proof from here. Can you see that it is also true that $(x,y)notin Atimes C$?






share|cite|improve this answer























  • I have actually done that I'm trying to prove the LHS and that's where I'm having troubles.
    – PotentialMemory
    Nov 24 at 19:34










  • Okay, then I misunderstood. But the proof is completely reversible. Try it!
    – MPW
    Nov 24 at 19:38














0












0








0






If I understand, you are trying to prove membership in RHS implies membership in LHS.



You should then start by assuming $(x,y)in Atimes B$ and $(x,y)notin Atimes C$, since that's what it means to have $(x,y)in (Atimes B)setminus (Atimes C)$.



Your goal is to prove that $xin A$ and $y in Bsetminus C$, since that's what it means to have $(x,y)in Atimes (Bsetminus C)$.



This implies that $xin A$, that $yin B$, and that $ynotin C$. Why? Because if $(x,y)notin Atimes C$, then either $xnotin A$ or $ynotin C$ -- but you already know that $xin A$, so it must be the case that $ynotin C$.



You should be able to proceed from here.




For the other direction, start by assuming that $xin A$ and $y in Bsetminus C$, since that's what it means to have $(x,y)in Atimes (Bsetminus C)$. Then $yin B$ and $ynotin C$, so surely "$xin A$ and $yin B$" is true, so $(x,y)in Atimes B$. You should be able to finish the proof from here. Can you see that it is also true that $(x,y)notin Atimes C$?






share|cite|improve this answer














If I understand, you are trying to prove membership in RHS implies membership in LHS.



You should then start by assuming $(x,y)in Atimes B$ and $(x,y)notin Atimes C$, since that's what it means to have $(x,y)in (Atimes B)setminus (Atimes C)$.



Your goal is to prove that $xin A$ and $y in Bsetminus C$, since that's what it means to have $(x,y)in Atimes (Bsetminus C)$.



This implies that $xin A$, that $yin B$, and that $ynotin C$. Why? Because if $(x,y)notin Atimes C$, then either $xnotin A$ or $ynotin C$ -- but you already know that $xin A$, so it must be the case that $ynotin C$.



You should be able to proceed from here.




For the other direction, start by assuming that $xin A$ and $y in Bsetminus C$, since that's what it means to have $(x,y)in Atimes (Bsetminus C)$. Then $yin B$ and $ynotin C$, so surely "$xin A$ and $yin B$" is true, so $(x,y)in Atimes B$. You should be able to finish the proof from here. Can you see that it is also true that $(x,y)notin Atimes C$?







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Nov 24 at 19:44

























answered Nov 24 at 19:30









MPW

29.8k11956




29.8k11956












  • I have actually done that I'm trying to prove the LHS and that's where I'm having troubles.
    – PotentialMemory
    Nov 24 at 19:34










  • Okay, then I misunderstood. But the proof is completely reversible. Try it!
    – MPW
    Nov 24 at 19:38


















  • I have actually done that I'm trying to prove the LHS and that's where I'm having troubles.
    – PotentialMemory
    Nov 24 at 19:34










  • Okay, then I misunderstood. But the proof is completely reversible. Try it!
    – MPW
    Nov 24 at 19:38
















I have actually done that I'm trying to prove the LHS and that's where I'm having troubles.
– PotentialMemory
Nov 24 at 19:34




I have actually done that I'm trying to prove the LHS and that's where I'm having troubles.
– PotentialMemory
Nov 24 at 19:34












Okay, then I misunderstood. But the proof is completely reversible. Try it!
– MPW
Nov 24 at 19:38




Okay, then I misunderstood. But the proof is completely reversible. Try it!
– MPW
Nov 24 at 19:38


















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