Set theory cartesian products and difference
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
edited Nov 24 at 19:24
MPW
29.8k11956
asked Nov 24 at 19:20
PotentialMemory
1
add a comment |
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
edited Nov 24 at 19:24
MPW
29.8k11956
asked Nov 24 at 19:20
PotentialMemory
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
add a comment |
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
edited Nov 24 at 19:24
MPW
29.8k11956
asked Nov 24 at 19:20
PotentialMemory
1
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
edited Nov 24 at 19:24
MPW
29.8k11956
asked Nov 24 at 19:20
PotentialMemory
1
edited Nov 24 at 19:24
MPW
29.8k11956
asked Nov 24 at 19:20
PotentialMemory
1
edited Nov 24 at 19:24
MPW
29.8k11956
edited Nov 24 at 19:24
MPW
29.8k11956
edited Nov 24 at 19:24
MPW
29.8k11956
29.8k11956
asked Nov 24 at 19:20
PotentialMemory
1
asked Nov 24 at 19:20
PotentialMemory
1
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
add a comment |
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
add a comment |
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 $(x,y)notin Atimes C$?
answered Nov 24 at 19:30
MPW
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
add a comment |
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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 $(x,y)notin Atimes C$?
answered Nov 24 at 19:30
MPW
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
add a comment |
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 $(x,y)notin Atimes C$?
answered Nov 24 at 19:30
MPW
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
add a comment |
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 $(x,y)notin Atimes C$?
answered Nov 24 at 19:30
MPW
29.8k11956
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 $(x,y)notin Atimes C$?
answered Nov 24 at 19:30
MPW
29.8k11956
edited Nov 24 at 19:44
answered Nov 24 at 19:30
MPW
29.8k11956
answered Nov 24 at 19:30
MPW
29.8k11956
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
add a comment |
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
add a comment |
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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