Suppose X,Y are sets, A,B ⊆ X and C,D ⊆ Y. Compare (A × B) − (C × D) with ((A − C) × B) ∪ (A ×...
$begingroup$
I'm a bit puzzled by this question. If A and B are subsets of X and C and D are subsets of Y, is it safe to assume that the elements of A and B are completely different from those of C and D? I feel like that's too simple and not a good assumption to make. If that were the case, then (B − D) would simply be B. This would certainly make the comparison much easier, but I don't have any reason as to why this assumption might be made.
Does anyone have any insight on how to compare these two arbitrary sets?
elementary-set-theory
asked Jan 15 at 16:20
NickNick
1
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add a comment |
$begingroup$
I'm a bit puzzled by this question. If A and B are subsets of X and C and D are subsets of Y, is it safe to assume that the elements of A and B are completely different from those of C and D? I feel like that's too simple and not a good assumption to make. If that were the case, then (B − D) would simply be B. This would certainly make the comparison much easier, but I don't have any reason as to why this assumption might be made.
Does anyone have any insight on how to compare these two arbitrary sets?
elementary-set-theory
asked Jan 15 at 16:20
NickNick
1
$endgroup$
add a comment |
$begingroup$
I'm a bit puzzled by this question. If A and B are subsets of X and C and D are subsets of Y, is it safe to assume that the elements of A and B are completely different from those of C and D? I feel like that's too simple and not a good assumption to make. If that were the case, then (B − D) would simply be B. This would certainly make the comparison much easier, but I don't have any reason as to why this assumption might be made.
Does anyone have any insight on how to compare these two arbitrary sets?
elementary-set-theory
asked Jan 15 at 16:20
NickNick
1
$endgroup$
I'm a bit puzzled by this question. If A and B are subsets of X and C and D are subsets of Y, is it safe to assume that the elements of A and B are completely different from those of C and D? I feel like that's too simple and not a good assumption to make. If that were the case, then (B − D) would simply be B. This would certainly make the comparison much easier, but I don't have any reason as to why this assumption might be made.
Does anyone have any insight on how to compare these two arbitrary sets?
elementary-set-theory
elementary-set-theory
asked Jan 15 at 16:20
NickNick
1
asked Jan 15 at 16:20
NickNick
1
asked Jan 15 at 16:20
NickNick
1
asked Jan 15 at 16:20
NickNick
1
asked Jan 15 at 16:20
NickNick
1
1
add a comment |
add a comment |
1 Answer
1
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$begingroup$
If $(x,y)$ is in $((A-C) times B) cup (A times (B-D))$, if in the LHS, then $x in A, x notin C, y in B$ and then $(x,y) in A times B$ and $(x,y) notin C times D$... The same can be said when it is in the RHS (check this..).
This shows at least one inclusion. The other also holds:
Now if $(x,y) in (A times B) - (C times D)$, we know $x in A, y in B$ and $xnotin C$ or $y notin D$. What can you say about these two last cases?
answered Jan 15 at 23:39
Henno BrandsmaHenno Brandsma
108k347114
$endgroup$
$begingroup$
I'm still very confused here. I'm having trouble following your train of thought. Does LHS refer to that which is to the left of the word 'with' or the left side of the union symbol? And I understand the statements you are making about the sets in which x and y reside, but I do not understand how you got to them.
$endgroup$
– Nick
Jan 16 at 3:17
$begingroup$
@Nick of the union. What exactly cannot you follow?
$endgroup$
– Henno Brandsma
Jan 16 at 16:29
$begingroup$
I just don’t know what the conclusion will be. I am especially confused that you never mentioned the sets X and Y. Do they not matter in the comparison?
$endgroup$
– Nick
Jan 16 at 18:47
$begingroup$
The only conclusion I can think is maybe RHS of the ‘with’ ends up being (A×B) ∪ (A×B) which is a subset of of the LHS??
$endgroup$
– Nick
Jan 16 at 18:50
$begingroup$
Because for both sides, x is in A, y is in B, x is NOT in C, and y is NOT in D
$endgroup$
– Nick
Jan 16 at 18:53
|
show 2 more comments
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1 Answer
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1 Answer
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$begingroup$
If $(x,y)$ is in $((A-C) times B) cup (A times (B-D))$, if in the LHS, then $x in A, x notin C, y in B$ and then $(x,y) in A times B$ and $(x,y) notin C times D$... The same can be said when it is in the RHS (check this..).
This shows at least one inclusion. The other also holds:
Now if $(x,y) in (A times B) - (C times D)$, we know $x in A, y in B$ and $xnotin C$ or $y notin D$. What can you say about these two last cases?
answered Jan 15 at 23:39
Henno BrandsmaHenno Brandsma
108k347114
$endgroup$
$begingroup$
I'm still very confused here. I'm having trouble following your train of thought. Does LHS refer to that which is to the left of the word 'with' or the left side of the union symbol? And I understand the statements you are making about the sets in which x and y reside, but I do not understand how you got to them.
$endgroup$
– Nick
Jan 16 at 3:17
$begingroup$
@Nick of the union. What exactly cannot you follow?
$endgroup$
– Henno Brandsma
Jan 16 at 16:29
$begingroup$
I just don’t know what the conclusion will be. I am especially confused that you never mentioned the sets X and Y. Do they not matter in the comparison?
$endgroup$
– Nick
Jan 16 at 18:47
$begingroup$
The only conclusion I can think is maybe RHS of the ‘with’ ends up being (A×B) ∪ (A×B) which is a subset of of the LHS??
$endgroup$
– Nick
Jan 16 at 18:50
$begingroup$
Because for both sides, x is in A, y is in B, x is NOT in C, and y is NOT in D
$endgroup$
– Nick
Jan 16 at 18:53
|
show 2 more comments
$begingroup$
If $(x,y)$ is in $((A-C) times B) cup (A times (B-D))$, if in the LHS, then $x in A, x notin C, y in B$ and then $(x,y) in A times B$ and $(x,y) notin C times D$... The same can be said when it is in the RHS (check this..).
This shows at least one inclusion. The other also holds:
Now if $(x,y) in (A times B) - (C times D)$, we know $x in A, y in B$ and $xnotin C$ or $y notin D$. What can you say about these two last cases?
answered Jan 15 at 23:39
Henno BrandsmaHenno Brandsma
108k347114
$endgroup$
$begingroup$
I'm still very confused here. I'm having trouble following your train of thought. Does LHS refer to that which is to the left of the word 'with' or the left side of the union symbol? And I understand the statements you are making about the sets in which x and y reside, but I do not understand how you got to them.
$endgroup$
– Nick
Jan 16 at 3:17
$begingroup$
@Nick of the union. What exactly cannot you follow?
$endgroup$
– Henno Brandsma
Jan 16 at 16:29
$begingroup$
I just don’t know what the conclusion will be. I am especially confused that you never mentioned the sets X and Y. Do they not matter in the comparison?
$endgroup$
– Nick
Jan 16 at 18:47
$begingroup$
The only conclusion I can think is maybe RHS of the ‘with’ ends up being (A×B) ∪ (A×B) which is a subset of of the LHS??
$endgroup$
– Nick
Jan 16 at 18:50
$begingroup$
Because for both sides, x is in A, y is in B, x is NOT in C, and y is NOT in D
$endgroup$
– Nick
Jan 16 at 18:53
|
show 2 more comments
$begingroup$
If $(x,y)$ is in $((A-C) times B) cup (A times (B-D))$, if in the LHS, then $x in A, x notin C, y in B$ and then $(x,y) in A times B$ and $(x,y) notin C times D$... The same can be said when it is in the RHS (check this..).
This shows at least one inclusion. The other also holds:
Now if $(x,y) in (A times B) - (C times D)$, we know $x in A, y in B$ and $xnotin C$ or $y notin D$. What can you say about these two last cases?
answered Jan 15 at 23:39
Henno BrandsmaHenno Brandsma
108k347114
$endgroup$
If $(x,y)$ is in $((A-C) times B) cup (A times (B-D))$, if in the LHS, then $x in A, x notin C, y in B$ and then $(x,y) in A times B$ and $(x,y) notin C times D$... The same can be said when it is in the RHS (check this..).
This shows at least one inclusion. The other also holds:
Now if $(x,y) in (A times B) - (C times D)$, we know $x in A, y in B$ and $xnotin C$ or $y notin D$. What can you say about these two last cases?
answered Jan 15 at 23:39
Henno BrandsmaHenno Brandsma
108k347114
edited Jan 16 at 16:55
answered Jan 15 at 23:39
Henno BrandsmaHenno Brandsma
108k347114
answered Jan 15 at 23:39
Henno BrandsmaHenno Brandsma
108k347114
answered Jan 15 at 23:39
Henno BrandsmaHenno Brandsma
108k347114
108k347114
$begingroup$
I'm still very confused here. I'm having trouble following your train of thought. Does LHS refer to that which is to the left of the word 'with' or the left side of the union symbol? And I understand the statements you are making about the sets in which x and y reside, but I do not understand how you got to them.
$endgroup$
– Nick
Jan 16 at 3:17
$begingroup$
@Nick of the union. What exactly cannot you follow?
$endgroup$
– Henno Brandsma
Jan 16 at 16:29
$begingroup$
I just don’t know what the conclusion will be. I am especially confused that you never mentioned the sets X and Y. Do they not matter in the comparison?
$endgroup$
– Nick
Jan 16 at 18:47
$begingroup$
The only conclusion I can think is maybe RHS of the ‘with’ ends up being (A×B) ∪ (A×B) which is a subset of of the LHS??
$endgroup$
– Nick
Jan 16 at 18:50
$begingroup$
Because for both sides, x is in A, y is in B, x is NOT in C, and y is NOT in D
$endgroup$
– Nick
Jan 16 at 18:53
|
show 2 more comments
$begingroup$
I'm still very confused here. I'm having trouble following your train of thought. Does LHS refer to that which is to the left of the word 'with' or the left side of the union symbol? And I understand the statements you are making about the sets in which x and y reside, but I do not understand how you got to them.
$endgroup$
– Nick
Jan 16 at 3:17
$begingroup$
@Nick of the union. What exactly cannot you follow?
$endgroup$
– Henno Brandsma
Jan 16 at 16:29
$begingroup$
I just don’t know what the conclusion will be. I am especially confused that you never mentioned the sets X and Y. Do they not matter in the comparison?
$endgroup$
– Nick
Jan 16 at 18:47
$begingroup$
The only conclusion I can think is maybe RHS of the ‘with’ ends up being (A×B) ∪ (A×B) which is a subset of of the LHS??
$endgroup$
– Nick
Jan 16 at 18:50
$begingroup$
Because for both sides, x is in A, y is in B, x is NOT in C, and y is NOT in D
$endgroup$
– Nick
Jan 16 at 18:53
$begingroup$
I'm still very confused here. I'm having trouble following your train of thought. Does LHS refer to that which is to the left of the word 'with' or the left side of the union symbol? And I understand the statements you are making about the sets in which x and y reside, but I do not understand how you got to them.
$endgroup$
– Nick
Jan 16 at 3:17
$begingroup$
I'm still very confused here. I'm having trouble following your train of thought. Does LHS refer to that which is to the left of the word 'with' or the left side of the union symbol? And I understand the statements you are making about the sets in which x and y reside, but I do not understand how you got to them.
$endgroup$
– Nick
Jan 16 at 3:17
$begingroup$
@Nick of the union. What exactly cannot you follow?
$endgroup$
– Henno Brandsma
Jan 16 at 16:29
$begingroup$
@Nick of the union. What exactly cannot you follow?
$endgroup$
– Henno Brandsma
Jan 16 at 16:29
$begingroup$
I just don’t know what the conclusion will be. I am especially confused that you never mentioned the sets X and Y. Do they not matter in the comparison?
$endgroup$
– Nick
Jan 16 at 18:47
$begingroup$
I just don’t know what the conclusion will be. I am especially confused that you never mentioned the sets X and Y. Do they not matter in the comparison?
$endgroup$
– Nick
Jan 16 at 18:47
$begingroup$
The only conclusion I can think is maybe RHS of the ‘with’ ends up being (A×B) ∪ (A×B) which is a subset of of the LHS??
$endgroup$
– Nick
Jan 16 at 18:50
$begingroup$
The only conclusion I can think is maybe RHS of the ‘with’ ends up being (A×B) ∪ (A×B) which is a subset of of the LHS??
$endgroup$
– Nick
Jan 16 at 18:50
$begingroup$
Because for both sides, x is in A, y is in B, x is NOT in C, and y is NOT in D
$endgroup$
– Nick
Jan 16 at 18:53
$begingroup$
Because for both sides, x is in A, y is in B, x is NOT in C, and y is NOT in D
$endgroup$
– Nick
Jan 16 at 18:53
|
show 2 more comments
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