Prove that sets are pairwise disjoint
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Prove for every two sets A and B that A − B, B − A and A ∩ B are pairwise disjoint.
I've been looking at this problem and keep thinking it isn't true every time I think I make progress.
My main issue is that A and B can be any sets, including themselves. If A = B then A-B={} but also B-A={} so they can't be disjoint.
Obviously I'm missing something, probably will feel dumb after I figure it out.
elementary-set-theory
asked Nov 19 at 2:37
T. Joe
62
add a comment |
up vote
0
down vote
favorite
Prove for every two sets A and B that A − B, B − A and A ∩ B are pairwise disjoint.
I've been looking at this problem and keep thinking it isn't true every time I think I make progress.
My main issue is that A and B can be any sets, including themselves. If A = B then A-B={} but also B-A={} so they can't be disjoint.
Obviously I'm missing something, probably will feel dumb after I figure it out.
elementary-set-theory
asked Nov 19 at 2:37
T. Joe
62
Empty sets are disjoint from anything, so A-B and B-A are disjoint when A=B.
– herb steinberg
Nov 19 at 2:44
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Prove for every two sets A and B that A − B, B − A and A ∩ B are pairwise disjoint.
I've been looking at this problem and keep thinking it isn't true every time I think I make progress.
My main issue is that A and B can be any sets, including themselves. If A = B then A-B={} but also B-A={} so they can't be disjoint.
Obviously I'm missing something, probably will feel dumb after I figure it out.
elementary-set-theory
asked Nov 19 at 2:37
T. Joe
62
Prove for every two sets A and B that A − B, B − A and A ∩ B are pairwise disjoint.
I've been looking at this problem and keep thinking it isn't true every time I think I make progress.
My main issue is that A and B can be any sets, including themselves. If A = B then A-B={} but also B-A={} so they can't be disjoint.
Obviously I'm missing something, probably will feel dumb after I figure it out.
elementary-set-theory
elementary-set-theory
asked Nov 19 at 2:37
T. Joe
62
asked Nov 19 at 2:37
T. Joe
62
asked Nov 19 at 2:37
T. Joe
62
asked Nov 19 at 2:37
T. Joe
62
asked Nov 19 at 2:37
T. Joe
62
62
Empty sets are disjoint from anything, so A-B and B-A are disjoint when A=B.
– herb steinberg
Nov 19 at 2:44
add a comment |
Empty sets are disjoint from anything, so A-B and B-A are disjoint when A=B.
– herb steinberg
Nov 19 at 2:44
Empty sets are disjoint from anything, so A-B and B-A are disjoint when A=B.
– herb steinberg
Nov 19 at 2:44
Empty sets are disjoint from anything, so A-B and B-A are disjoint when A=B.
– herb steinberg
Nov 19 at 2:44
add a comment |
1 Answer
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You seem to be forgetting the meaning to "pairwise disjoint." Two sets $A,B$ are disjoint if
$$A cap B = emptyset$$
(Note that "{}" is just an alternate way of denoting $emptyset$, and, further, that $emptyset cap emptyset = emptyset$, trivially.) More simply, two sets are disjoint if they have no elements in common.
Playing around with a few, small, trivial examples - or using Venn diagrams - might help illustrate what exactly is going on.
Note that $A$ is the entire left circle, $B$ is the entire right circle.
answered Nov 19 at 2:45
Eevee Trainer
2,179220
I absolutely did forget what "pairwise disjoint" meant. Just to clarify they are pairwise disjoint if it is more than two sets that are disjoint with each other? A-B is disjoint with B-A which is disjoint with A ∩ B. So the group is pairwise disjoint
– T. Joe
Nov 19 at 2:54
1
A group of multiple sets is pairwise disjoint if you can take any pair of them and they are disjoint. For example, if we want to say $A,B,C$ are pairwise disjoint, then $$A cap B = A cap C = B cap C = emptyset$$ I probably should've clarified that it is only "disjoint" if you're considering two sets ("$A$ and $B$ are disjoint"), and "pairwise disjoint" if you have a bunch of each, any two of which are disjoint
– Eevee Trainer
Nov 19 at 2:59
Yes, sorry I rushed my response. They are only "pairwise disjoint" when a group of sets (more than 2) have no common elements between them. Regular "disjoint" only refers to two sets
– T. Joe
Nov 19 at 3:04
It's fine, lol, I had rushed my own. But yup, that's exactly right. :)
– Eevee Trainer
Nov 19 at 3:06
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
You seem to be forgetting the meaning to "pairwise disjoint." Two sets $A,B$ are disjoint if
$$A cap B = emptyset$$
(Note that "{}" is just an alternate way of denoting $emptyset$, and, further, that $emptyset cap emptyset = emptyset$, trivially.) More simply, two sets are disjoint if they have no elements in common.
Playing around with a few, small, trivial examples - or using Venn diagrams - might help illustrate what exactly is going on.
Note that $A$ is the entire left circle, $B$ is the entire right circle.
answered Nov 19 at 2:45
Eevee Trainer
2,179220
I absolutely did forget what "pairwise disjoint" meant. Just to clarify they are pairwise disjoint if it is more than two sets that are disjoint with each other? A-B is disjoint with B-A which is disjoint with A ∩ B. So the group is pairwise disjoint
– T. Joe
Nov 19 at 2:54
1
A group of multiple sets is pairwise disjoint if you can take any pair of them and they are disjoint. For example, if we want to say $A,B,C$ are pairwise disjoint, then $$A cap B = A cap C = B cap C = emptyset$$ I probably should've clarified that it is only "disjoint" if you're considering two sets ("$A$ and $B$ are disjoint"), and "pairwise disjoint" if you have a bunch of each, any two of which are disjoint
– Eevee Trainer
Nov 19 at 2:59
Yes, sorry I rushed my response. They are only "pairwise disjoint" when a group of sets (more than 2) have no common elements between them. Regular "disjoint" only refers to two sets
– T. Joe
Nov 19 at 3:04
It's fine, lol, I had rushed my own. But yup, that's exactly right. :)
– Eevee Trainer
Nov 19 at 3:06
add a comment |
up vote
0
down vote
You seem to be forgetting the meaning to "pairwise disjoint." Two sets $A,B$ are disjoint if
$$A cap B = emptyset$$
(Note that "{}" is just an alternate way of denoting $emptyset$, and, further, that $emptyset cap emptyset = emptyset$, trivially.) More simply, two sets are disjoint if they have no elements in common.
Playing around with a few, small, trivial examples - or using Venn diagrams - might help illustrate what exactly is going on.
Note that $A$ is the entire left circle, $B$ is the entire right circle.
answered Nov 19 at 2:45
Eevee Trainer
2,179220
I absolutely did forget what "pairwise disjoint" meant. Just to clarify they are pairwise disjoint if it is more than two sets that are disjoint with each other? A-B is disjoint with B-A which is disjoint with A ∩ B. So the group is pairwise disjoint
– T. Joe
Nov 19 at 2:54
1
A group of multiple sets is pairwise disjoint if you can take any pair of them and they are disjoint. For example, if we want to say $A,B,C$ are pairwise disjoint, then $$A cap B = A cap C = B cap C = emptyset$$ I probably should've clarified that it is only "disjoint" if you're considering two sets ("$A$ and $B$ are disjoint"), and "pairwise disjoint" if you have a bunch of each, any two of which are disjoint
– Eevee Trainer
Nov 19 at 2:59
Yes, sorry I rushed my response. They are only "pairwise disjoint" when a group of sets (more than 2) have no common elements between them. Regular "disjoint" only refers to two sets
– T. Joe
Nov 19 at 3:04
It's fine, lol, I had rushed my own. But yup, that's exactly right. :)
– Eevee Trainer
Nov 19 at 3:06
add a comment |
up vote
0
down vote
up vote
0
down vote
You seem to be forgetting the meaning to "pairwise disjoint." Two sets $A,B$ are disjoint if
$$A cap B = emptyset$$
(Note that "{}" is just an alternate way of denoting $emptyset$, and, further, that $emptyset cap emptyset = emptyset$, trivially.) More simply, two sets are disjoint if they have no elements in common.
Playing around with a few, small, trivial examples - or using Venn diagrams - might help illustrate what exactly is going on.
Note that $A$ is the entire left circle, $B$ is the entire right circle.
answered Nov 19 at 2:45
Eevee Trainer
2,179220
You seem to be forgetting the meaning to "pairwise disjoint." Two sets $A,B$ are disjoint if
$$A cap B = emptyset$$
(Note that "{}" is just an alternate way of denoting $emptyset$, and, further, that $emptyset cap emptyset = emptyset$, trivially.) More simply, two sets are disjoint if they have no elements in common.
Playing around with a few, small, trivial examples - or using Venn diagrams - might help illustrate what exactly is going on.
Note that $A$ is the entire left circle, $B$ is the entire right circle.
answered Nov 19 at 2:45
Eevee Trainer
2,179220
edited Nov 19 at 2:59
answered Nov 19 at 2:45
Eevee Trainer
2,179220
answered Nov 19 at 2:45
Eevee Trainer
2,179220
answered Nov 19 at 2:45
Eevee Trainer
2,179220
2,179220
I absolutely did forget what "pairwise disjoint" meant. Just to clarify they are pairwise disjoint if it is more than two sets that are disjoint with each other? A-B is disjoint with B-A which is disjoint with A ∩ B. So the group is pairwise disjoint
– T. Joe
Nov 19 at 2:54
1
A group of multiple sets is pairwise disjoint if you can take any pair of them and they are disjoint. For example, if we want to say $A,B,C$ are pairwise disjoint, then $$A cap B = A cap C = B cap C = emptyset$$ I probably should've clarified that it is only "disjoint" if you're considering two sets ("$A$ and $B$ are disjoint"), and "pairwise disjoint" if you have a bunch of each, any two of which are disjoint
– Eevee Trainer
Nov 19 at 2:59
Yes, sorry I rushed my response. They are only "pairwise disjoint" when a group of sets (more than 2) have no common elements between them. Regular "disjoint" only refers to two sets
– T. Joe
Nov 19 at 3:04
It's fine, lol, I had rushed my own. But yup, that's exactly right. :)
– Eevee Trainer
Nov 19 at 3:06
add a comment |
I absolutely did forget what "pairwise disjoint" meant. Just to clarify they are pairwise disjoint if it is more than two sets that are disjoint with each other? A-B is disjoint with B-A which is disjoint with A ∩ B. So the group is pairwise disjoint
– T. Joe
Nov 19 at 2:54
1
A group of multiple sets is pairwise disjoint if you can take any pair of them and they are disjoint. For example, if we want to say $A,B,C$ are pairwise disjoint, then $$A cap B = A cap C = B cap C = emptyset$$ I probably should've clarified that it is only "disjoint" if you're considering two sets ("$A$ and $B$ are disjoint"), and "pairwise disjoint" if you have a bunch of each, any two of which are disjoint
– Eevee Trainer
Nov 19 at 2:59
Yes, sorry I rushed my response. They are only "pairwise disjoint" when a group of sets (more than 2) have no common elements between them. Regular "disjoint" only refers to two sets
– T. Joe
Nov 19 at 3:04
It's fine, lol, I had rushed my own. But yup, that's exactly right. :)
– Eevee Trainer
Nov 19 at 3:06
I absolutely did forget what "pairwise disjoint" meant. Just to clarify they are pairwise disjoint if it is more than two sets that are disjoint with each other? A-B is disjoint with B-A which is disjoint with A ∩ B. So the group is pairwise disjoint
– T. Joe
Nov 19 at 2:54
I absolutely did forget what "pairwise disjoint" meant. Just to clarify they are pairwise disjoint if it is more than two sets that are disjoint with each other? A-B is disjoint with B-A which is disjoint with A ∩ B. So the group is pairwise disjoint
– T. Joe
Nov 19 at 2:54
1
1
A group of multiple sets is pairwise disjoint if you can take any pair of them and they are disjoint. For example, if we want to say $A,B,C$ are pairwise disjoint, then $$A cap B = A cap C = B cap C = emptyset$$ I probably should've clarified that it is only "disjoint" if you're considering two sets ("$A$ and $B$ are disjoint"), and "pairwise disjoint" if you have a bunch of each, any two of which are disjoint
– Eevee Trainer
Nov 19 at 2:59
A group of multiple sets is pairwise disjoint if you can take any pair of them and they are disjoint. For example, if we want to say $A,B,C$ are pairwise disjoint, then $$A cap B = A cap C = B cap C = emptyset$$ I probably should've clarified that it is only "disjoint" if you're considering two sets ("$A$ and $B$ are disjoint"), and "pairwise disjoint" if you have a bunch of each, any two of which are disjoint
– Eevee Trainer
Nov 19 at 2:59
Yes, sorry I rushed my response. They are only "pairwise disjoint" when a group of sets (more than 2) have no common elements between them. Regular "disjoint" only refers to two sets
– T. Joe
Nov 19 at 3:04
Yes, sorry I rushed my response. They are only "pairwise disjoint" when a group of sets (more than 2) have no common elements between them. Regular "disjoint" only refers to two sets
– T. Joe
Nov 19 at 3:04
It's fine, lol, I had rushed my own. But yup, that's exactly right. :)
– Eevee Trainer
Nov 19 at 3:06
It's fine, lol, I had rushed my own. But yup, that's exactly right. :)
– Eevee Trainer
Nov 19 at 3:06
add a comment |
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Empty sets are disjoint from anything, so A-B and B-A are disjoint when A=B.
– herb steinberg
Nov 19 at 2:44