Finding disjoint sets
0
$begingroup$
Given any set $A$ what are some different ways I can construct a set $B$ such that $Acap B=emptyset$. I know there must exist many such sets, but I want to explictly construct one, not verify their existence. For example I have seen the disjoint union of any two sets $S$ and $Q$ wriiten as $Stimes {0}cup Qtimes {1}$, the person here explictly constructed two disjoint sets, what would be some other ways to do this?
elementary-set-theory logic
edited Dec 14 '18 at 5:48
Andrés E. Caicedo
65.9k8160252
asked Dec 14 '18 at 5:11
user3865391user3865391
6111215
$endgroup$
|
show 2 more comments
0
$begingroup$
Given any set $A$ what are some different ways I can construct a set $B$ such that $Acap B=emptyset$. I know there must exist many such sets, but I want to explictly construct one, not verify their existence. For example I have seen the disjoint union of any two sets $S$ and $Q$ wriiten as $Stimes {0}cup Qtimes {1}$, the person here explictly constructed two disjoint sets, what would be some other ways to do this?
elementary-set-theory logic
edited Dec 14 '18 at 5:48
Andrés E. Caicedo
65.9k8160252
asked Dec 14 '18 at 5:11
user3865391user3865391
6111215
$endgroup$
$begingroup$
Some easy solutions are $B=emptyset$ and (assuming the axiom of regularity) $B={A}$. Was there some other condition you wanted $B$ to satisfy?
$endgroup$
– bof
Dec 14 '18 at 5:53
$begingroup$
@bof Wait so $Acap {A}=emptyset$?
$endgroup$
– user3865391
Dec 14 '18 at 5:56
2
$begingroup$
The only element of ${A}$ is $A$, so $Acap{A}$ is empty unless $Ain A$. The "axiom of regularity" implies that $Ain A$ can't happen.
$endgroup$
– bof
Dec 14 '18 at 6:02
$begingroup$
Similar questions have been asked here, but usually they want a set $B$ which is disjoint from the given set $A$ and satisfies some other condition, for instance: Given a set $A$, construct a set $B$ such that $Acap B=emptyset$ and $|A|=|B|$.
$endgroup$
– bof
Dec 14 '18 at 6:05
$begingroup$
For instance, see this question: math.stackexchange.com/questions/2961610/…
$endgroup$
– bof
Dec 14 '18 at 6:06
|
show 2 more comments
0
0
0
$begingroup$
Given any set $A$ what are some different ways I can construct a set $B$ such that $Acap B=emptyset$. I know there must exist many such sets, but I want to explictly construct one, not verify their existence. For example I have seen the disjoint union of any two sets $S$ and $Q$ wriiten as $Stimes {0}cup Qtimes {1}$, the person here explictly constructed two disjoint sets, what would be some other ways to do this?
elementary-set-theory logic
edited Dec 14 '18 at 5:48
Andrés E. Caicedo
65.9k8160252
asked Dec 14 '18 at 5:11
user3865391user3865391
6111215
$endgroup$
Given any set $A$ what are some different ways I can construct a set $B$ such that $Acap B=emptyset$. I know there must exist many such sets, but I want to explictly construct one, not verify their existence. For example I have seen the disjoint union of any two sets $S$ and $Q$ wriiten as $Stimes {0}cup Qtimes {1}$, the person here explictly constructed two disjoint sets, what would be some other ways to do this?
elementary-set-theory logic
elementary-set-theory logic
edited Dec 14 '18 at 5:48
Andrés E. Caicedo
65.9k8160252
asked Dec 14 '18 at 5:11
user3865391user3865391
6111215
edited Dec 14 '18 at 5:48
Andrés E. Caicedo
65.9k8160252
asked Dec 14 '18 at 5:11
user3865391user3865391
6111215
edited Dec 14 '18 at 5:48
Andrés E. Caicedo
65.9k8160252
edited Dec 14 '18 at 5:48
Andrés E. Caicedo
65.9k8160252
edited Dec 14 '18 at 5:48
Andrés E. Caicedo
65.9k8160252
65.9k8160252
asked Dec 14 '18 at 5:11
user3865391user3865391
6111215
asked Dec 14 '18 at 5:11
user3865391user3865391
6111215
asked Dec 14 '18 at 5:11
user3865391user3865391
6111215
6111215
$begingroup$
Some easy solutions are $B=emptyset$ and (assuming the axiom of regularity) $B={A}$. Was there some other condition you wanted $B$ to satisfy?
$endgroup$
– bof
Dec 14 '18 at 5:53
$begingroup$
@bof Wait so $Acap {A}=emptyset$?
$endgroup$
– user3865391
Dec 14 '18 at 5:56
2
$begingroup$
The only element of ${A}$ is $A$, so $Acap{A}$ is empty unless $Ain A$. The "axiom of regularity" implies that $Ain A$ can't happen.
$endgroup$
– bof
Dec 14 '18 at 6:02
$begingroup$
Similar questions have been asked here, but usually they want a set $B$ which is disjoint from the given set $A$ and satisfies some other condition, for instance: Given a set $A$, construct a set $B$ such that $Acap B=emptyset$ and $|A|=|B|$.
$endgroup$
– bof
Dec 14 '18 at 6:05
$begingroup$
For instance, see this question: math.stackexchange.com/questions/2961610/…
$endgroup$
– bof
Dec 14 '18 at 6:06
|
show 2 more comments
$begingroup$
Some easy solutions are $B=emptyset$ and (assuming the axiom of regularity) $B={A}$. Was there some other condition you wanted $B$ to satisfy?
$endgroup$
– bof
Dec 14 '18 at 5:53
$begingroup$
@bof Wait so $Acap {A}=emptyset$?
$endgroup$
– user3865391
Dec 14 '18 at 5:56
2
$begingroup$
The only element of ${A}$ is $A$, so $Acap{A}$ is empty unless $Ain A$. The "axiom of regularity" implies that $Ain A$ can't happen.
$endgroup$
– bof
Dec 14 '18 at 6:02
$begingroup$
Similar questions have been asked here, but usually they want a set $B$ which is disjoint from the given set $A$ and satisfies some other condition, for instance: Given a set $A$, construct a set $B$ such that $Acap B=emptyset$ and $|A|=|B|$.
$endgroup$
– bof
Dec 14 '18 at 6:05
$begingroup$
For instance, see this question: math.stackexchange.com/questions/2961610/…
$endgroup$
– bof
Dec 14 '18 at 6:06
$begingroup$
Some easy solutions are $B=emptyset$ and (assuming the axiom of regularity) $B={A}$. Was there some other condition you wanted $B$ to satisfy?
$endgroup$
– bof
Dec 14 '18 at 5:53
$begingroup$
Some easy solutions are $B=emptyset$ and (assuming the axiom of regularity) $B={A}$. Was there some other condition you wanted $B$ to satisfy?
$endgroup$
– bof
Dec 14 '18 at 5:53
$begingroup$
@bof Wait so $Acap {A}=emptyset$?
$endgroup$
– user3865391
Dec 14 '18 at 5:56
$begingroup$
@bof Wait so $Acap {A}=emptyset$?
$endgroup$
– user3865391
Dec 14 '18 at 5:56
2
2
$begingroup$
The only element of ${A}$ is $A$, so $Acap{A}$ is empty unless $Ain A$. The "axiom of regularity" implies that $Ain A$ can't happen.
$endgroup$
– bof
Dec 14 '18 at 6:02
$begingroup$
The only element of ${A}$ is $A$, so $Acap{A}$ is empty unless $Ain A$. The "axiom of regularity" implies that $Ain A$ can't happen.
$endgroup$
– bof
Dec 14 '18 at 6:02
$begingroup$
Similar questions have been asked here, but usually they want a set $B$ which is disjoint from the given set $A$ and satisfies some other condition, for instance: Given a set $A$, construct a set $B$ such that $Acap B=emptyset$ and $|A|=|B|$.
$endgroup$
– bof
Dec 14 '18 at 6:05
$begingroup$
Similar questions have been asked here, but usually they want a set $B$ which is disjoint from the given set $A$ and satisfies some other condition, for instance: Given a set $A$, construct a set $B$ such that $Acap B=emptyset$ and $|A|=|B|$.
$endgroup$
– bof
Dec 14 '18 at 6:05
$begingroup$
For instance, see this question: math.stackexchange.com/questions/2961610/…
$endgroup$
– bof
Dec 14 '18 at 6:06
$begingroup$
For instance, see this question: math.stackexchange.com/questions/2961610/…
$endgroup$
– bof
Dec 14 '18 at 6:06
|
show 2 more comments
1 Answer
1
active
oldest
votes
0
$begingroup$
Your pick of options are from $mathcal{P}(A^c)$ (any subset of the complement of $A$).
EDIT: The disjoint union thing you are talking about is probably not something you need to worry about. It is a usually used as an almost artificial way to do a disjoint union when the sets are not actually disjoint. This is done mostly in more advanced mathematics, but based on the tags you've used I don't think this is what you need?
answered Dec 14 '18 at 5:50
SquirtleSquirtle
4,2181741
$endgroup$
$begingroup$
The complement isn't defined in terms of $A$ though, it requires an external set $Q$ and then we write $A^c=Qsetminus A$ for short hand. Is there some way I can change that?
$endgroup$
– user3865391
Dec 14 '18 at 5:54
$begingroup$
Can you please provide some more context. I'm completely lost what level of mathematics you are at. bof's comment suggests you're taking an advanced set theory course but I didn't think you were.
$endgroup$
– Squirtle
Dec 14 '18 at 5:57
$begingroup$
@Squirtle I don't see why the question would be out of place in a first course in set theory. What is so advanced about it?
$endgroup$
– bof
Dec 14 '18 at 6:09
$begingroup$
@bof It's not advanced. It's just not clear what the question is that is being asked.... plus a first course in set theory means different things to different people. It might help to know what textbook is being used and how much background the person has. Given that they are new and have a fairly low score, I just assume that "intro set theory" might as well be a course that introduces the notion of proofs. This is why I gave $mathcal{P}(A^c)$ as my solution, it seems like a perfectly rational answer in the absence of clarity.
$endgroup$
– Squirtle
Dec 14 '18 at 21:06
add a comment |
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
0
$begingroup$
Your pick of options are from $mathcal{P}(A^c)$ (any subset of the complement of $A$).
EDIT: The disjoint union thing you are talking about is probably not something you need to worry about. It is a usually used as an almost artificial way to do a disjoint union when the sets are not actually disjoint. This is done mostly in more advanced mathematics, but based on the tags you've used I don't think this is what you need?
answered Dec 14 '18 at 5:50
SquirtleSquirtle
4,2181741
$endgroup$
$begingroup$
The complement isn't defined in terms of $A$ though, it requires an external set $Q$ and then we write $A^c=Qsetminus A$ for short hand. Is there some way I can change that?
$endgroup$
– user3865391
Dec 14 '18 at 5:54
$begingroup$
Can you please provide some more context. I'm completely lost what level of mathematics you are at. bof's comment suggests you're taking an advanced set theory course but I didn't think you were.
$endgroup$
– Squirtle
Dec 14 '18 at 5:57
$begingroup$
@Squirtle I don't see why the question would be out of place in a first course in set theory. What is so advanced about it?
$endgroup$
– bof
Dec 14 '18 at 6:09
$begingroup$
@bof It's not advanced. It's just not clear what the question is that is being asked.... plus a first course in set theory means different things to different people. It might help to know what textbook is being used and how much background the person has. Given that they are new and have a fairly low score, I just assume that "intro set theory" might as well be a course that introduces the notion of proofs. This is why I gave $mathcal{P}(A^c)$ as my solution, it seems like a perfectly rational answer in the absence of clarity.
$endgroup$
– Squirtle
Dec 14 '18 at 21:06
add a comment |
0
$begingroup$
Your pick of options are from $mathcal{P}(A^c)$ (any subset of the complement of $A$).
EDIT: The disjoint union thing you are talking about is probably not something you need to worry about. It is a usually used as an almost artificial way to do a disjoint union when the sets are not actually disjoint. This is done mostly in more advanced mathematics, but based on the tags you've used I don't think this is what you need?
answered Dec 14 '18 at 5:50
SquirtleSquirtle
4,2181741
$endgroup$
$begingroup$
The complement isn't defined in terms of $A$ though, it requires an external set $Q$ and then we write $A^c=Qsetminus A$ for short hand. Is there some way I can change that?
$endgroup$
– user3865391
Dec 14 '18 at 5:54
$begingroup$
Can you please provide some more context. I'm completely lost what level of mathematics you are at. bof's comment suggests you're taking an advanced set theory course but I didn't think you were.
$endgroup$
– Squirtle
Dec 14 '18 at 5:57
$begingroup$
@Squirtle I don't see why the question would be out of place in a first course in set theory. What is so advanced about it?
$endgroup$
– bof
Dec 14 '18 at 6:09
$begingroup$
@bof It's not advanced. It's just not clear what the question is that is being asked.... plus a first course in set theory means different things to different people. It might help to know what textbook is being used and how much background the person has. Given that they are new and have a fairly low score, I just assume that "intro set theory" might as well be a course that introduces the notion of proofs. This is why I gave $mathcal{P}(A^c)$ as my solution, it seems like a perfectly rational answer in the absence of clarity.
$endgroup$
– Squirtle
Dec 14 '18 at 21:06
add a comment |
0
0
0
$begingroup$
Your pick of options are from $mathcal{P}(A^c)$ (any subset of the complement of $A$).
EDIT: The disjoint union thing you are talking about is probably not something you need to worry about. It is a usually used as an almost artificial way to do a disjoint union when the sets are not actually disjoint. This is done mostly in more advanced mathematics, but based on the tags you've used I don't think this is what you need?
answered Dec 14 '18 at 5:50
SquirtleSquirtle
4,2181741
$endgroup$
Your pick of options are from $mathcal{P}(A^c)$ (any subset of the complement of $A$).
EDIT: The disjoint union thing you are talking about is probably not something you need to worry about. It is a usually used as an almost artificial way to do a disjoint union when the sets are not actually disjoint. This is done mostly in more advanced mathematics, but based on the tags you've used I don't think this is what you need?
answered Dec 14 '18 at 5:50
SquirtleSquirtle
4,2181741
answered Dec 14 '18 at 5:50
SquirtleSquirtle
4,2181741
answered Dec 14 '18 at 5:50
SquirtleSquirtle
4,2181741
answered Dec 14 '18 at 5:50
SquirtleSquirtle
4,2181741
4,2181741
$begingroup$
The complement isn't defined in terms of $A$ though, it requires an external set $Q$ and then we write $A^c=Qsetminus A$ for short hand. Is there some way I can change that?
$endgroup$
– user3865391
Dec 14 '18 at 5:54
$begingroup$
Can you please provide some more context. I'm completely lost what level of mathematics you are at. bof's comment suggests you're taking an advanced set theory course but I didn't think you were.
$endgroup$
– Squirtle
Dec 14 '18 at 5:57
$begingroup$
@Squirtle I don't see why the question would be out of place in a first course in set theory. What is so advanced about it?
$endgroup$
– bof
Dec 14 '18 at 6:09
$begingroup$
@bof It's not advanced. It's just not clear what the question is that is being asked.... plus a first course in set theory means different things to different people. It might help to know what textbook is being used and how much background the person has. Given that they are new and have a fairly low score, I just assume that "intro set theory" might as well be a course that introduces the notion of proofs. This is why I gave $mathcal{P}(A^c)$ as my solution, it seems like a perfectly rational answer in the absence of clarity.
$endgroup$
– Squirtle
Dec 14 '18 at 21:06
add a comment |
$begingroup$
The complement isn't defined in terms of $A$ though, it requires an external set $Q$ and then we write $A^c=Qsetminus A$ for short hand. Is there some way I can change that?
$endgroup$
– user3865391
Dec 14 '18 at 5:54
$begingroup$
Can you please provide some more context. I'm completely lost what level of mathematics you are at. bof's comment suggests you're taking an advanced set theory course but I didn't think you were.
$endgroup$
– Squirtle
Dec 14 '18 at 5:57
$begingroup$
@Squirtle I don't see why the question would be out of place in a first course in set theory. What is so advanced about it?
$endgroup$
– bof
Dec 14 '18 at 6:09
$begingroup$
@bof It's not advanced. It's just not clear what the question is that is being asked.... plus a first course in set theory means different things to different people. It might help to know what textbook is being used and how much background the person has. Given that they are new and have a fairly low score, I just assume that "intro set theory" might as well be a course that introduces the notion of proofs. This is why I gave $mathcal{P}(A^c)$ as my solution, it seems like a perfectly rational answer in the absence of clarity.
$endgroup$
– Squirtle
Dec 14 '18 at 21:06
$begingroup$
The complement isn't defined in terms of $A$ though, it requires an external set $Q$ and then we write $A^c=Qsetminus A$ for short hand. Is there some way I can change that?
$endgroup$
– user3865391
Dec 14 '18 at 5:54
$begingroup$
The complement isn't defined in terms of $A$ though, it requires an external set $Q$ and then we write $A^c=Qsetminus A$ for short hand. Is there some way I can change that?
$endgroup$
– user3865391
Dec 14 '18 at 5:54
$begingroup$
Can you please provide some more context. I'm completely lost what level of mathematics you are at. bof's comment suggests you're taking an advanced set theory course but I didn't think you were.
$endgroup$
– Squirtle
Dec 14 '18 at 5:57
$begingroup$
Can you please provide some more context. I'm completely lost what level of mathematics you are at. bof's comment suggests you're taking an advanced set theory course but I didn't think you were.
$endgroup$
– Squirtle
Dec 14 '18 at 5:57
$begingroup$
@Squirtle I don't see why the question would be out of place in a first course in set theory. What is so advanced about it?
$endgroup$
– bof
Dec 14 '18 at 6:09
$begingroup$
@Squirtle I don't see why the question would be out of place in a first course in set theory. What is so advanced about it?
$endgroup$
– bof
Dec 14 '18 at 6:09
$begingroup$
@bof It's not advanced. It's just not clear what the question is that is being asked.... plus a first course in set theory means different things to different people. It might help to know what textbook is being used and how much background the person has. Given that they are new and have a fairly low score, I just assume that "intro set theory" might as well be a course that introduces the notion of proofs. This is why I gave $mathcal{P}(A^c)$ as my solution, it seems like a perfectly rational answer in the absence of clarity.
$endgroup$
– Squirtle
Dec 14 '18 at 21:06
$begingroup$
@bof It's not advanced. It's just not clear what the question is that is being asked.... plus a first course in set theory means different things to different people. It might help to know what textbook is being used and how much background the person has. Given that they are new and have a fairly low score, I just assume that "intro set theory" might as well be a course that introduces the notion of proofs. This is why I gave $mathcal{P}(A^c)$ as my solution, it seems like a perfectly rational answer in the absence of clarity.
$endgroup$
– Squirtle
Dec 14 '18 at 21:06
add a comment |
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$begingroup$
Some easy solutions are $B=emptyset$ and (assuming the axiom of regularity) $B={A}$. Was there some other condition you wanted $B$ to satisfy?
$endgroup$
– bof
Dec 14 '18 at 5:53
$begingroup$
@bof Wait so $Acap {A}=emptyset$?
$endgroup$
– user3865391
Dec 14 '18 at 5:56
2
$begingroup$
The only element of ${A}$ is $A$, so $Acap{A}$ is empty unless $Ain A$. The "axiom of regularity" implies that $Ain A$ can't happen.
$endgroup$
– bof
Dec 14 '18 at 6:02
$begingroup$
Similar questions have been asked here, but usually they want a set $B$ which is disjoint from the given set $A$ and satisfies some other condition, for instance: Given a set $A$, construct a set $B$ such that $Acap B=emptyset$ and $|A|=|B|$.
$endgroup$
– bof
Dec 14 '18 at 6:05
$begingroup$
For instance, see this question: math.stackexchange.com/questions/2961610/…
$endgroup$
– bof
Dec 14 '18 at 6:06