Why do we say 'Pairwise Disjoint', rather than 'Disjoint'?If $A$ is infinite then it has two infinite subsets $B, C$ which are pairwise disjoint.Pairwise disjoint proofDistinction between the notions of pairwise disjointPairwise and Mutually disjoint setsIs the empty family of sets pairwise disjoint?Do Kolmogorov's axioms really need only disjointness rather than pairwise disjointness?Proof containing pairwise disjoint setsA set whose power set is pairwise disjoint?pairwise disjoint , or disjointProve that sets are pairwise disjoint
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Why do we say 'Pairwise Disjoint', rather than 'Disjoint'?
If $A$ is infinite then it has two infinite subsets $B, C$ which are pairwise disjoint.Pairwise disjoint proofDistinction between the notions of pairwise disjointPairwise and Mutually disjoint setsIs the empty family of sets pairwise disjoint?Do Kolmogorov's axioms really need only disjointness rather than pairwise disjointness?Proof containing pairwise disjoint setsA set whose power set is pairwise disjoint?pairwise disjoint , or disjointProve that sets are pairwise disjoint
$begingroup$
I don't see the ambiguity that 'Pairwise' resolves.
Surely if A,B,C are disjoint sets then they are pairwise disjoint and vice versa?
Or am I being dim?
elementary-set-theory
asked 10 hours ago
John Lawrence AspdenJohn Lawrence Aspden
31818
$endgroup$
add a comment |
$begingroup$
I don't see the ambiguity that 'Pairwise' resolves.
Surely if A,B,C are disjoint sets then they are pairwise disjoint and vice versa?
Or am I being dim?
elementary-set-theory
asked 10 hours ago
John Lawrence AspdenJohn Lawrence Aspden
31818
$endgroup$
add a comment |
$begingroup$
I don't see the ambiguity that 'Pairwise' resolves.
Surely if A,B,C are disjoint sets then they are pairwise disjoint and vice versa?
Or am I being dim?
elementary-set-theory
asked 10 hours ago
John Lawrence AspdenJohn Lawrence Aspden
31818
$endgroup$
I don't see the ambiguity that 'Pairwise' resolves.
Surely if A,B,C are disjoint sets then they are pairwise disjoint and vice versa?
Or am I being dim?
elementary-set-theory
elementary-set-theory
asked 10 hours ago
John Lawrence AspdenJohn Lawrence Aspden
31818
asked 10 hours ago
John Lawrence AspdenJohn Lawrence Aspden
31818
asked 10 hours ago
John Lawrence AspdenJohn Lawrence Aspden
31818
asked 10 hours ago
John Lawrence AspdenJohn Lawrence Aspden
31818
asked 10 hours ago
John Lawrence AspdenJohn Lawrence Aspden
31818
31818
add a comment |
add a comment |
7 Answers
7
active
oldest
votes
$begingroup$
$1,2,2,3,1,3$ are disjoint but not pairwise disjoint.
answered 10 hours ago
saulspatzsaulspatz
16.6k31333
$endgroup$
5
$begingroup$
Really? Who would call those disjoint sets?
$endgroup$
– John Lawrence Aspden
10 hours ago
12
$begingroup$
Everyone. Disjoint means their intersection is empty.
$endgroup$
– saulspatz
10 hours ago
7
$begingroup$
If sets $A_1,A_2,dots,A_n$ are said to be disjoint then usually it is meant that the sets are pairwise disjoint. See Wolfram for instance. In the other case one says simply that the sets have an empty intersection.
$endgroup$
– drhab
10 hours ago
9
$begingroup$
So be it. But actually I am the living proof that "everyone" in your first comment is not correct. Lots of times I heard "them" say things like: "we have $mu(bigcup_n=1^inftyA_n)=sum_n=1^inftymu(A_n)$ if the sets are disjoint" (so leaving out pairwise). I do that myself too. It is a good thing however to have a proper definition of this excluding ambiguities. So let's look at this as abuse of language.
$endgroup$
– drhab
9 hours ago
7
$begingroup$
These sets are not disjoint, at least not by the modern usage of the word “disjoint,” which afaik is used interchangeably with “pairwise disjoint.”
$endgroup$
– Shalop
8 hours ago
|
show 8 more comments
$begingroup$
As evidenced by the answers and comments on this page, the term "disjoint" is ambiguous - some use it to mean "pairwise disjoint", others use it to mean "empty intersection".
Thus, for the sake of clarity, I'd recommend avoiding "disjoint" and using "pairwise disjoint".
answered 5 hours ago
BlueRaja - Danny PflughoeftBlueRaja - Danny Pflughoeft
5,61532843
$endgroup$
add a comment |
$begingroup$
In this context disjoint means $A cap B cap C = emptyset$.
answered 10 hours ago
Umberto P.Umberto P.
39.8k13267
$endgroup$
7
$begingroup$
Is that a standard meaning? I never saw the term formally defined that way in four years of undergrad math classes.
$endgroup$
– Connor Harris
10 hours ago
$begingroup$
me neither, but four people have answered the question this way in four minutes!
$endgroup$
– John Lawrence Aspden
10 hours ago
$begingroup$
If you define "disjoint" to mean "empty intersection" (which is the standard definition) then formally for a family of sets "disjoint" would mean the intersection of the entire family is empty unless stated otherwise. The use of the pleonastic term "pairwise" helps to avoid confusion.
$endgroup$
– Umberto P.
10 hours ago
$begingroup$
This is not the standard definition of disjoint. Google it if you like.
$endgroup$
– Marc van Leeuwen
6 hours ago
$begingroup$
@MarcvanLeeuwen Be warned that Google likes to tell people what they want to hear. If you use it as support in an argument, there is a chance that Google will tell your opponent that he's right, while still telling you that you're right.
$endgroup$
– Brilliand
3 hours ago
add a comment |
$begingroup$
The word "pairwise" in "pairwise disjoint" is superfluous: a collection of sets is disjoint if no element appears in more than one of the sets at a time, and this means that every pair of distinct sets in the collection has an empty intersection. However including the "pairwise" emphasizes that the property can be checked at the level of pairs from the collection (unlike for instance linear independence of vectors in linear algebra). A "disjoint union" is a union of pairwise disjoint sets; one does not say "pairwise disjoint union".
answered 6 hours ago
Marc van LeeuwenMarc van Leeuwen
88k5111226
$endgroup$
add a comment |
$begingroup$
More generally, sets are disjoint when their intersection is empty, but pairwise disjoint when any two of them are disjoint.
answered 10 hours ago
J.G.J.G.
29.1k22845
$endgroup$
1
$begingroup$
I have never seen this sense of disjoint for more than two sets been defined or used in mathematics. The Wikipedia article for disjoint sets does not mention it, and the notion of "disjoint union" definitely refers to pairwise disjoint sets.
$endgroup$
– Marc van Leeuwen
6 hours ago
$begingroup$
@MarcvanLeeuwen Three-or-more sets that are disjoint-but-not-pairwise-disjoint in the sense I describe probably warrant some name, but as you note it's not usually the nomenclature I mentioned, probably because it makes "pairwise" compulsory more than anyone wants to write it.
$endgroup$
– J.G.
6 hours ago
add a comment |
$begingroup$
Consider the sets $A = 1,2$, $B = 2,3$, $C = 3, 1$. Then $Acap Bcap C = varnothing$, but $A,B,C$ are not pairwise disjoint.
answered 10 hours ago
Kyle DuffyKyle Duffy
112
New contributor
Kyle Duffy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
Let $A=1,2, B=2,3,C=3,4$. Then the sets are disjoint because $Acap Bcap C=emptyset$, but not pairwise disjoint because you have pairs such as $A,B$ such that $Acap Bnot =emptyset$.
edited 10 hours ago
J.G.
29.1k22845
answered 10 hours ago
Oscar LanziOscar Lanzi
13k12136
$endgroup$
$begingroup$
Rats! What is the notation for an empty set?
$endgroup$
– Oscar Lanzi
10 hours ago
$begingroup$
Thank you, @jg.
$endgroup$
– Oscar Lanzi
10 hours ago
add a comment |
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7 Answers
7
active
oldest
votes
7 Answers
7
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
$1,2,2,3,1,3$ are disjoint but not pairwise disjoint.
answered 10 hours ago
saulspatzsaulspatz
16.6k31333
$endgroup$
5
$begingroup$
Really? Who would call those disjoint sets?
$endgroup$
– John Lawrence Aspden
10 hours ago
12
$begingroup$
Everyone. Disjoint means their intersection is empty.
$endgroup$
– saulspatz
10 hours ago
7
$begingroup$
If sets $A_1,A_2,dots,A_n$ are said to be disjoint then usually it is meant that the sets are pairwise disjoint. See Wolfram for instance. In the other case one says simply that the sets have an empty intersection.
$endgroup$
– drhab
10 hours ago
9
$begingroup$
So be it. But actually I am the living proof that "everyone" in your first comment is not correct. Lots of times I heard "them" say things like: "we have $mu(bigcup_n=1^inftyA_n)=sum_n=1^inftymu(A_n)$ if the sets are disjoint" (so leaving out pairwise). I do that myself too. It is a good thing however to have a proper definition of this excluding ambiguities. So let's look at this as abuse of language.
$endgroup$
– drhab
9 hours ago
7
$begingroup$
These sets are not disjoint, at least not by the modern usage of the word “disjoint,” which afaik is used interchangeably with “pairwise disjoint.”
$endgroup$
– Shalop
8 hours ago
|
show 8 more comments
$begingroup$
$1,2,2,3,1,3$ are disjoint but not pairwise disjoint.
answered 10 hours ago
saulspatzsaulspatz
16.6k31333
$endgroup$
5
$begingroup$
Really? Who would call those disjoint sets?
$endgroup$
– John Lawrence Aspden
10 hours ago
12
$begingroup$
Everyone. Disjoint means their intersection is empty.
$endgroup$
– saulspatz
10 hours ago
7
$begingroup$
If sets $A_1,A_2,dots,A_n$ are said to be disjoint then usually it is meant that the sets are pairwise disjoint. See Wolfram for instance. In the other case one says simply that the sets have an empty intersection.
$endgroup$
– drhab
10 hours ago
9
$begingroup$
So be it. But actually I am the living proof that "everyone" in your first comment is not correct. Lots of times I heard "them" say things like: "we have $mu(bigcup_n=1^inftyA_n)=sum_n=1^inftymu(A_n)$ if the sets are disjoint" (so leaving out pairwise). I do that myself too. It is a good thing however to have a proper definition of this excluding ambiguities. So let's look at this as abuse of language.
$endgroup$
– drhab
9 hours ago
7
$begingroup$
These sets are not disjoint, at least not by the modern usage of the word “disjoint,” which afaik is used interchangeably with “pairwise disjoint.”
$endgroup$
– Shalop
8 hours ago
|
show 8 more comments
$begingroup$
$1,2,2,3,1,3$ are disjoint but not pairwise disjoint.
answered 10 hours ago
saulspatzsaulspatz
16.6k31333
$endgroup$
$1,2,2,3,1,3$ are disjoint but not pairwise disjoint.
answered 10 hours ago
saulspatzsaulspatz
16.6k31333
answered 10 hours ago
saulspatzsaulspatz
16.6k31333
answered 10 hours ago
saulspatzsaulspatz
16.6k31333
answered 10 hours ago
saulspatzsaulspatz
16.6k31333
16.6k31333
5
$begingroup$
Really? Who would call those disjoint sets?
$endgroup$
– John Lawrence Aspden
10 hours ago
12
$begingroup$
Everyone. Disjoint means their intersection is empty.
$endgroup$
– saulspatz
10 hours ago
7
$begingroup$
If sets $A_1,A_2,dots,A_n$ are said to be disjoint then usually it is meant that the sets are pairwise disjoint. See Wolfram for instance. In the other case one says simply that the sets have an empty intersection.
$endgroup$
– drhab
10 hours ago
9
$begingroup$
So be it. But actually I am the living proof that "everyone" in your first comment is not correct. Lots of times I heard "them" say things like: "we have $mu(bigcup_n=1^inftyA_n)=sum_n=1^inftymu(A_n)$ if the sets are disjoint" (so leaving out pairwise). I do that myself too. It is a good thing however to have a proper definition of this excluding ambiguities. So let's look at this as abuse of language.
$endgroup$
– drhab
9 hours ago
7
$begingroup$
These sets are not disjoint, at least not by the modern usage of the word “disjoint,” which afaik is used interchangeably with “pairwise disjoint.”
$endgroup$
– Shalop
8 hours ago
|
show 8 more comments
5
$begingroup$
Really? Who would call those disjoint sets?
$endgroup$
– John Lawrence Aspden
10 hours ago
12
$begingroup$
Everyone. Disjoint means their intersection is empty.
$endgroup$
– saulspatz
10 hours ago
7
$begingroup$
If sets $A_1,A_2,dots,A_n$ are said to be disjoint then usually it is meant that the sets are pairwise disjoint. See Wolfram for instance. In the other case one says simply that the sets have an empty intersection.
$endgroup$
– drhab
10 hours ago
9
$begingroup$
So be it. But actually I am the living proof that "everyone" in your first comment is not correct. Lots of times I heard "them" say things like: "we have $mu(bigcup_n=1^inftyA_n)=sum_n=1^inftymu(A_n)$ if the sets are disjoint" (so leaving out pairwise). I do that myself too. It is a good thing however to have a proper definition of this excluding ambiguities. So let's look at this as abuse of language.
$endgroup$
– drhab
9 hours ago
7
$begingroup$
These sets are not disjoint, at least not by the modern usage of the word “disjoint,” which afaik is used interchangeably with “pairwise disjoint.”
$endgroup$
– Shalop
8 hours ago
5
5
$begingroup$
Really? Who would call those disjoint sets?
$endgroup$
– John Lawrence Aspden
10 hours ago
$begingroup$
Really? Who would call those disjoint sets?
$endgroup$
– John Lawrence Aspden
10 hours ago
12
12
$begingroup$
Everyone. Disjoint means their intersection is empty.
$endgroup$
– saulspatz
10 hours ago
$begingroup$
Everyone. Disjoint means their intersection is empty.
$endgroup$
– saulspatz
10 hours ago
7
7
$begingroup$
If sets $A_1,A_2,dots,A_n$ are said to be disjoint then usually it is meant that the sets are pairwise disjoint. See Wolfram for instance. In the other case one says simply that the sets have an empty intersection.
$endgroup$
– drhab
10 hours ago
$begingroup$
If sets $A_1,A_2,dots,A_n$ are said to be disjoint then usually it is meant that the sets are pairwise disjoint. See Wolfram for instance. In the other case one says simply that the sets have an empty intersection.
$endgroup$
– drhab
10 hours ago
9
9
$begingroup$
So be it. But actually I am the living proof that "everyone" in your first comment is not correct. Lots of times I heard "them" say things like: "we have $mu(bigcup_n=1^inftyA_n)=sum_n=1^inftymu(A_n)$ if the sets are disjoint" (so leaving out pairwise). I do that myself too. It is a good thing however to have a proper definition of this excluding ambiguities. So let's look at this as abuse of language.
$endgroup$
– drhab
9 hours ago
$begingroup$
So be it. But actually I am the living proof that "everyone" in your first comment is not correct. Lots of times I heard "them" say things like: "we have $mu(bigcup_n=1^inftyA_n)=sum_n=1^inftymu(A_n)$ if the sets are disjoint" (so leaving out pairwise). I do that myself too. It is a good thing however to have a proper definition of this excluding ambiguities. So let's look at this as abuse of language.
$endgroup$
– drhab
9 hours ago
7
7
$begingroup$
These sets are not disjoint, at least not by the modern usage of the word “disjoint,” which afaik is used interchangeably with “pairwise disjoint.”
$endgroup$
– Shalop
8 hours ago
$begingroup$
These sets are not disjoint, at least not by the modern usage of the word “disjoint,” which afaik is used interchangeably with “pairwise disjoint.”
$endgroup$
– Shalop
8 hours ago
|
show 8 more comments
$begingroup$
As evidenced by the answers and comments on this page, the term "disjoint" is ambiguous - some use it to mean "pairwise disjoint", others use it to mean "empty intersection".
Thus, for the sake of clarity, I'd recommend avoiding "disjoint" and using "pairwise disjoint".
answered 5 hours ago
BlueRaja - Danny PflughoeftBlueRaja - Danny Pflughoeft
5,61532843
$endgroup$
add a comment |
$begingroup$
As evidenced by the answers and comments on this page, the term "disjoint" is ambiguous - some use it to mean "pairwise disjoint", others use it to mean "empty intersection".
Thus, for the sake of clarity, I'd recommend avoiding "disjoint" and using "pairwise disjoint".
answered 5 hours ago
BlueRaja - Danny PflughoeftBlueRaja - Danny Pflughoeft
5,61532843
$endgroup$
add a comment |
$begingroup$
As evidenced by the answers and comments on this page, the term "disjoint" is ambiguous - some use it to mean "pairwise disjoint", others use it to mean "empty intersection".
Thus, for the sake of clarity, I'd recommend avoiding "disjoint" and using "pairwise disjoint".
answered 5 hours ago
BlueRaja - Danny PflughoeftBlueRaja - Danny Pflughoeft
5,61532843
$endgroup$
As evidenced by the answers and comments on this page, the term "disjoint" is ambiguous - some use it to mean "pairwise disjoint", others use it to mean "empty intersection".
Thus, for the sake of clarity, I'd recommend avoiding "disjoint" and using "pairwise disjoint".
answered 5 hours ago
BlueRaja - Danny PflughoeftBlueRaja - Danny Pflughoeft
5,61532843
answered 5 hours ago
BlueRaja - Danny PflughoeftBlueRaja - Danny Pflughoeft
5,61532843
answered 5 hours ago
BlueRaja - Danny PflughoeftBlueRaja - Danny Pflughoeft
5,61532843
answered 5 hours ago
BlueRaja - Danny PflughoeftBlueRaja - Danny Pflughoeft
5,61532843
5,61532843
add a comment |
add a comment |
$begingroup$
In this context disjoint means $A cap B cap C = emptyset$.
answered 10 hours ago
Umberto P.Umberto P.
39.8k13267
$endgroup$
7
$begingroup$
Is that a standard meaning? I never saw the term formally defined that way in four years of undergrad math classes.
$endgroup$
– Connor Harris
10 hours ago
$begingroup$
me neither, but four people have answered the question this way in four minutes!
$endgroup$
– John Lawrence Aspden
10 hours ago
$begingroup$
If you define "disjoint" to mean "empty intersection" (which is the standard definition) then formally for a family of sets "disjoint" would mean the intersection of the entire family is empty unless stated otherwise. The use of the pleonastic term "pairwise" helps to avoid confusion.
$endgroup$
– Umberto P.
10 hours ago
$begingroup$
This is not the standard definition of disjoint. Google it if you like.
$endgroup$
– Marc van Leeuwen
6 hours ago
$begingroup$
@MarcvanLeeuwen Be warned that Google likes to tell people what they want to hear. If you use it as support in an argument, there is a chance that Google will tell your opponent that he's right, while still telling you that you're right.
$endgroup$
– Brilliand
3 hours ago
add a comment |
$begingroup$
In this context disjoint means $A cap B cap C = emptyset$.
answered 10 hours ago
Umberto P.Umberto P.
39.8k13267
$endgroup$
7
$begingroup$
Is that a standard meaning? I never saw the term formally defined that way in four years of undergrad math classes.
$endgroup$
– Connor Harris
10 hours ago
$begingroup$
me neither, but four people have answered the question this way in four minutes!
$endgroup$
– John Lawrence Aspden
10 hours ago
$begingroup$
If you define "disjoint" to mean "empty intersection" (which is the standard definition) then formally for a family of sets "disjoint" would mean the intersection of the entire family is empty unless stated otherwise. The use of the pleonastic term "pairwise" helps to avoid confusion.
$endgroup$
– Umberto P.
10 hours ago
$begingroup$
This is not the standard definition of disjoint. Google it if you like.
$endgroup$
– Marc van Leeuwen
6 hours ago
$begingroup$
@MarcvanLeeuwen Be warned that Google likes to tell people what they want to hear. If you use it as support in an argument, there is a chance that Google will tell your opponent that he's right, while still telling you that you're right.
$endgroup$
– Brilliand
3 hours ago
add a comment |
$begingroup$
In this context disjoint means $A cap B cap C = emptyset$.
answered 10 hours ago
Umberto P.Umberto P.
39.8k13267
$endgroup$
In this context disjoint means $A cap B cap C = emptyset$.
answered 10 hours ago
Umberto P.Umberto P.
39.8k13267
answered 10 hours ago
Umberto P.Umberto P.
39.8k13267
answered 10 hours ago
Umberto P.Umberto P.
39.8k13267
answered 10 hours ago
Umberto P.Umberto P.
39.8k13267
39.8k13267
7
$begingroup$
Is that a standard meaning? I never saw the term formally defined that way in four years of undergrad math classes.
$endgroup$
– Connor Harris
10 hours ago
$begingroup$
me neither, but four people have answered the question this way in four minutes!
$endgroup$
– John Lawrence Aspden
10 hours ago
$begingroup$
If you define "disjoint" to mean "empty intersection" (which is the standard definition) then formally for a family of sets "disjoint" would mean the intersection of the entire family is empty unless stated otherwise. The use of the pleonastic term "pairwise" helps to avoid confusion.
$endgroup$
– Umberto P.
10 hours ago
$begingroup$
This is not the standard definition of disjoint. Google it if you like.
$endgroup$
– Marc van Leeuwen
6 hours ago
$begingroup$
@MarcvanLeeuwen Be warned that Google likes to tell people what they want to hear. If you use it as support in an argument, there is a chance that Google will tell your opponent that he's right, while still telling you that you're right.
$endgroup$
– Brilliand
3 hours ago
add a comment |
7
$begingroup$
Is that a standard meaning? I never saw the term formally defined that way in four years of undergrad math classes.
$endgroup$
– Connor Harris
10 hours ago
$begingroup$
me neither, but four people have answered the question this way in four minutes!
$endgroup$
– John Lawrence Aspden
10 hours ago
$begingroup$
If you define "disjoint" to mean "empty intersection" (which is the standard definition) then formally for a family of sets "disjoint" would mean the intersection of the entire family is empty unless stated otherwise. The use of the pleonastic term "pairwise" helps to avoid confusion.
$endgroup$
– Umberto P.
10 hours ago
$begingroup$
This is not the standard definition of disjoint. Google it if you like.
$endgroup$
– Marc van Leeuwen
6 hours ago
$begingroup$
@MarcvanLeeuwen Be warned that Google likes to tell people what they want to hear. If you use it as support in an argument, there is a chance that Google will tell your opponent that he's right, while still telling you that you're right.
$endgroup$
– Brilliand
3 hours ago
7
7
$begingroup$
Is that a standard meaning? I never saw the term formally defined that way in four years of undergrad math classes.
$endgroup$
– Connor Harris
10 hours ago
$begingroup$
Is that a standard meaning? I never saw the term formally defined that way in four years of undergrad math classes.
$endgroup$
– Connor Harris
10 hours ago
$begingroup$
me neither, but four people have answered the question this way in four minutes!
$endgroup$
– John Lawrence Aspden
10 hours ago
$begingroup$
me neither, but four people have answered the question this way in four minutes!
$endgroup$
– John Lawrence Aspden
10 hours ago
$begingroup$
If you define "disjoint" to mean "empty intersection" (which is the standard definition) then formally for a family of sets "disjoint" would mean the intersection of the entire family is empty unless stated otherwise. The use of the pleonastic term "pairwise" helps to avoid confusion.
$endgroup$
– Umberto P.
10 hours ago
$begingroup$
If you define "disjoint" to mean "empty intersection" (which is the standard definition) then formally for a family of sets "disjoint" would mean the intersection of the entire family is empty unless stated otherwise. The use of the pleonastic term "pairwise" helps to avoid confusion.
$endgroup$
– Umberto P.
10 hours ago
$begingroup$
This is not the standard definition of disjoint. Google it if you like.
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– Marc van Leeuwen
6 hours ago
$begingroup$
This is not the standard definition of disjoint. Google it if you like.
$endgroup$
– Marc van Leeuwen
6 hours ago
$begingroup$
@MarcvanLeeuwen Be warned that Google likes to tell people what they want to hear. If you use it as support in an argument, there is a chance that Google will tell your opponent that he's right, while still telling you that you're right.
$endgroup$
– Brilliand
3 hours ago
$begingroup$
@MarcvanLeeuwen Be warned that Google likes to tell people what they want to hear. If you use it as support in an argument, there is a chance that Google will tell your opponent that he's right, while still telling you that you're right.
$endgroup$
– Brilliand
3 hours ago
add a comment |
$begingroup$
The word "pairwise" in "pairwise disjoint" is superfluous: a collection of sets is disjoint if no element appears in more than one of the sets at a time, and this means that every pair of distinct sets in the collection has an empty intersection. However including the "pairwise" emphasizes that the property can be checked at the level of pairs from the collection (unlike for instance linear independence of vectors in linear algebra). A "disjoint union" is a union of pairwise disjoint sets; one does not say "pairwise disjoint union".
answered 6 hours ago
Marc van LeeuwenMarc van Leeuwen
88k5111226
$endgroup$
add a comment |
$begingroup$
The word "pairwise" in "pairwise disjoint" is superfluous: a collection of sets is disjoint if no element appears in more than one of the sets at a time, and this means that every pair of distinct sets in the collection has an empty intersection. However including the "pairwise" emphasizes that the property can be checked at the level of pairs from the collection (unlike for instance linear independence of vectors in linear algebra). A "disjoint union" is a union of pairwise disjoint sets; one does not say "pairwise disjoint union".
answered 6 hours ago
Marc van LeeuwenMarc van Leeuwen
88k5111226
$endgroup$
add a comment |
$begingroup$
The word "pairwise" in "pairwise disjoint" is superfluous: a collection of sets is disjoint if no element appears in more than one of the sets at a time, and this means that every pair of distinct sets in the collection has an empty intersection. However including the "pairwise" emphasizes that the property can be checked at the level of pairs from the collection (unlike for instance linear independence of vectors in linear algebra). A "disjoint union" is a union of pairwise disjoint sets; one does not say "pairwise disjoint union".
answered 6 hours ago
Marc van LeeuwenMarc van Leeuwen
88k5111226
$endgroup$
The word "pairwise" in "pairwise disjoint" is superfluous: a collection of sets is disjoint if no element appears in more than one of the sets at a time, and this means that every pair of distinct sets in the collection has an empty intersection. However including the "pairwise" emphasizes that the property can be checked at the level of pairs from the collection (unlike for instance linear independence of vectors in linear algebra). A "disjoint union" is a union of pairwise disjoint sets; one does not say "pairwise disjoint union".
answered 6 hours ago
Marc van LeeuwenMarc van Leeuwen
88k5111226
answered 6 hours ago
Marc van LeeuwenMarc van Leeuwen
88k5111226
answered 6 hours ago
Marc van LeeuwenMarc van Leeuwen
88k5111226
answered 6 hours ago
Marc van LeeuwenMarc van Leeuwen
88k5111226
88k5111226
add a comment |
add a comment |
$begingroup$
More generally, sets are disjoint when their intersection is empty, but pairwise disjoint when any two of them are disjoint.
answered 10 hours ago
J.G.J.G.
29.1k22845
$endgroup$
1
$begingroup$
I have never seen this sense of disjoint for more than two sets been defined or used in mathematics. The Wikipedia article for disjoint sets does not mention it, and the notion of "disjoint union" definitely refers to pairwise disjoint sets.
$endgroup$
– Marc van Leeuwen
6 hours ago
$begingroup$
@MarcvanLeeuwen Three-or-more sets that are disjoint-but-not-pairwise-disjoint in the sense I describe probably warrant some name, but as you note it's not usually the nomenclature I mentioned, probably because it makes "pairwise" compulsory more than anyone wants to write it.
$endgroup$
– J.G.
6 hours ago
add a comment |
$begingroup$
More generally, sets are disjoint when their intersection is empty, but pairwise disjoint when any two of them are disjoint.
answered 10 hours ago
J.G.J.G.
29.1k22845
$endgroup$
1
$begingroup$
I have never seen this sense of disjoint for more than two sets been defined or used in mathematics. The Wikipedia article for disjoint sets does not mention it, and the notion of "disjoint union" definitely refers to pairwise disjoint sets.
$endgroup$
– Marc van Leeuwen
6 hours ago
$begingroup$
@MarcvanLeeuwen Three-or-more sets that are disjoint-but-not-pairwise-disjoint in the sense I describe probably warrant some name, but as you note it's not usually the nomenclature I mentioned, probably because it makes "pairwise" compulsory more than anyone wants to write it.
$endgroup$
– J.G.
6 hours ago
add a comment |
$begingroup$
More generally, sets are disjoint when their intersection is empty, but pairwise disjoint when any two of them are disjoint.
answered 10 hours ago
J.G.J.G.
29.1k22845
$endgroup$
More generally, sets are disjoint when their intersection is empty, but pairwise disjoint when any two of them are disjoint.
answered 10 hours ago
J.G.J.G.
29.1k22845
answered 10 hours ago
J.G.J.G.
29.1k22845
answered 10 hours ago
J.G.J.G.
29.1k22845
answered 10 hours ago
J.G.J.G.
29.1k22845
29.1k22845
1
$begingroup$
I have never seen this sense of disjoint for more than two sets been defined or used in mathematics. The Wikipedia article for disjoint sets does not mention it, and the notion of "disjoint union" definitely refers to pairwise disjoint sets.
$endgroup$
– Marc van Leeuwen
6 hours ago
$begingroup$
@MarcvanLeeuwen Three-or-more sets that are disjoint-but-not-pairwise-disjoint in the sense I describe probably warrant some name, but as you note it's not usually the nomenclature I mentioned, probably because it makes "pairwise" compulsory more than anyone wants to write it.
$endgroup$
– J.G.
6 hours ago
add a comment |
1
$begingroup$
I have never seen this sense of disjoint for more than two sets been defined or used in mathematics. The Wikipedia article for disjoint sets does not mention it, and the notion of "disjoint union" definitely refers to pairwise disjoint sets.
$endgroup$
– Marc van Leeuwen
6 hours ago
$begingroup$
@MarcvanLeeuwen Three-or-more sets that are disjoint-but-not-pairwise-disjoint in the sense I describe probably warrant some name, but as you note it's not usually the nomenclature I mentioned, probably because it makes "pairwise" compulsory more than anyone wants to write it.
$endgroup$
– J.G.
6 hours ago
1
1
$begingroup$
I have never seen this sense of disjoint for more than two sets been defined or used in mathematics. The Wikipedia article for disjoint sets does not mention it, and the notion of "disjoint union" definitely refers to pairwise disjoint sets.
$endgroup$
– Marc van Leeuwen
6 hours ago
$begingroup$
I have never seen this sense of disjoint for more than two sets been defined or used in mathematics. The Wikipedia article for disjoint sets does not mention it, and the notion of "disjoint union" definitely refers to pairwise disjoint sets.
$endgroup$
– Marc van Leeuwen
6 hours ago
$begingroup$
@MarcvanLeeuwen Three-or-more sets that are disjoint-but-not-pairwise-disjoint in the sense I describe probably warrant some name, but as you note it's not usually the nomenclature I mentioned, probably because it makes "pairwise" compulsory more than anyone wants to write it.
$endgroup$
– J.G.
6 hours ago
$begingroup$
@MarcvanLeeuwen Three-or-more sets that are disjoint-but-not-pairwise-disjoint in the sense I describe probably warrant some name, but as you note it's not usually the nomenclature I mentioned, probably because it makes "pairwise" compulsory more than anyone wants to write it.
$endgroup$
– J.G.
6 hours ago
add a comment |
$begingroup$
Consider the sets $A = 1,2$, $B = 2,3$, $C = 3, 1$. Then $Acap Bcap C = varnothing$, but $A,B,C$ are not pairwise disjoint.
answered 10 hours ago
Kyle DuffyKyle Duffy
112
New contributor
Kyle Duffy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
Consider the sets $A = 1,2$, $B = 2,3$, $C = 3, 1$. Then $Acap Bcap C = varnothing$, but $A,B,C$ are not pairwise disjoint.
answered 10 hours ago
Kyle DuffyKyle Duffy
112
New contributor
Kyle Duffy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
Consider the sets $A = 1,2$, $B = 2,3$, $C = 3, 1$. Then $Acap Bcap C = varnothing$, but $A,B,C$ are not pairwise disjoint.
answered 10 hours ago
Kyle DuffyKyle Duffy
112
New contributor
Kyle Duffy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
Consider the sets $A = 1,2$, $B = 2,3$, $C = 3, 1$. Then $Acap Bcap C = varnothing$, but $A,B,C$ are not pairwise disjoint.
answered 10 hours ago
Kyle DuffyKyle Duffy
112
New contributor
Kyle Duffy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 10 hours ago
Kyle DuffyKyle Duffy
112
New contributor
Kyle Duffy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 10 hours ago
Kyle DuffyKyle Duffy
112
answered 10 hours ago
Kyle DuffyKyle Duffy
112
112
New contributor
Kyle Duffy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Kyle Duffy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Kyle Duffy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
add a comment |
$begingroup$
Let $A=1,2, B=2,3,C=3,4$. Then the sets are disjoint because $Acap Bcap C=emptyset$, but not pairwise disjoint because you have pairs such as $A,B$ such that $Acap Bnot =emptyset$.
edited 10 hours ago
J.G.
29.1k22845
answered 10 hours ago
Oscar LanziOscar Lanzi
13k12136
$endgroup$
$begingroup$
Rats! What is the notation for an empty set?
$endgroup$
– Oscar Lanzi
10 hours ago
$begingroup$
Thank you, @jg.
$endgroup$
– Oscar Lanzi
10 hours ago
add a comment |
$begingroup$
Let $A=1,2, B=2,3,C=3,4$. Then the sets are disjoint because $Acap Bcap C=emptyset$, but not pairwise disjoint because you have pairs such as $A,B$ such that $Acap Bnot =emptyset$.
edited 10 hours ago
J.G.
29.1k22845
answered 10 hours ago
Oscar LanziOscar Lanzi
13k12136
$endgroup$
$begingroup$
Rats! What is the notation for an empty set?
$endgroup$
– Oscar Lanzi
10 hours ago
$begingroup$
Thank you, @jg.
$endgroup$
– Oscar Lanzi
10 hours ago
add a comment |
$begingroup$
Let $A=1,2, B=2,3,C=3,4$. Then the sets are disjoint because $Acap Bcap C=emptyset$, but not pairwise disjoint because you have pairs such as $A,B$ such that $Acap Bnot =emptyset$.
edited 10 hours ago
J.G.
29.1k22845
answered 10 hours ago
Oscar LanziOscar Lanzi
13k12136
$endgroup$
Let $A=1,2, B=2,3,C=3,4$. Then the sets are disjoint because $Acap Bcap C=emptyset$, but not pairwise disjoint because you have pairs such as $A,B$ such that $Acap Bnot =emptyset$.
edited 10 hours ago
J.G.
29.1k22845
answered 10 hours ago
Oscar LanziOscar Lanzi
13k12136
edited 10 hours ago
J.G.
29.1k22845
edited 10 hours ago
J.G.
29.1k22845
edited 10 hours ago
J.G.
29.1k22845
29.1k22845
answered 10 hours ago
Oscar LanziOscar Lanzi
13k12136
answered 10 hours ago
Oscar LanziOscar Lanzi
13k12136
answered 10 hours ago
Oscar LanziOscar Lanzi
13k12136
13k12136
$begingroup$
Rats! What is the notation for an empty set?
$endgroup$
– Oscar Lanzi
10 hours ago
$begingroup$
Thank you, @jg.
$endgroup$
– Oscar Lanzi
10 hours ago
add a comment |
$begingroup$
Rats! What is the notation for an empty set?
$endgroup$
– Oscar Lanzi
10 hours ago
$begingroup$
Thank you, @jg.
$endgroup$
– Oscar Lanzi
10 hours ago
$begingroup$
Rats! What is the notation for an empty set?
$endgroup$
– Oscar Lanzi
10 hours ago
$begingroup$
Rats! What is the notation for an empty set?
$endgroup$
– Oscar Lanzi
10 hours ago
$begingroup$
Thank you, @jg.
$endgroup$
– Oscar Lanzi
10 hours ago
$begingroup$
Thank you, @jg.
$endgroup$
– Oscar Lanzi
10 hours ago
add a comment |
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