Why do we say 'Pairwise Disjoint', rather than '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 12 hours ago
John Lawrence AspdenJohn Lawrence Aspden
32418
$endgroup$
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 12 hours ago
John Lawrence AspdenJohn Lawrence Aspden
32418
$endgroup$
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 12 hours ago
John Lawrence AspdenJohn Lawrence Aspden
32418
$endgroup$
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 12 hours ago
John Lawrence AspdenJohn Lawrence Aspden
32418
asked 12 hours ago
John Lawrence AspdenJohn Lawrence Aspden
32418
asked 12 hours ago
John Lawrence AspdenJohn Lawrence Aspden
32418
asked 12 hours ago
John Lawrence AspdenJohn Lawrence Aspden
32418
asked 12 hours ago
John Lawrence AspdenJohn Lawrence Aspden
32418
32418
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 12 hours ago
saulspatzsaulspatz
16.6k31333
$endgroup$
6
$begingroup$
Really? Who would call those disjoint sets?
$endgroup$
– John Lawrence Aspden
12 hours ago
15
$begingroup$
Everyone. Disjoint means their intersection is empty.
$endgroup$
– saulspatz
12 hours ago
9
$begingroup$
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
12 hours ago
10
$begingroup$
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 $mu(bigcup_{n=1}^{infty}A_n)=sum_{n=1}^{infty}mu(A_n)$ 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
12 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
11 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 8 hours ago
BlueRaja - Danny PflughoeftBlueRaja - Danny Pflughoeft
5,64532843
$endgroup$
add a comment |
$begingroup$
In this context disjoint means $A cap B cap C = emptyset$.
answered 12 hours ago
Umberto P.Umberto P.
39.8k13267
$endgroup$
8
$begingroup$
Is that a standard meaning? I never saw the term formally defined that way in four years of undergrad math classes.
$endgroup$
– Connor Harris
12 hours ago
$begingroup$
me neither, but four people have answered the question this way in four minutes!
$endgroup$
– John Lawrence Aspden
12 hours ago
1
$begingroup$
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.
12 hours ago
$begingroup$
This is not the standard definition of disjoint. Google it if you like.
$endgroup$
– Marc van Leeuwen
8 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
5 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 8 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 12 hours ago
J.G.J.G.
29.1k22845
$endgroup$
2
$begingroup$
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
8 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.
8 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 12 hours ago
Kyle DuffyKyle Duffy
312
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 12 hours ago
J.G.
29.1k22845
answered 12 hours ago
Oscar LanziOscar Lanzi
13k12136
$endgroup$
$begingroup$
Rats! What is the notation for an empty set?
$endgroup$
– Oscar Lanzi
12 hours ago
$begingroup$
Thank you, @jg.
$endgroup$
– Oscar Lanzi
12 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 12 hours ago
saulspatzsaulspatz
16.6k31333
$endgroup$
6
$begingroup$
Really? Who would call those disjoint sets?
$endgroup$
– John Lawrence Aspden
12 hours ago
15
$begingroup$
Everyone. Disjoint means their intersection is empty.
$endgroup$
– saulspatz
12 hours ago
9
$begingroup$
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
12 hours ago
10
$begingroup$
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 $mu(bigcup_{n=1}^{infty}A_n)=sum_{n=1}^{infty}mu(A_n)$ 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
12 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
11 hours ago
|
show 8 more comments
$begingroup$
${1,2},{2,3},{1,3}$ are disjoint but not pairwise disjoint.
answered 12 hours ago
saulspatzsaulspatz
16.6k31333
$endgroup$
6
$begingroup$
Really? Who would call those disjoint sets?
$endgroup$
– John Lawrence Aspden
12 hours ago
15
$begingroup$
Everyone. Disjoint means their intersection is empty.
$endgroup$
– saulspatz
12 hours ago
9
$begingroup$
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
12 hours ago
10
$begingroup$
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 $mu(bigcup_{n=1}^{infty}A_n)=sum_{n=1}^{infty}mu(A_n)$ 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
12 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
11 hours ago
|
show 8 more comments
$begingroup$
${1,2},{2,3},{1,3}$ are disjoint but not pairwise disjoint.
answered 12 hours ago
saulspatzsaulspatz
16.6k31333
$endgroup$
${1,2},{2,3},{1,3}$ are disjoint but not pairwise disjoint.
answered 12 hours ago
saulspatzsaulspatz
16.6k31333
answered 12 hours ago
saulspatzsaulspatz
16.6k31333
answered 12 hours ago
saulspatzsaulspatz
16.6k31333
answered 12 hours ago
saulspatzsaulspatz
16.6k31333
16.6k31333
6
$begingroup$
Really? Who would call those disjoint sets?
$endgroup$
– John Lawrence Aspden
12 hours ago
15
$begingroup$
Everyone. Disjoint means their intersection is empty.
$endgroup$
– saulspatz
12 hours ago
9
$begingroup$
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
12 hours ago
10
$begingroup$
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 $mu(bigcup_{n=1}^{infty}A_n)=sum_{n=1}^{infty}mu(A_n)$ 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
12 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
11 hours ago
|
show 8 more comments
6
$begingroup$
Really? Who would call those disjoint sets?
$endgroup$
– John Lawrence Aspden
12 hours ago
15
$begingroup$
Everyone. Disjoint means their intersection is empty.
$endgroup$
– saulspatz
12 hours ago
9
$begingroup$
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
12 hours ago
10
$begingroup$
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 $mu(bigcup_{n=1}^{infty}A_n)=sum_{n=1}^{infty}mu(A_n)$ 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
12 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
11 hours ago
6
6
$begingroup$
Really? Who would call those disjoint sets?
$endgroup$
– John Lawrence Aspden
12 hours ago
$begingroup$
Really? Who would call those disjoint sets?
$endgroup$
– John Lawrence Aspden
12 hours ago
15
15
$begingroup$
Everyone. Disjoint means their intersection is empty.
$endgroup$
– saulspatz
12 hours ago
$begingroup$
Everyone. Disjoint means their intersection is empty.
$endgroup$
– saulspatz
12 hours ago
9
9
$begingroup$
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
12 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
12 hours ago
10
10
$begingroup$
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 $mu(bigcup_{n=1}^{infty}A_n)=sum_{n=1}^{infty}mu(A_n)$ 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
12 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 $mu(bigcup_{n=1}^{infty}A_n)=sum_{n=1}^{infty}mu(A_n)$ 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
12 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
11 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
11 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 8 hours ago
BlueRaja - Danny PflughoeftBlueRaja - Danny Pflughoeft
5,64532843
$endgroup$
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 8 hours ago
BlueRaja - Danny PflughoeftBlueRaja - Danny Pflughoeft
5,64532843
$endgroup$
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 8 hours ago
BlueRaja - Danny PflughoeftBlueRaja - Danny Pflughoeft
5,64532843
$endgroup$
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 8 hours ago
BlueRaja - Danny PflughoeftBlueRaja - Danny Pflughoeft
5,64532843
answered 8 hours ago
BlueRaja - Danny PflughoeftBlueRaja - Danny Pflughoeft
5,64532843
answered 8 hours ago
BlueRaja - Danny PflughoeftBlueRaja - Danny Pflughoeft
5,64532843
answered 8 hours ago
BlueRaja - Danny PflughoeftBlueRaja - Danny Pflughoeft
5,64532843
5,64532843
add a comment |
add a comment |
$begingroup$
In this context disjoint means $A cap B cap C = emptyset$.
answered 12 hours ago
Umberto P.Umberto P.
39.8k13267
$endgroup$
8
$begingroup$
Is that a standard meaning? I never saw the term formally defined that way in four years of undergrad math classes.
$endgroup$
– Connor Harris
12 hours ago
$begingroup$
me neither, but four people have answered the question this way in four minutes!
$endgroup$
– John Lawrence Aspden
12 hours ago
1
$begingroup$
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.
12 hours ago
$begingroup$
This is not the standard definition of disjoint. Google it if you like.
$endgroup$
– Marc van Leeuwen
8 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
5 hours ago
add a comment |
$begingroup$
In this context disjoint means $A cap B cap C = emptyset$.
answered 12 hours ago
Umberto P.Umberto P.
39.8k13267
$endgroup$
8
$begingroup$
Is that a standard meaning? I never saw the term formally defined that way in four years of undergrad math classes.
$endgroup$
– Connor Harris
12 hours ago
$begingroup$
me neither, but four people have answered the question this way in four minutes!
$endgroup$
– John Lawrence Aspden
12 hours ago
1
$begingroup$
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.
12 hours ago
$begingroup$
This is not the standard definition of disjoint. Google it if you like.
$endgroup$
– Marc van Leeuwen
8 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
5 hours ago
add a comment |
$begingroup$
In this context disjoint means $A cap B cap C = emptyset$.
answered 12 hours ago
Umberto P.Umberto P.
39.8k13267
$endgroup$
In this context disjoint means $A cap B cap C = emptyset$.
answered 12 hours ago
Umberto P.Umberto P.
39.8k13267
answered 12 hours ago
Umberto P.Umberto P.
39.8k13267
answered 12 hours ago
Umberto P.Umberto P.
39.8k13267
answered 12 hours ago
Umberto P.Umberto P.
39.8k13267
39.8k13267
8
$begingroup$
Is that a standard meaning? I never saw the term formally defined that way in four years of undergrad math classes.
$endgroup$
– Connor Harris
12 hours ago
$begingroup$
me neither, but four people have answered the question this way in four minutes!
$endgroup$
– John Lawrence Aspden
12 hours ago
1
$begingroup$
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.
12 hours ago
$begingroup$
This is not the standard definition of disjoint. Google it if you like.
$endgroup$
– Marc van Leeuwen
8 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
5 hours ago
add a comment |
8
$begingroup$
Is that a standard meaning? I never saw the term formally defined that way in four years of undergrad math classes.
$endgroup$
– Connor Harris
12 hours ago
$begingroup$
me neither, but four people have answered the question this way in four minutes!
$endgroup$
– John Lawrence Aspden
12 hours ago
1
$begingroup$
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.
12 hours ago
$begingroup$
This is not the standard definition of disjoint. Google it if you like.
$endgroup$
– Marc van Leeuwen
8 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
5 hours ago
8
8
$begingroup$
Is that a standard meaning? I never saw the term formally defined that way in four years of undergrad math classes.
$endgroup$
– Connor Harris
12 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
12 hours ago
$begingroup$
me neither, but four people have answered the question this way in four minutes!
$endgroup$
– John Lawrence Aspden
12 hours ago
$begingroup$
me neither, but four people have answered the question this way in four minutes!
$endgroup$
– John Lawrence Aspden
12 hours ago
1
1
$begingroup$
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.
12 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.
12 hours ago
$begingroup$
This is not the standard definition of disjoint. Google it if you like.
$endgroup$
– Marc van Leeuwen
8 hours ago
$begingroup$
This is not the standard definition of disjoint. Google it if you like.
$endgroup$
– Marc van Leeuwen
8 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
5 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.
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– Brilliand
5 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 8 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 8 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 8 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 8 hours ago
Marc van LeeuwenMarc van Leeuwen
88k5111226
answered 8 hours ago
Marc van LeeuwenMarc van Leeuwen
88k5111226
answered 8 hours ago
Marc van LeeuwenMarc van Leeuwen
88k5111226
answered 8 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 12 hours ago
J.G.J.G.
29.1k22845
$endgroup$
2
$begingroup$
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.
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– Marc van Leeuwen
8 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.
8 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 12 hours ago
J.G.J.G.
29.1k22845
$endgroup$
2
$begingroup$
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
8 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.
8 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 12 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 12 hours ago
J.G.J.G.
29.1k22845
answered 12 hours ago
J.G.J.G.
29.1k22845
answered 12 hours ago
J.G.J.G.
29.1k22845
answered 12 hours ago
J.G.J.G.
29.1k22845
29.1k22845
2
$begingroup$
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
8 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.
8 hours ago
add a comment |
2
$begingroup$
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
8 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.
8 hours ago
2
2
$begingroup$
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
8 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
8 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.
8 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.
8 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 12 hours ago
Kyle DuffyKyle Duffy
312
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 12 hours ago
Kyle DuffyKyle Duffy
312
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 12 hours ago
Kyle DuffyKyle Duffy
312
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 12 hours ago
Kyle DuffyKyle Duffy
312
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 12 hours ago
Kyle DuffyKyle Duffy
312
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 12 hours ago
Kyle DuffyKyle Duffy
312
answered 12 hours ago
Kyle DuffyKyle Duffy
312
312
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 12 hours ago
J.G.
29.1k22845
answered 12 hours ago
Oscar LanziOscar Lanzi
13k12136
$endgroup$
$begingroup$
Rats! What is the notation for an empty set?
$endgroup$
– Oscar Lanzi
12 hours ago
$begingroup$
Thank you, @jg.
$endgroup$
– Oscar Lanzi
12 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 12 hours ago
J.G.
29.1k22845
answered 12 hours ago
Oscar LanziOscar Lanzi
13k12136
$endgroup$
$begingroup$
Rats! What is the notation for an empty set?
$endgroup$
– Oscar Lanzi
12 hours ago
$begingroup$
Thank you, @jg.
$endgroup$
– Oscar Lanzi
12 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 12 hours ago
J.G.
29.1k22845
answered 12 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 12 hours ago
J.G.
29.1k22845
answered 12 hours ago
Oscar LanziOscar Lanzi
13k12136
edited 12 hours ago
J.G.
29.1k22845
edited 12 hours ago
J.G.
29.1k22845
edited 12 hours ago
J.G.
29.1k22845
29.1k22845
answered 12 hours ago
Oscar LanziOscar Lanzi
13k12136
answered 12 hours ago
Oscar LanziOscar Lanzi
13k12136
answered 12 hours ago
Oscar LanziOscar Lanzi
13k12136
13k12136
$begingroup$
Rats! What is the notation for an empty set?
$endgroup$
– Oscar Lanzi
12 hours ago
$begingroup$
Thank you, @jg.
$endgroup$
– Oscar Lanzi
12 hours ago
add a comment |
$begingroup$
Rats! What is the notation for an empty set?
$endgroup$
– Oscar Lanzi
12 hours ago
$begingroup$
Thank you, @jg.
$endgroup$
– Oscar Lanzi
12 hours ago
$begingroup$
Rats! What is the notation for an empty set?
$endgroup$
– Oscar Lanzi
12 hours ago
$begingroup$
Rats! What is the notation for an empty set?
$endgroup$
– Oscar Lanzi
12 hours ago
$begingroup$
Thank you, @jg.
$endgroup$
– Oscar Lanzi
12 hours ago
$begingroup$
Thank you, @jg.
$endgroup$
– Oscar Lanzi
12 hours ago
add a comment |
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