Why do we say ‘pairwise disjoint’, rather than ‘disjoint’?
Clash Royale CLAN TAG#URR8PPP
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I don’t see the ambiguity that ‘pairwise’ resolves.
Surely if $A$, $B$ and $C$ are disjoint sets then they are pairwise disjoint and vice versa?
Or am I being dim?
elementary-set-theory terminology definition
edited Mar 10 at 14:00
Rócherz
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asked Mar 8 at 15:48
John Lawrence AspdenJohn Lawrence Aspden
416410
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add a comment |
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I don’t see the ambiguity that ‘pairwise’ resolves.
Surely if $A$, $B$ and $C$ are disjoint sets then they are pairwise disjoint and vice versa?
Or am I being dim?
elementary-set-theory terminology definition
edited Mar 10 at 14:00
Rócherz
3,0263822
asked Mar 8 at 15:48
John Lawrence AspdenJohn Lawrence Aspden
416410
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1
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Possible duplicate of pairwise disjoint , or disjoint
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– YuiTo Cheng
Mar 9 at 9:35
5
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Note to Reviewers: the target post provided by YuiTo Cheng has much less content. Here the question post is poor but there are some good answers demonstrating the lack of consensus on this terminology.
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– Lee David Chung Lin
Mar 10 at 0:24
2
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This comment is not directly about sets, but about a similar situation that arises with relative primality of integers. The numbers 6, 10, and 15 are relatively prime as a triple ($gcd(6,10,15) = 1$) but they are not pairwise relatively prime. For positive integers $a_1, ldots, a_n$ where $n geq 3$, there is a distinction between being pairwise relatively prime ($gcd(a_i,a_j) = 1$ for all $i not= j$) and being relatively prime as an $n$-tuple ($gcd(a_1, ldots, a_n) = 1$). If you want to say $n$ positive integers are "relatively prime" and $n geq 3$ then you should be more precise.
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– KCd
Mar 10 at 14:06
add a comment |
$begingroup$
I don’t see the ambiguity that ‘pairwise’ resolves.
Surely if $A$, $B$ and $C$ are disjoint sets then they are pairwise disjoint and vice versa?
Or am I being dim?
elementary-set-theory terminology definition
edited Mar 10 at 14:00
Rócherz
3,0263822
asked Mar 8 at 15:48
John Lawrence AspdenJohn Lawrence Aspden
416410
$endgroup$
I don’t see the ambiguity that ‘pairwise’ resolves.
Surely if $A$, $B$ and $C$ are disjoint sets then they are pairwise disjoint and vice versa?
Or am I being dim?
elementary-set-theory terminology definition
elementary-set-theory terminology definition
edited Mar 10 at 14:00
Rócherz
3,0263822
asked Mar 8 at 15:48
John Lawrence AspdenJohn Lawrence Aspden
416410
edited Mar 10 at 14:00
Rócherz
3,0263822
asked Mar 8 at 15:48
John Lawrence AspdenJohn Lawrence Aspden
416410
edited Mar 10 at 14:00
Rócherz
3,0263822
edited Mar 10 at 14:00
Rócherz
3,0263822
edited Mar 10 at 14:00
Rócherz
3,0263822
3,0263822
asked Mar 8 at 15:48
John Lawrence AspdenJohn Lawrence Aspden
416410
asked Mar 8 at 15:48
John Lawrence AspdenJohn Lawrence Aspden
416410
asked Mar 8 at 15:48
John Lawrence AspdenJohn Lawrence Aspden
416410
416410
1
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Possible duplicate of pairwise disjoint , or disjoint
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– YuiTo Cheng
Mar 9 at 9:35
5
$begingroup$
Note to Reviewers: the target post provided by YuiTo Cheng has much less content. Here the question post is poor but there are some good answers demonstrating the lack of consensus on this terminology.
$endgroup$
– Lee David Chung Lin
Mar 10 at 0:24
2
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This comment is not directly about sets, but about a similar situation that arises with relative primality of integers. The numbers 6, 10, and 15 are relatively prime as a triple ($gcd(6,10,15) = 1$) but they are not pairwise relatively prime. For positive integers $a_1, ldots, a_n$ where $n geq 3$, there is a distinction between being pairwise relatively prime ($gcd(a_i,a_j) = 1$ for all $i not= j$) and being relatively prime as an $n$-tuple ($gcd(a_1, ldots, a_n) = 1$). If you want to say $n$ positive integers are "relatively prime" and $n geq 3$ then you should be more precise.
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– KCd
Mar 10 at 14:06
add a comment |
1
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Possible duplicate of pairwise disjoint , or disjoint
$endgroup$
– YuiTo Cheng
Mar 9 at 9:35
5
$begingroup$
Note to Reviewers: the target post provided by YuiTo Cheng has much less content. Here the question post is poor but there are some good answers demonstrating the lack of consensus on this terminology.
$endgroup$
– Lee David Chung Lin
Mar 10 at 0:24
2
$begingroup$
This comment is not directly about sets, but about a similar situation that arises with relative primality of integers. The numbers 6, 10, and 15 are relatively prime as a triple ($gcd(6,10,15) = 1$) but they are not pairwise relatively prime. For positive integers $a_1, ldots, a_n$ where $n geq 3$, there is a distinction between being pairwise relatively prime ($gcd(a_i,a_j) = 1$ for all $i not= j$) and being relatively prime as an $n$-tuple ($gcd(a_1, ldots, a_n) = 1$). If you want to say $n$ positive integers are "relatively prime" and $n geq 3$ then you should be more precise.
$endgroup$
– KCd
Mar 10 at 14:06
1
1
$begingroup$
Possible duplicate of pairwise disjoint , or disjoint
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– YuiTo Cheng
Mar 9 at 9:35
$begingroup$
Possible duplicate of pairwise disjoint , or disjoint
$endgroup$
– YuiTo Cheng
Mar 9 at 9:35
5
5
$begingroup$
Note to Reviewers: the target post provided by YuiTo Cheng has much less content. Here the question post is poor but there are some good answers demonstrating the lack of consensus on this terminology.
$endgroup$
– Lee David Chung Lin
Mar 10 at 0:24
$begingroup$
Note to Reviewers: the target post provided by YuiTo Cheng has much less content. Here the question post is poor but there are some good answers demonstrating the lack of consensus on this terminology.
$endgroup$
– Lee David Chung Lin
Mar 10 at 0:24
2
2
$begingroup$
This comment is not directly about sets, but about a similar situation that arises with relative primality of integers. The numbers 6, 10, and 15 are relatively prime as a triple ($gcd(6,10,15) = 1$) but they are not pairwise relatively prime. For positive integers $a_1, ldots, a_n$ where $n geq 3$, there is a distinction between being pairwise relatively prime ($gcd(a_i,a_j) = 1$ for all $i not= j$) and being relatively prime as an $n$-tuple ($gcd(a_1, ldots, a_n) = 1$). If you want to say $n$ positive integers are "relatively prime" and $n geq 3$ then you should be more precise.
$endgroup$
– KCd
Mar 10 at 14:06
$begingroup$
This comment is not directly about sets, but about a similar situation that arises with relative primality of integers. The numbers 6, 10, and 15 are relatively prime as a triple ($gcd(6,10,15) = 1$) but they are not pairwise relatively prime. For positive integers $a_1, ldots, a_n$ where $n geq 3$, there is a distinction between being pairwise relatively prime ($gcd(a_i,a_j) = 1$ for all $i not= j$) and being relatively prime as an $n$-tuple ($gcd(a_1, ldots, a_n) = 1$). If you want to say $n$ positive integers are "relatively prime" and $n geq 3$ then you should be more precise.
$endgroup$
– KCd
Mar 10 at 14:06
add a comment |
9 Answers
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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".
To corroborate my point of view, here is a citation from Halmos:
Pairs of sets with an empty intersection occur frequently enough to justify
the use of a special word: if $Acap B=emptyset$, the sets $A$ and $B$ are
called disjoint. The same word is sometimes applied to a collection of sets
to indicate that any two distinct sets of the collection are disjoint;
alternatively we may speak in such a situation of a pairwise disjoint
collection.
By the way, it is amazing how this site contains lots of answers saying disjoint (for any collection of sets) means empty intersection, like at this question and questions linked from there, and lots of comments saying that is wrong. Which was my motivation to post this as an answer.
answered Mar 8 at 20:03
Marc van LeeuwenMarc van Leeuwen
88.7k5111230
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This answer shows another case where the same terminology means something different to different people. To me a disjoint union is a specific formulation of the coproduct in $mathbfSet$ like the definition on Wikpedia rather than a union of pairwise disjoint sets. But this difference is a bit more immaterial because they are isomorphic when the sets involved are pairwise disjoint.
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– Potato44
Mar 9 at 4:48
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@Potato44: Yes I know, but I did not want to bring up the distinction internal/external disjoint union, which is similar to internal/external direct sum, which has nothing to do with the current question and would be confusing. My point is that nobody says, or should have to say, "pairwise disjoint union".
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– Marc van Leeuwen
Mar 9 at 9:15
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I just checked Royden (4th edition). It uses disjoint to denote what others here call "pairwise disjoint." Same as Karatzas and Shreve, Pugh in his "Real Mathematical Analysis" and Abbott in his "Understanding Analysis." These are the texts I have in a pdf form so I could search quickly. I have never heard or seen anybody calling a collection of sets disjoint to mean empty intersection.
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– Mdoc
Mar 28 at 2:17
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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 Mar 8 at 20:42
BlueRaja - Danny PflughoeftBlueRaja - Danny Pflughoeft
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Could you back up the claim that 'others use it to mean "empty intersection"' (in the case of more than two sets) with any mathematical text (not with answers on this site, which are indeed east to find)?
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– Marc van Leeuwen
Mar 9 at 12:30
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I agree with your conclusion, and with @MarcvanLeeuwen's comment: I don't think I've ever seen "disjoint" alone for more than $2$ sets, and I think it would be really weird.
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– Arnaud Mortier
Mar 10 at 0:39
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$1,2,2,3,1,3$ are disjoint but not pairwise disjoint.
answered Mar 8 at 15:51
saulspatzsaulspatz
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Really? Who would call those disjoint sets?
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– John Lawrence Aspden
Mar 8 at 15:52
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Everyone. Disjoint means their intersection is empty.
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– saulspatz
Mar 8 at 15:53
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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.
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– drhab
Mar 8 at 16:16
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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.
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– drhab
Mar 8 at 16:37
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These sets are not disjoint, at least not by the modern usage of the word “disjoint,” which afaik is used interchangeably with “pairwise disjoint.”
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– Shalop
Mar 8 at 17:33
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show 8 more comments
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Two sets are disjoint when their intersection is empty. Sets are pairwise disjoint when any two of them are disjoint. Most if not all mathematicians also call such sets disjoint, making pairwise a superfluous term for emphasis.
answered Mar 8 at 15:53
J.G.J.G.
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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
Mar 8 at 19:45
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@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.
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– J.G.
Mar 8 at 19:51
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@MarcvanLeeuwen From the second sentence in the Wikipedia article: "disjoint sets are sets whose intersection is the empty set." If this definition was solely meant for pairs of sets, I would expect it to say "two sets are disjoint if..." rather than "disjoint sets are...".
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– Ryan
Mar 9 at 4:49
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@Ryan: Thank you for that comment, it allowed me to find the citation of Halmos that I added to my answer (the one provided in the footnote to your WP quote) and which clearly shows that the WP quote is incorrectly phrased. Elsewhere in the WP article (like in the opening phrase) they are careful to always specify intersection of two sets. I'll correct the WP intro.
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– Marc van Leeuwen
Mar 9 at 7:50
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@MarcvanLeeuwen I've amended my answer to reflect the usage consensus.
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– J.G.
Mar 9 at 8:17
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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 Mar 8 at 15:54
J.G.
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answered Mar 8 at 15:53
Oscar LanziOscar Lanzi
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Rats! What is the notation for an empty set?
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– Oscar Lanzi
Mar 8 at 15:54
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Thank you, @jg.
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– Oscar Lanzi
Mar 8 at 15:58
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I use varnothing, "$varnothing$".
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– Eric Towers
Mar 9 at 22:14
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This is a question about terminology and its usage in practice, so the basis for an answer should come from real quotations and the historical record rather than unsupported opinion. Looking at early known uses of the term, the few sources that I examined all used "disjoint" in its pairwise sense rather than in the sense of having empty intersection.
The earliest published use (known to me) of the term "disjoint" for its mathematical meaning is in a paper "The Thesis of Modern Logistic" (1909), which is the earliest such use found in a JSTOR search and the earliest such use listed in Earliest Known Uses of
Some of the Words of Mathematics. (Note: Here and throughout, I do not claim that the quoted sources are truly the earliest sources, but merely the earliest ones I was able to find. It is clear that readily available search tools today cover only a small part of historical record. Especially missing from my research is German papers and books, which surely are a large part of the early history of set theory.)
This first paper uses "disjoint" in its description of a construction similar to what we now call an $n$-ary Cartesian product, but where the members of the product are sets rather than ordered $n$-tuples:
Multiplication of cardinals is also defined in purely logical terms. This is done by means of the concept (due to Whitehead) of multiplicative class, which is itself given in terms of logical constants: $k$ being a class of disjoint classes, the multiplicative class of $k$ is the class of all the classes each of which contains one and but one term of each class in $k$. [Italics present in the original; bold emphasis added.]
Although this paper does not state an explicit definition of "disjoint" for more than two "classes", the construction described in the quote requires "disjoint" to mean pairwise disjoint.
The second earliest use I found is in a paper "Differentiation with Respect to a Function of Limited Variation" (1918). This gives an explicit definition of "disjoint intervals" as "intervals with no points common to any two".
When "pairwise disjoint" appeared, it was not meant to change the meaning of "disjoint", but rather to clarify that the author's meaning of "disjoint" was always pairwise. The earliest use I found is "Integral Forms and Variational Orthogonality" (1938):
"Let $U = U_epsilon = sum_j=1^infty [u_j, u'_j)$ be an enumerable set of pairwise disjoint, half-open intervals which contains $S$."
Here, "pairwise disjoint" has the modern meaning we expect. Yet the same paper also uses "disjoint" by itself to mean the same thing as "pairwise disjoint":
if $S$ is a set consisting of a finite number of disjoint half-open intervals $[u_j, u'_j), j = 1, ..., s$
This usage of "disjoint" occurs earlier in the paper than "pairwise disjoint". To this author, the two terms are not contrasting, but synonymous. This view is the same view that you expressed in your question: "pairwise disjoint" is already the default meaning of "disjoint", but can optionally be added as extra clarification in case the reader believes in the other potential interpretation of "disjoint".
answered Mar 10 at 20:51
echinodermataechinodermata
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Thanks for the great research work. I do not agree with your conclusion about the first citation ("The Thesis of Modern Logistic" by C. J. Keyser). What is being described is defining multiplication of cardinals by taking essentially the cardinal of a Cartesian product of "factors" (sets), but using sets (classes) rather than tuples to form the product. For a set containing exactly one element from each factor to faithfully represent an element of the Cartesian product of the factors, it is essential that no two factors have an element in common. So: the factors must be pairwise disjoint.
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– Marc van Leeuwen
Mar 11 at 11:04
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... What I just said is not quite exact: if two or more factors share an element, then the problem is not so much that a set containing exactly one element from each factor fails to identify an element of their Cartesian product, the problem is that some elements of the Cartesian product (i.e., some tuples) cannot be represented by such a set. This is because the set of components of the tuple would have more than one element in common with certain factors, so fail the requirement. But it remains: the factors need to be pairwise disjoint for the construction to define cardinal multiplication.
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– Marc van Leeuwen
Mar 11 at 11:14
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For concreteness, $k=0,0,1,1$ is not a multiplicative class, though $bigcap k =emptyset$. Indeed the "class of all the classes each of which contains one and but one term of each class in $k$" is empty, as one easily sees; however the cardinal multiplication should result in a class with $1times2times1=2$ members. For a disjoint class of classes, one would get the right cardinal for the multiplicative class. Final remark: maybe only multiplication of $2$ cardinals is intended here; then of course "disjoint" only applies to two sets and one gets no verdict at all from it
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– Marc van Leeuwen
Mar 11 at 11:36
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@MarcvanLeeuwen I agree - thanks for the important correction. It definitely means pairwise in the first quote. I edited the answer.
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– echinodermata
Mar 15 at 5:48
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In this context disjoint means $A cap B cap C = emptyset$.
answered Mar 8 at 15:52
Umberto P.Umberto P.
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Is that a standard meaning? I never saw the term formally defined that way in four years of undergrad math classes.
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– Connor Harris
Mar 8 at 15:52
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me neither, but four people have answered the question this way in four minutes!
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– John Lawrence Aspden
Mar 8 at 15:57
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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.
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– Umberto P.
Mar 8 at 16:06
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This is not the standard definition of disjoint. Google it if you like.
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– Marc van Leeuwen
Mar 8 at 19:47
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@Brilliand Indeed, somebody trying to find evidence for the fact that disjoint means empty intersection might land on the current page and find plenty of evidence to support it. But if you bluntly proclaim as this answer does that this is what the term means than it does not hurt to check that you can find evidence of a text using it that way, and in spite of the multitude of answers here in the same sense, nobody has been able to do that.
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– Marc van Leeuwen
Mar 9 at 5:54
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show 3 more comments
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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 Mar 8 at 15:53
Kyle DuffyKyle Duffy
843
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If disjoint is understood as having an empty intersection, then disjoint sets are not necessarily pairwise disjoint. Pairwise disjoint avoid any doubt, at no cost.
answered Mar 14 at 17:19
Yves DaoustYves Daoust
132k676230
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9 Answers
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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".
To corroborate my point of view, here is a citation from Halmos:
Pairs of sets with an empty intersection occur frequently enough to justify
the use of a special word: if $Acap B=emptyset$, the sets $A$ and $B$ are
called disjoint. The same word is sometimes applied to a collection of sets
to indicate that any two distinct sets of the collection are disjoint;
alternatively we may speak in such a situation of a pairwise disjoint
collection.
By the way, it is amazing how this site contains lots of answers saying disjoint (for any collection of sets) means empty intersection, like at this question and questions linked from there, and lots of comments saying that is wrong. Which was my motivation to post this as an answer.
answered Mar 8 at 20:03
Marc van LeeuwenMarc van Leeuwen
88.7k5111230
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2
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This answer shows another case where the same terminology means something different to different people. To me a disjoint union is a specific formulation of the coproduct in $mathbfSet$ like the definition on Wikpedia rather than a union of pairwise disjoint sets. But this difference is a bit more immaterial because they are isomorphic when the sets involved are pairwise disjoint.
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– Potato44
Mar 9 at 4:48
1
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@Potato44: Yes I know, but I did not want to bring up the distinction internal/external disjoint union, which is similar to internal/external direct sum, which has nothing to do with the current question and would be confusing. My point is that nobody says, or should have to say, "pairwise disjoint union".
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– Marc van Leeuwen
Mar 9 at 9:15
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I just checked Royden (4th edition). It uses disjoint to denote what others here call "pairwise disjoint." Same as Karatzas and Shreve, Pugh in his "Real Mathematical Analysis" and Abbott in his "Understanding Analysis." These are the texts I have in a pdf form so I could search quickly. I have never heard or seen anybody calling a collection of sets disjoint to mean empty intersection.
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– Mdoc
Mar 28 at 2:17
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".
To corroborate my point of view, here is a citation from Halmos:
Pairs of sets with an empty intersection occur frequently enough to justify
the use of a special word: if $Acap B=emptyset$, the sets $A$ and $B$ are
called disjoint. The same word is sometimes applied to a collection of sets
to indicate that any two distinct sets of the collection are disjoint;
alternatively we may speak in such a situation of a pairwise disjoint
collection.
By the way, it is amazing how this site contains lots of answers saying disjoint (for any collection of sets) means empty intersection, like at this question and questions linked from there, and lots of comments saying that is wrong. Which was my motivation to post this as an answer.
answered Mar 8 at 20:03
Marc van LeeuwenMarc van Leeuwen
88.7k5111230
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2
$begingroup$
This answer shows another case where the same terminology means something different to different people. To me a disjoint union is a specific formulation of the coproduct in $mathbfSet$ like the definition on Wikpedia rather than a union of pairwise disjoint sets. But this difference is a bit more immaterial because they are isomorphic when the sets involved are pairwise disjoint.
$endgroup$
– Potato44
Mar 9 at 4:48
1
$begingroup$
@Potato44: Yes I know, but I did not want to bring up the distinction internal/external disjoint union, which is similar to internal/external direct sum, which has nothing to do with the current question and would be confusing. My point is that nobody says, or should have to say, "pairwise disjoint union".
$endgroup$
– Marc van Leeuwen
Mar 9 at 9:15
$begingroup$
I just checked Royden (4th edition). It uses disjoint to denote what others here call "pairwise disjoint." Same as Karatzas and Shreve, Pugh in his "Real Mathematical Analysis" and Abbott in his "Understanding Analysis." These are the texts I have in a pdf form so I could search quickly. I have never heard or seen anybody calling a collection of sets disjoint to mean empty intersection.
$endgroup$
– Mdoc
Mar 28 at 2:17
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".
To corroborate my point of view, here is a citation from Halmos:
Pairs of sets with an empty intersection occur frequently enough to justify
the use of a special word: if $Acap B=emptyset$, the sets $A$ and $B$ are
called disjoint. The same word is sometimes applied to a collection of sets
to indicate that any two distinct sets of the collection are disjoint;
alternatively we may speak in such a situation of a pairwise disjoint
collection.
By the way, it is amazing how this site contains lots of answers saying disjoint (for any collection of sets) means empty intersection, like at this question and questions linked from there, and lots of comments saying that is wrong. Which was my motivation to post this as an answer.
answered Mar 8 at 20:03
Marc van LeeuwenMarc van Leeuwen
88.7k5111230
$endgroup$
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".
To corroborate my point of view, here is a citation from Halmos:
Pairs of sets with an empty intersection occur frequently enough to justify
the use of a special word: if $Acap B=emptyset$, the sets $A$ and $B$ are
called disjoint. The same word is sometimes applied to a collection of sets
to indicate that any two distinct sets of the collection are disjoint;
alternatively we may speak in such a situation of a pairwise disjoint
collection.
By the way, it is amazing how this site contains lots of answers saying disjoint (for any collection of sets) means empty intersection, like at this question and questions linked from there, and lots of comments saying that is wrong. Which was my motivation to post this as an answer.
answered Mar 8 at 20:03
Marc van LeeuwenMarc van Leeuwen
88.7k5111230
edited Mar 9 at 12:41
answered Mar 8 at 20:03
Marc van LeeuwenMarc van Leeuwen
88.7k5111230
answered Mar 8 at 20:03
Marc van LeeuwenMarc van Leeuwen
88.7k5111230
answered Mar 8 at 20:03
Marc van LeeuwenMarc van Leeuwen
88.7k5111230
88.7k5111230
2
$begingroup$
This answer shows another case where the same terminology means something different to different people. To me a disjoint union is a specific formulation of the coproduct in $mathbfSet$ like the definition on Wikpedia rather than a union of pairwise disjoint sets. But this difference is a bit more immaterial because they are isomorphic when the sets involved are pairwise disjoint.
$endgroup$
– Potato44
Mar 9 at 4:48
1
$begingroup$
@Potato44: Yes I know, but I did not want to bring up the distinction internal/external disjoint union, which is similar to internal/external direct sum, which has nothing to do with the current question and would be confusing. My point is that nobody says, or should have to say, "pairwise disjoint union".
$endgroup$
– Marc van Leeuwen
Mar 9 at 9:15
$begingroup$
I just checked Royden (4th edition). It uses disjoint to denote what others here call "pairwise disjoint." Same as Karatzas and Shreve, Pugh in his "Real Mathematical Analysis" and Abbott in his "Understanding Analysis." These are the texts I have in a pdf form so I could search quickly. I have never heard or seen anybody calling a collection of sets disjoint to mean empty intersection.
$endgroup$
– Mdoc
Mar 28 at 2:17
add a comment |
2
$begingroup$
This answer shows another case where the same terminology means something different to different people. To me a disjoint union is a specific formulation of the coproduct in $mathbfSet$ like the definition on Wikpedia rather than a union of pairwise disjoint sets. But this difference is a bit more immaterial because they are isomorphic when the sets involved are pairwise disjoint.
$endgroup$
– Potato44
Mar 9 at 4:48
1
$begingroup$
@Potato44: Yes I know, but I did not want to bring up the distinction internal/external disjoint union, which is similar to internal/external direct sum, which has nothing to do with the current question and would be confusing. My point is that nobody says, or should have to say, "pairwise disjoint union".
$endgroup$
– Marc van Leeuwen
Mar 9 at 9:15
$begingroup$
I just checked Royden (4th edition). It uses disjoint to denote what others here call "pairwise disjoint." Same as Karatzas and Shreve, Pugh in his "Real Mathematical Analysis" and Abbott in his "Understanding Analysis." These are the texts I have in a pdf form so I could search quickly. I have never heard or seen anybody calling a collection of sets disjoint to mean empty intersection.
$endgroup$
– Mdoc
Mar 28 at 2:17
2
2
$begingroup$
This answer shows another case where the same terminology means something different to different people. To me a disjoint union is a specific formulation of the coproduct in $mathbfSet$ like the definition on Wikpedia rather than a union of pairwise disjoint sets. But this difference is a bit more immaterial because they are isomorphic when the sets involved are pairwise disjoint.
$endgroup$
– Potato44
Mar 9 at 4:48
$begingroup$
This answer shows another case where the same terminology means something different to different people. To me a disjoint union is a specific formulation of the coproduct in $mathbfSet$ like the definition on Wikpedia rather than a union of pairwise disjoint sets. But this difference is a bit more immaterial because they are isomorphic when the sets involved are pairwise disjoint.
$endgroup$
– Potato44
Mar 9 at 4:48
1
1
$begingroup$
@Potato44: Yes I know, but I did not want to bring up the distinction internal/external disjoint union, which is similar to internal/external direct sum, which has nothing to do with the current question and would be confusing. My point is that nobody says, or should have to say, "pairwise disjoint union".
$endgroup$
– Marc van Leeuwen
Mar 9 at 9:15
$begingroup$
@Potato44: Yes I know, but I did not want to bring up the distinction internal/external disjoint union, which is similar to internal/external direct sum, which has nothing to do with the current question and would be confusing. My point is that nobody says, or should have to say, "pairwise disjoint union".
$endgroup$
– Marc van Leeuwen
Mar 9 at 9:15
$begingroup$
I just checked Royden (4th edition). It uses disjoint to denote what others here call "pairwise disjoint." Same as Karatzas and Shreve, Pugh in his "Real Mathematical Analysis" and Abbott in his "Understanding Analysis." These are the texts I have in a pdf form so I could search quickly. I have never heard or seen anybody calling a collection of sets disjoint to mean empty intersection.
$endgroup$
– Mdoc
Mar 28 at 2:17
$begingroup$
I just checked Royden (4th edition). It uses disjoint to denote what others here call "pairwise disjoint." Same as Karatzas and Shreve, Pugh in his "Real Mathematical Analysis" and Abbott in his "Understanding Analysis." These are the texts I have in a pdf form so I could search quickly. I have never heard or seen anybody calling a collection of sets disjoint to mean empty intersection.
$endgroup$
– Mdoc
Mar 28 at 2:17
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 Mar 8 at 20:42
BlueRaja - Danny PflughoeftBlueRaja - Danny Pflughoeft
5,87342943
$endgroup$
4
$begingroup$
Could you back up the claim that 'others use it to mean "empty intersection"' (in the case of more than two sets) with any mathematical text (not with answers on this site, which are indeed east to find)?
$endgroup$
– Marc van Leeuwen
Mar 9 at 12:30
1
$begingroup$
I agree with your conclusion, and with @MarcvanLeeuwen's comment: I don't think I've ever seen "disjoint" alone for more than $2$ sets, and I think it would be really weird.
$endgroup$
– Arnaud Mortier
Mar 10 at 0:39
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 Mar 8 at 20:42
BlueRaja - Danny PflughoeftBlueRaja - Danny Pflughoeft
5,87342943
$endgroup$
4
$begingroup$
Could you back up the claim that 'others use it to mean "empty intersection"' (in the case of more than two sets) with any mathematical text (not with answers on this site, which are indeed east to find)?
$endgroup$
– Marc van Leeuwen
Mar 9 at 12:30
1
$begingroup$
I agree with your conclusion, and with @MarcvanLeeuwen's comment: I don't think I've ever seen "disjoint" alone for more than $2$ sets, and I think it would be really weird.
$endgroup$
– Arnaud Mortier
Mar 10 at 0:39
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 Mar 8 at 20:42
BlueRaja - Danny PflughoeftBlueRaja - Danny Pflughoeft
5,87342943
$endgroup$
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 Mar 8 at 20:42
BlueRaja - Danny PflughoeftBlueRaja - Danny Pflughoeft
5,87342943
answered Mar 8 at 20:42
BlueRaja - Danny PflughoeftBlueRaja - Danny Pflughoeft
5,87342943
answered Mar 8 at 20:42
BlueRaja - Danny PflughoeftBlueRaja - Danny Pflughoeft
5,87342943
answered Mar 8 at 20:42
BlueRaja - Danny PflughoeftBlueRaja - Danny Pflughoeft
5,87342943
5,87342943
4
$begingroup$
Could you back up the claim that 'others use it to mean "empty intersection"' (in the case of more than two sets) with any mathematical text (not with answers on this site, which are indeed east to find)?
$endgroup$
– Marc van Leeuwen
Mar 9 at 12:30
1
$begingroup$
I agree with your conclusion, and with @MarcvanLeeuwen's comment: I don't think I've ever seen "disjoint" alone for more than $2$ sets, and I think it would be really weird.
$endgroup$
– Arnaud Mortier
Mar 10 at 0:39
add a comment |
4
$begingroup$
Could you back up the claim that 'others use it to mean "empty intersection"' (in the case of more than two sets) with any mathematical text (not with answers on this site, which are indeed east to find)?
$endgroup$
– Marc van Leeuwen
Mar 9 at 12:30
1
$begingroup$
I agree with your conclusion, and with @MarcvanLeeuwen's comment: I don't think I've ever seen "disjoint" alone for more than $2$ sets, and I think it would be really weird.
$endgroup$
– Arnaud Mortier
Mar 10 at 0:39
4
4
$begingroup$
Could you back up the claim that 'others use it to mean "empty intersection"' (in the case of more than two sets) with any mathematical text (not with answers on this site, which are indeed east to find)?
$endgroup$
– Marc van Leeuwen
Mar 9 at 12:30
$begingroup$
Could you back up the claim that 'others use it to mean "empty intersection"' (in the case of more than two sets) with any mathematical text (not with answers on this site, which are indeed east to find)?
$endgroup$
– Marc van Leeuwen
Mar 9 at 12:30
1
1
$begingroup$
I agree with your conclusion, and with @MarcvanLeeuwen's comment: I don't think I've ever seen "disjoint" alone for more than $2$ sets, and I think it would be really weird.
$endgroup$
– Arnaud Mortier
Mar 10 at 0:39
$begingroup$
I agree with your conclusion, and with @MarcvanLeeuwen's comment: I don't think I've ever seen "disjoint" alone for more than $2$ sets, and I think it would be really weird.
$endgroup$
– Arnaud Mortier
Mar 10 at 0:39
add a comment |
$begingroup$
$1,2,2,3,1,3$ are disjoint but not pairwise disjoint.
answered Mar 8 at 15:51
saulspatzsaulspatz
17.2k31435
$endgroup$
20
$begingroup$
Really? Who would call those disjoint sets?
$endgroup$
– John Lawrence Aspden
Mar 8 at 15:52
38
$begingroup$
Everyone. Disjoint means their intersection is empty.
$endgroup$
– saulspatz
Mar 8 at 15:53
16
$begingroup$
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
Mar 8 at 16:16
22
$begingroup$
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
Mar 8 at 16:37
17
$begingroup$
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
Mar 8 at 17:33
|
show 8 more comments
$begingroup$
$1,2,2,3,1,3$ are disjoint but not pairwise disjoint.
answered Mar 8 at 15:51
saulspatzsaulspatz
17.2k31435
$endgroup$
20
$begingroup$
Really? Who would call those disjoint sets?
$endgroup$
– John Lawrence Aspden
Mar 8 at 15:52
38
$begingroup$
Everyone. Disjoint means their intersection is empty.
$endgroup$
– saulspatz
Mar 8 at 15:53
16
$begingroup$
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
Mar 8 at 16:16
22
$begingroup$
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
Mar 8 at 16:37
17
$begingroup$
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
Mar 8 at 17:33
|
show 8 more comments
$begingroup$
$1,2,2,3,1,3$ are disjoint but not pairwise disjoint.
answered Mar 8 at 15:51
saulspatzsaulspatz
17.2k31435
$endgroup$
$1,2,2,3,1,3$ are disjoint but not pairwise disjoint.
answered Mar 8 at 15:51
saulspatzsaulspatz
17.2k31435
answered Mar 8 at 15:51
saulspatzsaulspatz
17.2k31435
answered Mar 8 at 15:51
saulspatzsaulspatz
17.2k31435
answered Mar 8 at 15:51
saulspatzsaulspatz
17.2k31435
17.2k31435
20
$begingroup$
Really? Who would call those disjoint sets?
$endgroup$
– John Lawrence Aspden
Mar 8 at 15:52
38
$begingroup$
Everyone. Disjoint means their intersection is empty.
$endgroup$
– saulspatz
Mar 8 at 15:53
16
$begingroup$
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
Mar 8 at 16:16
22
$begingroup$
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
Mar 8 at 16:37
17
$begingroup$
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
Mar 8 at 17:33
|
show 8 more comments
20
$begingroup$
Really? Who would call those disjoint sets?
$endgroup$
– John Lawrence Aspden
Mar 8 at 15:52
38
$begingroup$
Everyone. Disjoint means their intersection is empty.
$endgroup$
– saulspatz
Mar 8 at 15:53
16
$begingroup$
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
Mar 8 at 16:16
22
$begingroup$
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
Mar 8 at 16:37
17
$begingroup$
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
Mar 8 at 17:33
20
20
$begingroup$
Really? Who would call those disjoint sets?
$endgroup$
– John Lawrence Aspden
Mar 8 at 15:52
$begingroup$
Really? Who would call those disjoint sets?
$endgroup$
– John Lawrence Aspden
Mar 8 at 15:52
38
38
$begingroup$
Everyone. Disjoint means their intersection is empty.
$endgroup$
– saulspatz
Mar 8 at 15:53
$begingroup$
Everyone. Disjoint means their intersection is empty.
$endgroup$
– saulspatz
Mar 8 at 15:53
16
16
$begingroup$
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
Mar 8 at 16:16
$begingroup$
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
Mar 8 at 16:16
22
22
$begingroup$
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
Mar 8 at 16:37
$begingroup$
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
Mar 8 at 16:37
17
17
$begingroup$
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
Mar 8 at 17:33
$begingroup$
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
Mar 8 at 17:33
|
show 8 more comments
$begingroup$
Two sets are disjoint when their intersection is empty. Sets are pairwise disjoint when any two of them are disjoint. Most if not all mathematicians also call such sets disjoint, making pairwise a superfluous term for emphasis.
answered Mar 8 at 15:53
J.G.J.G.
32.6k23250
$endgroup$
6
$begingroup$
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
Mar 8 at 19:45
1
$begingroup$
@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.
Mar 8 at 19:51
4
$begingroup$
@MarcvanLeeuwen From the second sentence in the Wikipedia article: "disjoint sets are sets whose intersection is the empty set." If this definition was solely meant for pairs of sets, I would expect it to say "two sets are disjoint if..." rather than "disjoint sets are...".
$endgroup$
– Ryan
Mar 9 at 4:49
1
$begingroup$
@Ryan: Thank you for that comment, it allowed me to find the citation of Halmos that I added to my answer (the one provided in the footnote to your WP quote) and which clearly shows that the WP quote is incorrectly phrased. Elsewhere in the WP article (like in the opening phrase) they are careful to always specify intersection of two sets. I'll correct the WP intro.
$endgroup$
– Marc van Leeuwen
Mar 9 at 7:50
1
$begingroup$
@MarcvanLeeuwen I've amended my answer to reflect the usage consensus.
$endgroup$
– J.G.
Mar 9 at 8:17
|
show 1 more comment
$begingroup$
Two sets are disjoint when their intersection is empty. Sets are pairwise disjoint when any two of them are disjoint. Most if not all mathematicians also call such sets disjoint, making pairwise a superfluous term for emphasis.
answered Mar 8 at 15:53
J.G.J.G.
32.6k23250
$endgroup$
6
$begingroup$
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
Mar 8 at 19:45
1
$begingroup$
@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.
Mar 8 at 19:51
4
$begingroup$
@MarcvanLeeuwen From the second sentence in the Wikipedia article: "disjoint sets are sets whose intersection is the empty set." If this definition was solely meant for pairs of sets, I would expect it to say "two sets are disjoint if..." rather than "disjoint sets are...".
$endgroup$
– Ryan
Mar 9 at 4:49
1
$begingroup$
@Ryan: Thank you for that comment, it allowed me to find the citation of Halmos that I added to my answer (the one provided in the footnote to your WP quote) and which clearly shows that the WP quote is incorrectly phrased. Elsewhere in the WP article (like in the opening phrase) they are careful to always specify intersection of two sets. I'll correct the WP intro.
$endgroup$
– Marc van Leeuwen
Mar 9 at 7:50
1
$begingroup$
@MarcvanLeeuwen I've amended my answer to reflect the usage consensus.
$endgroup$
– J.G.
Mar 9 at 8:17
|
show 1 more comment
$begingroup$
Two sets are disjoint when their intersection is empty. Sets are pairwise disjoint when any two of them are disjoint. Most if not all mathematicians also call such sets disjoint, making pairwise a superfluous term for emphasis.
answered Mar 8 at 15:53
J.G.J.G.
32.6k23250
$endgroup$
Two sets are disjoint when their intersection is empty. Sets are pairwise disjoint when any two of them are disjoint. Most if not all mathematicians also call such sets disjoint, making pairwise a superfluous term for emphasis.
answered Mar 8 at 15:53
J.G.J.G.
32.6k23250
edited Mar 9 at 8:16
answered Mar 8 at 15:53
J.G.J.G.
32.6k23250
answered Mar 8 at 15:53
J.G.J.G.
32.6k23250
answered Mar 8 at 15:53
J.G.J.G.
32.6k23250
32.6k23250
6
$begingroup$
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
Mar 8 at 19:45
1
$begingroup$
@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.
Mar 8 at 19:51
4
$begingroup$
@MarcvanLeeuwen From the second sentence in the Wikipedia article: "disjoint sets are sets whose intersection is the empty set." If this definition was solely meant for pairs of sets, I would expect it to say "two sets are disjoint if..." rather than "disjoint sets are...".
$endgroup$
– Ryan
Mar 9 at 4:49
1
$begingroup$
@Ryan: Thank you for that comment, it allowed me to find the citation of Halmos that I added to my answer (the one provided in the footnote to your WP quote) and which clearly shows that the WP quote is incorrectly phrased. Elsewhere in the WP article (like in the opening phrase) they are careful to always specify intersection of two sets. I'll correct the WP intro.
$endgroup$
– Marc van Leeuwen
Mar 9 at 7:50
1
$begingroup$
@MarcvanLeeuwen I've amended my answer to reflect the usage consensus.
$endgroup$
– J.G.
Mar 9 at 8:17
|
show 1 more comment
6
$begingroup$
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
Mar 8 at 19:45
1
$begingroup$
@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.
Mar 8 at 19:51
4
$begingroup$
@MarcvanLeeuwen From the second sentence in the Wikipedia article: "disjoint sets are sets whose intersection is the empty set." If this definition was solely meant for pairs of sets, I would expect it to say "two sets are disjoint if..." rather than "disjoint sets are...".
$endgroup$
– Ryan
Mar 9 at 4:49
1
$begingroup$
@Ryan: Thank you for that comment, it allowed me to find the citation of Halmos that I added to my answer (the one provided in the footnote to your WP quote) and which clearly shows that the WP quote is incorrectly phrased. Elsewhere in the WP article (like in the opening phrase) they are careful to always specify intersection of two sets. I'll correct the WP intro.
$endgroup$
– Marc van Leeuwen
Mar 9 at 7:50
1
$begingroup$
@MarcvanLeeuwen I've amended my answer to reflect the usage consensus.
$endgroup$
– J.G.
Mar 9 at 8:17
6
6
$begingroup$
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
Mar 8 at 19:45
$begingroup$
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
Mar 8 at 19:45
1
1
$begingroup$
@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.
Mar 8 at 19:51
$begingroup$
@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.
Mar 8 at 19:51
4
4
$begingroup$
@MarcvanLeeuwen From the second sentence in the Wikipedia article: "disjoint sets are sets whose intersection is the empty set." If this definition was solely meant for pairs of sets, I would expect it to say "two sets are disjoint if..." rather than "disjoint sets are...".
$endgroup$
– Ryan
Mar 9 at 4:49
$begingroup$
@MarcvanLeeuwen From the second sentence in the Wikipedia article: "disjoint sets are sets whose intersection is the empty set." If this definition was solely meant for pairs of sets, I would expect it to say "two sets are disjoint if..." rather than "disjoint sets are...".
$endgroup$
– Ryan
Mar 9 at 4:49
1
1
$begingroup$
@Ryan: Thank you for that comment, it allowed me to find the citation of Halmos that I added to my answer (the one provided in the footnote to your WP quote) and which clearly shows that the WP quote is incorrectly phrased. Elsewhere in the WP article (like in the opening phrase) they are careful to always specify intersection of two sets. I'll correct the WP intro.
$endgroup$
– Marc van Leeuwen
Mar 9 at 7:50
$begingroup$
@Ryan: Thank you for that comment, it allowed me to find the citation of Halmos that I added to my answer (the one provided in the footnote to your WP quote) and which clearly shows that the WP quote is incorrectly phrased. Elsewhere in the WP article (like in the opening phrase) they are careful to always specify intersection of two sets. I'll correct the WP intro.
$endgroup$
– Marc van Leeuwen
Mar 9 at 7:50
1
1
$begingroup$
@MarcvanLeeuwen I've amended my answer to reflect the usage consensus.
$endgroup$
– J.G.
Mar 9 at 8:17
$begingroup$
@MarcvanLeeuwen I've amended my answer to reflect the usage consensus.
$endgroup$
– J.G.
Mar 9 at 8:17
|
show 1 more 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 Mar 8 at 15:54
J.G.
32.6k23250
answered Mar 8 at 15:53
Oscar LanziOscar Lanzi
13.5k12136
$endgroup$
$begingroup$
Rats! What is the notation for an empty set?
$endgroup$
– Oscar Lanzi
Mar 8 at 15:54
$begingroup$
Thank you, @jg.
$endgroup$
– Oscar Lanzi
Mar 8 at 15:58
1
$begingroup$
I use varnothing, "$varnothing$".
$endgroup$
– Eric Towers
Mar 9 at 22:14
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 Mar 8 at 15:54
J.G.
32.6k23250
answered Mar 8 at 15:53
Oscar LanziOscar Lanzi
13.5k12136
$endgroup$
$begingroup$
Rats! What is the notation for an empty set?
$endgroup$
– Oscar Lanzi
Mar 8 at 15:54
$begingroup$
Thank you, @jg.
$endgroup$
– Oscar Lanzi
Mar 8 at 15:58
1
$begingroup$
I use varnothing, "$varnothing$".
$endgroup$
– Eric Towers
Mar 9 at 22:14
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 Mar 8 at 15:54
J.G.
32.6k23250
answered Mar 8 at 15:53
Oscar LanziOscar Lanzi
13.5k12136
$endgroup$
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 Mar 8 at 15:54
J.G.
32.6k23250
answered Mar 8 at 15:53
Oscar LanziOscar Lanzi
13.5k12136
edited Mar 8 at 15:54
J.G.
32.6k23250
edited Mar 8 at 15:54
J.G.
32.6k23250
edited Mar 8 at 15:54
J.G.
32.6k23250
32.6k23250
answered Mar 8 at 15:53
Oscar LanziOscar Lanzi
13.5k12136
answered Mar 8 at 15:53
Oscar LanziOscar Lanzi
13.5k12136
answered Mar 8 at 15:53
Oscar LanziOscar Lanzi
13.5k12136
13.5k12136
$begingroup$
Rats! What is the notation for an empty set?
$endgroup$
– Oscar Lanzi
Mar 8 at 15:54
$begingroup$
Thank you, @jg.
$endgroup$
– Oscar Lanzi
Mar 8 at 15:58
1
$begingroup$
I use varnothing, "$varnothing$".
$endgroup$
– Eric Towers
Mar 9 at 22:14
add a comment |
$begingroup$
Rats! What is the notation for an empty set?
$endgroup$
– Oscar Lanzi
Mar 8 at 15:54
$begingroup$
Thank you, @jg.
$endgroup$
– Oscar Lanzi
Mar 8 at 15:58
1
$begingroup$
I use varnothing, "$varnothing$".
$endgroup$
– Eric Towers
Mar 9 at 22:14
$begingroup$
Rats! What is the notation for an empty set?
$endgroup$
– Oscar Lanzi
Mar 8 at 15:54
$begingroup$
Rats! What is the notation for an empty set?
$endgroup$
– Oscar Lanzi
Mar 8 at 15:54
$begingroup$
Thank you, @jg.
$endgroup$
– Oscar Lanzi
Mar 8 at 15:58
$begingroup$
Thank you, @jg.
$endgroup$
– Oscar Lanzi
Mar 8 at 15:58
1
1
$begingroup$
I use varnothing, "$varnothing$".
$endgroup$
– Eric Towers
Mar 9 at 22:14
$begingroup$
I use varnothing, "$varnothing$".
$endgroup$
– Eric Towers
Mar 9 at 22:14
add a comment |
$begingroup$
This is a question about terminology and its usage in practice, so the basis for an answer should come from real quotations and the historical record rather than unsupported opinion. Looking at early known uses of the term, the few sources that I examined all used "disjoint" in its pairwise sense rather than in the sense of having empty intersection.
The earliest published use (known to me) of the term "disjoint" for its mathematical meaning is in a paper "The Thesis of Modern Logistic" (1909), which is the earliest such use found in a JSTOR search and the earliest such use listed in Earliest Known Uses of
Some of the Words of Mathematics. (Note: Here and throughout, I do not claim that the quoted sources are truly the earliest sources, but merely the earliest ones I was able to find. It is clear that readily available search tools today cover only a small part of historical record. Especially missing from my research is German papers and books, which surely are a large part of the early history of set theory.)
This first paper uses "disjoint" in its description of a construction similar to what we now call an $n$-ary Cartesian product, but where the members of the product are sets rather than ordered $n$-tuples:
Multiplication of cardinals is also defined in purely logical terms. This is done by means of the concept (due to Whitehead) of multiplicative class, which is itself given in terms of logical constants: $k$ being a class of disjoint classes, the multiplicative class of $k$ is the class of all the classes each of which contains one and but one term of each class in $k$. [Italics present in the original; bold emphasis added.]
Although this paper does not state an explicit definition of "disjoint" for more than two "classes", the construction described in the quote requires "disjoint" to mean pairwise disjoint.
The second earliest use I found is in a paper "Differentiation with Respect to a Function of Limited Variation" (1918). This gives an explicit definition of "disjoint intervals" as "intervals with no points common to any two".
When "pairwise disjoint" appeared, it was not meant to change the meaning of "disjoint", but rather to clarify that the author's meaning of "disjoint" was always pairwise. The earliest use I found is "Integral Forms and Variational Orthogonality" (1938):
"Let $U = U_epsilon = sum_j=1^infty [u_j, u'_j)$ be an enumerable set of pairwise disjoint, half-open intervals which contains $S$."
Here, "pairwise disjoint" has the modern meaning we expect. Yet the same paper also uses "disjoint" by itself to mean the same thing as "pairwise disjoint":
if $S$ is a set consisting of a finite number of disjoint half-open intervals $[u_j, u'_j), j = 1, ..., s$
This usage of "disjoint" occurs earlier in the paper than "pairwise disjoint". To this author, the two terms are not contrasting, but synonymous. This view is the same view that you expressed in your question: "pairwise disjoint" is already the default meaning of "disjoint", but can optionally be added as extra clarification in case the reader believes in the other potential interpretation of "disjoint".
answered Mar 10 at 20:51
echinodermataechinodermata
2,52911228
$endgroup$
1
$begingroup$
Thanks for the great research work. I do not agree with your conclusion about the first citation ("The Thesis of Modern Logistic" by C. J. Keyser). What is being described is defining multiplication of cardinals by taking essentially the cardinal of a Cartesian product of "factors" (sets), but using sets (classes) rather than tuples to form the product. For a set containing exactly one element from each factor to faithfully represent an element of the Cartesian product of the factors, it is essential that no two factors have an element in common. So: the factors must be pairwise disjoint.
$endgroup$
– Marc van Leeuwen
Mar 11 at 11:04
$begingroup$
... What I just said is not quite exact: if two or more factors share an element, then the problem is not so much that a set containing exactly one element from each factor fails to identify an element of their Cartesian product, the problem is that some elements of the Cartesian product (i.e., some tuples) cannot be represented by such a set. This is because the set of components of the tuple would have more than one element in common with certain factors, so fail the requirement. But it remains: the factors need to be pairwise disjoint for the construction to define cardinal multiplication.
$endgroup$
– Marc van Leeuwen
Mar 11 at 11:14
$begingroup$
For concreteness, $k=0,0,1,1$ is not a multiplicative class, though $bigcap k =emptyset$. Indeed the "class of all the classes each of which contains one and but one term of each class in $k$" is empty, as one easily sees; however the cardinal multiplication should result in a class with $1times2times1=2$ members. For a disjoint class of classes, one would get the right cardinal for the multiplicative class. Final remark: maybe only multiplication of $2$ cardinals is intended here; then of course "disjoint" only applies to two sets and one gets no verdict at all from it
$endgroup$
– Marc van Leeuwen
Mar 11 at 11:36
$begingroup$
@MarcvanLeeuwen I agree - thanks for the important correction. It definitely means pairwise in the first quote. I edited the answer.
$endgroup$
– echinodermata
Mar 15 at 5:48
add a comment |
$begingroup$
This is a question about terminology and its usage in practice, so the basis for an answer should come from real quotations and the historical record rather than unsupported opinion. Looking at early known uses of the term, the few sources that I examined all used "disjoint" in its pairwise sense rather than in the sense of having empty intersection.
The earliest published use (known to me) of the term "disjoint" for its mathematical meaning is in a paper "The Thesis of Modern Logistic" (1909), which is the earliest such use found in a JSTOR search and the earliest such use listed in Earliest Known Uses of
Some of the Words of Mathematics. (Note: Here and throughout, I do not claim that the quoted sources are truly the earliest sources, but merely the earliest ones I was able to find. It is clear that readily available search tools today cover only a small part of historical record. Especially missing from my research is German papers and books, which surely are a large part of the early history of set theory.)
This first paper uses "disjoint" in its description of a construction similar to what we now call an $n$-ary Cartesian product, but where the members of the product are sets rather than ordered $n$-tuples:
Multiplication of cardinals is also defined in purely logical terms. This is done by means of the concept (due to Whitehead) of multiplicative class, which is itself given in terms of logical constants: $k$ being a class of disjoint classes, the multiplicative class of $k$ is the class of all the classes each of which contains one and but one term of each class in $k$. [Italics present in the original; bold emphasis added.]
Although this paper does not state an explicit definition of "disjoint" for more than two "classes", the construction described in the quote requires "disjoint" to mean pairwise disjoint.
The second earliest use I found is in a paper "Differentiation with Respect to a Function of Limited Variation" (1918). This gives an explicit definition of "disjoint intervals" as "intervals with no points common to any two".
When "pairwise disjoint" appeared, it was not meant to change the meaning of "disjoint", but rather to clarify that the author's meaning of "disjoint" was always pairwise. The earliest use I found is "Integral Forms and Variational Orthogonality" (1938):
"Let $U = U_epsilon = sum_j=1^infty [u_j, u'_j)$ be an enumerable set of pairwise disjoint, half-open intervals which contains $S$."
Here, "pairwise disjoint" has the modern meaning we expect. Yet the same paper also uses "disjoint" by itself to mean the same thing as "pairwise disjoint":
if $S$ is a set consisting of a finite number of disjoint half-open intervals $[u_j, u'_j), j = 1, ..., s$
This usage of "disjoint" occurs earlier in the paper than "pairwise disjoint". To this author, the two terms are not contrasting, but synonymous. This view is the same view that you expressed in your question: "pairwise disjoint" is already the default meaning of "disjoint", but can optionally be added as extra clarification in case the reader believes in the other potential interpretation of "disjoint".
answered Mar 10 at 20:51
echinodermataechinodermata
2,52911228
$endgroup$
1
$begingroup$
Thanks for the great research work. I do not agree with your conclusion about the first citation ("The Thesis of Modern Logistic" by C. J. Keyser). What is being described is defining multiplication of cardinals by taking essentially the cardinal of a Cartesian product of "factors" (sets), but using sets (classes) rather than tuples to form the product. For a set containing exactly one element from each factor to faithfully represent an element of the Cartesian product of the factors, it is essential that no two factors have an element in common. So: the factors must be pairwise disjoint.
$endgroup$
– Marc van Leeuwen
Mar 11 at 11:04
$begingroup$
... What I just said is not quite exact: if two or more factors share an element, then the problem is not so much that a set containing exactly one element from each factor fails to identify an element of their Cartesian product, the problem is that some elements of the Cartesian product (i.e., some tuples) cannot be represented by such a set. This is because the set of components of the tuple would have more than one element in common with certain factors, so fail the requirement. But it remains: the factors need to be pairwise disjoint for the construction to define cardinal multiplication.
$endgroup$
– Marc van Leeuwen
Mar 11 at 11:14
$begingroup$
For concreteness, $k=0,0,1,1$ is not a multiplicative class, though $bigcap k =emptyset$. Indeed the "class of all the classes each of which contains one and but one term of each class in $k$" is empty, as one easily sees; however the cardinal multiplication should result in a class with $1times2times1=2$ members. For a disjoint class of classes, one would get the right cardinal for the multiplicative class. Final remark: maybe only multiplication of $2$ cardinals is intended here; then of course "disjoint" only applies to two sets and one gets no verdict at all from it
$endgroup$
– Marc van Leeuwen
Mar 11 at 11:36
$begingroup$
@MarcvanLeeuwen I agree - thanks for the important correction. It definitely means pairwise in the first quote. I edited the answer.
$endgroup$
– echinodermata
Mar 15 at 5:48
add a comment |
$begingroup$
This is a question about terminology and its usage in practice, so the basis for an answer should come from real quotations and the historical record rather than unsupported opinion. Looking at early known uses of the term, the few sources that I examined all used "disjoint" in its pairwise sense rather than in the sense of having empty intersection.
The earliest published use (known to me) of the term "disjoint" for its mathematical meaning is in a paper "The Thesis of Modern Logistic" (1909), which is the earliest such use found in a JSTOR search and the earliest such use listed in Earliest Known Uses of
Some of the Words of Mathematics. (Note: Here and throughout, I do not claim that the quoted sources are truly the earliest sources, but merely the earliest ones I was able to find. It is clear that readily available search tools today cover only a small part of historical record. Especially missing from my research is German papers and books, which surely are a large part of the early history of set theory.)
This first paper uses "disjoint" in its description of a construction similar to what we now call an $n$-ary Cartesian product, but where the members of the product are sets rather than ordered $n$-tuples:
Multiplication of cardinals is also defined in purely logical terms. This is done by means of the concept (due to Whitehead) of multiplicative class, which is itself given in terms of logical constants: $k$ being a class of disjoint classes, the multiplicative class of $k$ is the class of all the classes each of which contains one and but one term of each class in $k$. [Italics present in the original; bold emphasis added.]
Although this paper does not state an explicit definition of "disjoint" for more than two "classes", the construction described in the quote requires "disjoint" to mean pairwise disjoint.
The second earliest use I found is in a paper "Differentiation with Respect to a Function of Limited Variation" (1918). This gives an explicit definition of "disjoint intervals" as "intervals with no points common to any two".
When "pairwise disjoint" appeared, it was not meant to change the meaning of "disjoint", but rather to clarify that the author's meaning of "disjoint" was always pairwise. The earliest use I found is "Integral Forms and Variational Orthogonality" (1938):
"Let $U = U_epsilon = sum_j=1^infty [u_j, u'_j)$ be an enumerable set of pairwise disjoint, half-open intervals which contains $S$."
Here, "pairwise disjoint" has the modern meaning we expect. Yet the same paper also uses "disjoint" by itself to mean the same thing as "pairwise disjoint":
if $S$ is a set consisting of a finite number of disjoint half-open intervals $[u_j, u'_j), j = 1, ..., s$
This usage of "disjoint" occurs earlier in the paper than "pairwise disjoint". To this author, the two terms are not contrasting, but synonymous. This view is the same view that you expressed in your question: "pairwise disjoint" is already the default meaning of "disjoint", but can optionally be added as extra clarification in case the reader believes in the other potential interpretation of "disjoint".
answered Mar 10 at 20:51
echinodermataechinodermata
2,52911228
$endgroup$
This is a question about terminology and its usage in practice, so the basis for an answer should come from real quotations and the historical record rather than unsupported opinion. Looking at early known uses of the term, the few sources that I examined all used "disjoint" in its pairwise sense rather than in the sense of having empty intersection.
The earliest published use (known to me) of the term "disjoint" for its mathematical meaning is in a paper "The Thesis of Modern Logistic" (1909), which is the earliest such use found in a JSTOR search and the earliest such use listed in Earliest Known Uses of
Some of the Words of Mathematics. (Note: Here and throughout, I do not claim that the quoted sources are truly the earliest sources, but merely the earliest ones I was able to find. It is clear that readily available search tools today cover only a small part of historical record. Especially missing from my research is German papers and books, which surely are a large part of the early history of set theory.)
This first paper uses "disjoint" in its description of a construction similar to what we now call an $n$-ary Cartesian product, but where the members of the product are sets rather than ordered $n$-tuples:
Multiplication of cardinals is also defined in purely logical terms. This is done by means of the concept (due to Whitehead) of multiplicative class, which is itself given in terms of logical constants: $k$ being a class of disjoint classes, the multiplicative class of $k$ is the class of all the classes each of which contains one and but one term of each class in $k$. [Italics present in the original; bold emphasis added.]
Although this paper does not state an explicit definition of "disjoint" for more than two "classes", the construction described in the quote requires "disjoint" to mean pairwise disjoint.
The second earliest use I found is in a paper "Differentiation with Respect to a Function of Limited Variation" (1918). This gives an explicit definition of "disjoint intervals" as "intervals with no points common to any two".
When "pairwise disjoint" appeared, it was not meant to change the meaning of "disjoint", but rather to clarify that the author's meaning of "disjoint" was always pairwise. The earliest use I found is "Integral Forms and Variational Orthogonality" (1938):
"Let $U = U_epsilon = sum_j=1^infty [u_j, u'_j)$ be an enumerable set of pairwise disjoint, half-open intervals which contains $S$."
Here, "pairwise disjoint" has the modern meaning we expect. Yet the same paper also uses "disjoint" by itself to mean the same thing as "pairwise disjoint":
if $S$ is a set consisting of a finite number of disjoint half-open intervals $[u_j, u'_j), j = 1, ..., s$
This usage of "disjoint" occurs earlier in the paper than "pairwise disjoint". To this author, the two terms are not contrasting, but synonymous. This view is the same view that you expressed in your question: "pairwise disjoint" is already the default meaning of "disjoint", but can optionally be added as extra clarification in case the reader believes in the other potential interpretation of "disjoint".
answered Mar 10 at 20:51
echinodermataechinodermata
2,52911228
edited Mar 15 at 5:20
answered Mar 10 at 20:51
echinodermataechinodermata
2,52911228
answered Mar 10 at 20:51
echinodermataechinodermata
2,52911228
answered Mar 10 at 20:51
echinodermataechinodermata
2,52911228
2,52911228
1
$begingroup$
Thanks for the great research work. I do not agree with your conclusion about the first citation ("The Thesis of Modern Logistic" by C. J. Keyser). What is being described is defining multiplication of cardinals by taking essentially the cardinal of a Cartesian product of "factors" (sets), but using sets (classes) rather than tuples to form the product. For a set containing exactly one element from each factor to faithfully represent an element of the Cartesian product of the factors, it is essential that no two factors have an element in common. So: the factors must be pairwise disjoint.
$endgroup$
– Marc van Leeuwen
Mar 11 at 11:04
$begingroup$
... What I just said is not quite exact: if two or more factors share an element, then the problem is not so much that a set containing exactly one element from each factor fails to identify an element of their Cartesian product, the problem is that some elements of the Cartesian product (i.e., some tuples) cannot be represented by such a set. This is because the set of components of the tuple would have more than one element in common with certain factors, so fail the requirement. But it remains: the factors need to be pairwise disjoint for the construction to define cardinal multiplication.
$endgroup$
– Marc van Leeuwen
Mar 11 at 11:14
$begingroup$
For concreteness, $k=0,0,1,1$ is not a multiplicative class, though $bigcap k =emptyset$. Indeed the "class of all the classes each of which contains one and but one term of each class in $k$" is empty, as one easily sees; however the cardinal multiplication should result in a class with $1times2times1=2$ members. For a disjoint class of classes, one would get the right cardinal for the multiplicative class. Final remark: maybe only multiplication of $2$ cardinals is intended here; then of course "disjoint" only applies to two sets and one gets no verdict at all from it
$endgroup$
– Marc van Leeuwen
Mar 11 at 11:36
$begingroup$
@MarcvanLeeuwen I agree - thanks for the important correction. It definitely means pairwise in the first quote. I edited the answer.
$endgroup$
– echinodermata
Mar 15 at 5:48
add a comment |
1
$begingroup$
Thanks for the great research work. I do not agree with your conclusion about the first citation ("The Thesis of Modern Logistic" by C. J. Keyser). What is being described is defining multiplication of cardinals by taking essentially the cardinal of a Cartesian product of "factors" (sets), but using sets (classes) rather than tuples to form the product. For a set containing exactly one element from each factor to faithfully represent an element of the Cartesian product of the factors, it is essential that no two factors have an element in common. So: the factors must be pairwise disjoint.
$endgroup$
– Marc van Leeuwen
Mar 11 at 11:04
$begingroup$
... What I just said is not quite exact: if two or more factors share an element, then the problem is not so much that a set containing exactly one element from each factor fails to identify an element of their Cartesian product, the problem is that some elements of the Cartesian product (i.e., some tuples) cannot be represented by such a set. This is because the set of components of the tuple would have more than one element in common with certain factors, so fail the requirement. But it remains: the factors need to be pairwise disjoint for the construction to define cardinal multiplication.
$endgroup$
– Marc van Leeuwen
Mar 11 at 11:14
$begingroup$
For concreteness, $k=0,0,1,1$ is not a multiplicative class, though $bigcap k =emptyset$. Indeed the "class of all the classes each of which contains one and but one term of each class in $k$" is empty, as one easily sees; however the cardinal multiplication should result in a class with $1times2times1=2$ members. For a disjoint class of classes, one would get the right cardinal for the multiplicative class. Final remark: maybe only multiplication of $2$ cardinals is intended here; then of course "disjoint" only applies to two sets and one gets no verdict at all from it
$endgroup$
– Marc van Leeuwen
Mar 11 at 11:36
$begingroup$
@MarcvanLeeuwen I agree - thanks for the important correction. It definitely means pairwise in the first quote. I edited the answer.
$endgroup$
– echinodermata
Mar 15 at 5:48
1
1
$begingroup$
Thanks for the great research work. I do not agree with your conclusion about the first citation ("The Thesis of Modern Logistic" by C. J. Keyser). What is being described is defining multiplication of cardinals by taking essentially the cardinal of a Cartesian product of "factors" (sets), but using sets (classes) rather than tuples to form the product. For a set containing exactly one element from each factor to faithfully represent an element of the Cartesian product of the factors, it is essential that no two factors have an element in common. So: the factors must be pairwise disjoint.
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– Marc van Leeuwen
Mar 11 at 11:04
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Thanks for the great research work. I do not agree with your conclusion about the first citation ("The Thesis of Modern Logistic" by C. J. Keyser). What is being described is defining multiplication of cardinals by taking essentially the cardinal of a Cartesian product of "factors" (sets), but using sets (classes) rather than tuples to form the product. For a set containing exactly one element from each factor to faithfully represent an element of the Cartesian product of the factors, it is essential that no two factors have an element in common. So: the factors must be pairwise disjoint.
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– Marc van Leeuwen
Mar 11 at 11:04
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... What I just said is not quite exact: if two or more factors share an element, then the problem is not so much that a set containing exactly one element from each factor fails to identify an element of their Cartesian product, the problem is that some elements of the Cartesian product (i.e., some tuples) cannot be represented by such a set. This is because the set of components of the tuple would have more than one element in common with certain factors, so fail the requirement. But it remains: the factors need to be pairwise disjoint for the construction to define cardinal multiplication.
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– Marc van Leeuwen
Mar 11 at 11:14
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... What I just said is not quite exact: if two or more factors share an element, then the problem is not so much that a set containing exactly one element from each factor fails to identify an element of their Cartesian product, the problem is that some elements of the Cartesian product (i.e., some tuples) cannot be represented by such a set. This is because the set of components of the tuple would have more than one element in common with certain factors, so fail the requirement. But it remains: the factors need to be pairwise disjoint for the construction to define cardinal multiplication.
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– Marc van Leeuwen
Mar 11 at 11:14
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For concreteness, $k=0,0,1,1$ is not a multiplicative class, though $bigcap k =emptyset$. Indeed the "class of all the classes each of which contains one and but one term of each class in $k$" is empty, as one easily sees; however the cardinal multiplication should result in a class with $1times2times1=2$ members. For a disjoint class of classes, one would get the right cardinal for the multiplicative class. Final remark: maybe only multiplication of $2$ cardinals is intended here; then of course "disjoint" only applies to two sets and one gets no verdict at all from it
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– Marc van Leeuwen
Mar 11 at 11:36
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For concreteness, $k=0,0,1,1$ is not a multiplicative class, though $bigcap k =emptyset$. Indeed the "class of all the classes each of which contains one and but one term of each class in $k$" is empty, as one easily sees; however the cardinal multiplication should result in a class with $1times2times1=2$ members. For a disjoint class of classes, one would get the right cardinal for the multiplicative class. Final remark: maybe only multiplication of $2$ cardinals is intended here; then of course "disjoint" only applies to two sets and one gets no verdict at all from it
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– Marc van Leeuwen
Mar 11 at 11:36
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@MarcvanLeeuwen I agree - thanks for the important correction. It definitely means pairwise in the first quote. I edited the answer.
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– echinodermata
Mar 15 at 5:48
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@MarcvanLeeuwen I agree - thanks for the important correction. It definitely means pairwise in the first quote. I edited the answer.
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– echinodermata
Mar 15 at 5:48
add a comment |
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In this context disjoint means $A cap B cap C = emptyset$.
answered Mar 8 at 15:52
Umberto P.Umberto P.
40.3k13370
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12
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Is that a standard meaning? I never saw the term formally defined that way in four years of undergrad math classes.
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– Connor Harris
Mar 8 at 15:52
1
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me neither, but four people have answered the question this way in four minutes!
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– John Lawrence Aspden
Mar 8 at 15:57
2
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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.
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– Umberto P.
Mar 8 at 16:06
1
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This is not the standard definition of disjoint. Google it if you like.
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– Marc van Leeuwen
Mar 8 at 19:47
5
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@Brilliand Indeed, somebody trying to find evidence for the fact that disjoint means empty intersection might land on the current page and find plenty of evidence to support it. But if you bluntly proclaim as this answer does that this is what the term means than it does not hurt to check that you can find evidence of a text using it that way, and in spite of the multitude of answers here in the same sense, nobody has been able to do that.
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– Marc van Leeuwen
Mar 9 at 5:54
|
show 3 more comments
$begingroup$
In this context disjoint means $A cap B cap C = emptyset$.
answered Mar 8 at 15:52
Umberto P.Umberto P.
40.3k13370
$endgroup$
12
$begingroup$
Is that a standard meaning? I never saw the term formally defined that way in four years of undergrad math classes.
$endgroup$
– Connor Harris
Mar 8 at 15:52
1
$begingroup$
me neither, but four people have answered the question this way in four minutes!
$endgroup$
– John Lawrence Aspden
Mar 8 at 15:57
2
$begingroup$
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.
Mar 8 at 16:06
1
$begingroup$
This is not the standard definition of disjoint. Google it if you like.
$endgroup$
– Marc van Leeuwen
Mar 8 at 19:47
5
$begingroup$
@Brilliand Indeed, somebody trying to find evidence for the fact that disjoint means empty intersection might land on the current page and find plenty of evidence to support it. But if you bluntly proclaim as this answer does that this is what the term means than it does not hurt to check that you can find evidence of a text using it that way, and in spite of the multitude of answers here in the same sense, nobody has been able to do that.
$endgroup$
– Marc van Leeuwen
Mar 9 at 5:54
|
show 3 more comments
$begingroup$
In this context disjoint means $A cap B cap C = emptyset$.
answered Mar 8 at 15:52
Umberto P.Umberto P.
40.3k13370
$endgroup$
In this context disjoint means $A cap B cap C = emptyset$.
answered Mar 8 at 15:52
Umberto P.Umberto P.
40.3k13370
answered Mar 8 at 15:52
Umberto P.Umberto P.
40.3k13370
answered Mar 8 at 15:52
Umberto P.Umberto P.
40.3k13370
answered Mar 8 at 15:52
Umberto P.Umberto P.
40.3k13370
40.3k13370
12
$begingroup$
Is that a standard meaning? I never saw the term formally defined that way in four years of undergrad math classes.
$endgroup$
– Connor Harris
Mar 8 at 15:52
1
$begingroup$
me neither, but four people have answered the question this way in four minutes!
$endgroup$
– John Lawrence Aspden
Mar 8 at 15:57
2
$begingroup$
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.
Mar 8 at 16:06
1
$begingroup$
This is not the standard definition of disjoint. Google it if you like.
$endgroup$
– Marc van Leeuwen
Mar 8 at 19:47
5
$begingroup$
@Brilliand Indeed, somebody trying to find evidence for the fact that disjoint means empty intersection might land on the current page and find plenty of evidence to support it. But if you bluntly proclaim as this answer does that this is what the term means than it does not hurt to check that you can find evidence of a text using it that way, and in spite of the multitude of answers here in the same sense, nobody has been able to do that.
$endgroup$
– Marc van Leeuwen
Mar 9 at 5:54
|
show 3 more comments
12
$begingroup$
Is that a standard meaning? I never saw the term formally defined that way in four years of undergrad math classes.
$endgroup$
– Connor Harris
Mar 8 at 15:52
1
$begingroup$
me neither, but four people have answered the question this way in four minutes!
$endgroup$
– John Lawrence Aspden
Mar 8 at 15:57
2
$begingroup$
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.
Mar 8 at 16:06
1
$begingroup$
This is not the standard definition of disjoint. Google it if you like.
$endgroup$
– Marc van Leeuwen
Mar 8 at 19:47
5
$begingroup$
@Brilliand Indeed, somebody trying to find evidence for the fact that disjoint means empty intersection might land on the current page and find plenty of evidence to support it. But if you bluntly proclaim as this answer does that this is what the term means than it does not hurt to check that you can find evidence of a text using it that way, and in spite of the multitude of answers here in the same sense, nobody has been able to do that.
$endgroup$
– Marc van Leeuwen
Mar 9 at 5:54
12
12
$begingroup$
Is that a standard meaning? I never saw the term formally defined that way in four years of undergrad math classes.
$endgroup$
– Connor Harris
Mar 8 at 15:52
$begingroup$
Is that a standard meaning? I never saw the term formally defined that way in four years of undergrad math classes.
$endgroup$
– Connor Harris
Mar 8 at 15:52
1
1
$begingroup$
me neither, but four people have answered the question this way in four minutes!
$endgroup$
– John Lawrence Aspden
Mar 8 at 15:57
$begingroup$
me neither, but four people have answered the question this way in four minutes!
$endgroup$
– John Lawrence Aspden
Mar 8 at 15:57
2
2
$begingroup$
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.
Mar 8 at 16:06
$begingroup$
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.
Mar 8 at 16:06
1
1
$begingroup$
This is not the standard definition of disjoint. Google it if you like.
$endgroup$
– Marc van Leeuwen
Mar 8 at 19:47
$begingroup$
This is not the standard definition of disjoint. Google it if you like.
$endgroup$
– Marc van Leeuwen
Mar 8 at 19:47
5
5
$begingroup$
@Brilliand Indeed, somebody trying to find evidence for the fact that disjoint means empty intersection might land on the current page and find plenty of evidence to support it. But if you bluntly proclaim as this answer does that this is what the term means than it does not hurt to check that you can find evidence of a text using it that way, and in spite of the multitude of answers here in the same sense, nobody has been able to do that.
$endgroup$
– Marc van Leeuwen
Mar 9 at 5:54
$begingroup$
@Brilliand Indeed, somebody trying to find evidence for the fact that disjoint means empty intersection might land on the current page and find plenty of evidence to support it. But if you bluntly proclaim as this answer does that this is what the term means than it does not hurt to check that you can find evidence of a text using it that way, and in spite of the multitude of answers here in the same sense, nobody has been able to do that.
$endgroup$
– Marc van Leeuwen
Mar 9 at 5:54
|
show 3 more comments
$begingroup$
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 Mar 8 at 15:53
Kyle DuffyKyle Duffy
843
$endgroup$
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 Mar 8 at 15:53
Kyle DuffyKyle Duffy
843
$endgroup$
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 Mar 8 at 15:53
Kyle DuffyKyle Duffy
843
$endgroup$
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 Mar 8 at 15:53
Kyle DuffyKyle Duffy
843
answered Mar 8 at 15:53
Kyle DuffyKyle Duffy
843
answered Mar 8 at 15:53
Kyle DuffyKyle Duffy
843
answered Mar 8 at 15:53
Kyle DuffyKyle Duffy
843
843
add a comment |
add a comment |
$begingroup$
If disjoint is understood as having an empty intersection, then disjoint sets are not necessarily pairwise disjoint. Pairwise disjoint avoid any doubt, at no cost.
answered Mar 14 at 17:19
Yves DaoustYves Daoust
132k676230
$endgroup$
add a comment |
$begingroup$
If disjoint is understood as having an empty intersection, then disjoint sets are not necessarily pairwise disjoint. Pairwise disjoint avoid any doubt, at no cost.
answered Mar 14 at 17:19
Yves DaoustYves Daoust
132k676230
$endgroup$
add a comment |
$begingroup$
If disjoint is understood as having an empty intersection, then disjoint sets are not necessarily pairwise disjoint. Pairwise disjoint avoid any doubt, at no cost.
answered Mar 14 at 17:19
Yves DaoustYves Daoust
132k676230
$endgroup$
If disjoint is understood as having an empty intersection, then disjoint sets are not necessarily pairwise disjoint. Pairwise disjoint avoid any doubt, at no cost.
answered Mar 14 at 17:19
Yves DaoustYves Daoust
132k676230
answered Mar 14 at 17:19
Yves DaoustYves Daoust
132k676230
answered Mar 14 at 17:19
Yves DaoustYves Daoust
132k676230
answered Mar 14 at 17:19
Yves DaoustYves Daoust
132k676230
132k676230
add a comment |
add a comment |
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Possible duplicate of pairwise disjoint , or disjoint
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– YuiTo Cheng
Mar 9 at 9:35
5
$begingroup$
Note to Reviewers: the target post provided by YuiTo Cheng has much less content. Here the question post is poor but there are some good answers demonstrating the lack of consensus on this terminology.
$endgroup$
– Lee David Chung Lin
Mar 10 at 0:24
2
$begingroup$
This comment is not directly about sets, but about a similar situation that arises with relative primality of integers. The numbers 6, 10, and 15 are relatively prime as a triple ($gcd(6,10,15) = 1$) but they are not pairwise relatively prime. For positive integers $a_1, ldots, a_n$ where $n geq 3$, there is a distinction between being pairwise relatively prime ($gcd(a_i,a_j) = 1$ for all $i not= j$) and being relatively prime as an $n$-tuple ($gcd(a_1, ldots, a_n) = 1$). If you want to say $n$ positive integers are "relatively prime" and $n geq 3$ then you should be more precise.
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– KCd
Mar 10 at 14:06