# Why do we say ‘pairwise disjoint’, rather than ‘disjoint’?

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?

• Possible duplicate of pairwise disjoint , or disjoint – YuiTo Cheng Mar 9 '19 at 9:35
• 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. – Lee David Chung Lin Mar 10 '19 at 0:24
• 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. – KCd Mar 10 '19 at 14:06

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 $$A\cap B=\emptyset$$, 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.

• 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 $\mathbf{Set}$ 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. – Potato44 Mar 9 '19 at 4:48
• @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". – Marc van Leeuwen Mar 9 '19 at 9:15
• 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. – Mdoc Mar 28 '19 at 2:17

$$\{1,2\},\{2,3\},\{1,3\}$$ are disjoint but not pairwise disjoint.

• Really? Who would call those disjoint sets? – John Lawrence Aspden Mar 8 '19 at 15:52
• Everyone. Disjoint means their intersection is empty. – saulspatz Mar 8 '19 at 15:53
• 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. – drhab Mar 8 '19 at 16:16
• So be it. But actually I am the living proof that "everyone" in your first comment is not correct. Lots of times I heard "them" say things like: "we have $\mu(\bigcup_{n=1}^{\infty}A_n)=\sum_{n=1}^{\infty}\mu(A_n)$ if the sets are disjoint" (so leaving out pairwise). I do that myself too. It is a good thing however to have a proper definition of this excluding ambiguities. So let's look at this as abuse of language. – drhab Mar 8 '19 at 16:37
• These sets are not disjoint, at least not by the modern usage of the word “disjoint,” which afaik is used interchangeably with “pairwise disjoint.” – Shalop Mar 8 '19 at 17:33

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".

• 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)? – Marc van Leeuwen Mar 9 '19 at 12:30
• 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. – Arnaud Mortier Mar 10 '19 at 0:39

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.

• 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. – Marc van Leeuwen Mar 8 '19 at 19:45
• @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. – J.G. Mar 8 '19 at 19:51
• @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...". – Ryan Mar 9 '19 at 4:49
• @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. – Marc van Leeuwen Mar 9 '19 at 7:50
• @MarcvanLeeuwen I've amended my answer to reflect the usage consensus. – J.G. Mar 9 '19 at 8:17

In this context disjoint means $$A \cap B \cap C = \emptyset$$.

• Is that a standard meaning? I never saw the term formally defined that way in four years of undergrad math classes. – Connor Harris Mar 8 '19 at 15:52
• me neither, but four people have answered the question this way in four minutes! – John Lawrence Aspden Mar 8 '19 at 15:57
• 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. – Umberto P. Mar 8 '19 at 16:06
• This is not the standard definition of disjoint. Google it if you like. – Marc van Leeuwen Mar 8 '19 at 19:47
• @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. – Marc van Leeuwen Mar 9 '19 at 5:54

Let $$A=\{1,2\}, B=\{2,3\},C=\{3,4\}$$. Then the sets are disjoint because $$A\cap B\cap C=\emptyset$$, but not pairwise disjoint because you have pairs such as $$A,B$$ such that $$A\cap B\not =\emptyset$$.

• Rats! What is the notation for an empty set? – Oscar Lanzi Mar 8 '19 at 15:54
• Thank you, @jg. – Oscar Lanzi Mar 8 '19 at 15:58
• I use \varnothing, "$\varnothing$". – Eric Towers Mar 9 '19 at 22:14

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".

• 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. – Marc van Leeuwen Mar 11 '19 at 11:04
• ... 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. – Marc van Leeuwen Mar 11 '19 at 11:14
• 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 $1\times2\times1=2$ 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 – Marc van Leeuwen Mar 11 '19 at 11:36
• @MarcvanLeeuwen I agree - thanks for the important correction. It definitely means pairwise in the first quote. I edited the answer. – echinodermata Mar 15 '19 at 5:48

Consider the sets $$A = \{1,2\}$$, $$B = \{2,3\}$$, $$C = \{3, 1\}$$. Then $$A\cap B\cap C = \varnothing$$, but $$A,B,C$$ are not pairwise disjoint.

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.