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According to Wolfram MathWorld, a collection of sets $A_1, A_2, \ldots, A_n$ is said to be disjoint if $A_i \cap A_j = \emptyset$ for all $i \ne j$. In other words, 'disjoint' refers only to 'pairwise disjoint'.

I am looking for a name for a collection of sets where $A_1 \cap A_2 \ldots \cap A_n = \emptyset$ but the sets are not necessarily pairwise disjoint. I was hoping there would be a term like 'qualifier disjoint' to refer to this.

For example, $\{0,1\}, \{0,2\}$ and $\{1,2\}$ are not pairwise disjoint, but the intersection of all three sets is empty.

If there's not an accepted name for this, how should I best express the concept in writing (given that I will need to refer to it many times)?

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  • $\begingroup$ Why don't you make up a term? Say.. "totally disjoint" or something? $\endgroup$ – Cameron Williams Jul 25 '13 at 1:50
  • $\begingroup$ I've heard teachers consciously use the awkward term "jointly disjoint" to contrast with the stronger property of pairwise disjointness. It's usually easier just to state the sets "have empty intersection" and not make a phrase that parallels the pairwise disjoint case. $\endgroup$ – hardmath Jul 25 '13 at 1:55
  • $\begingroup$ a collection of sets $\{A_i\mid i\in I\}$ is said to have the finite intersection property (FIP) if the intersection of a finite subcollection is always non-empty. The term is pretty standard, and it is used in the context of compactness (Every closed family with FIP has non-empty intersection) or filters, which by definition have this property. Since your collection is finite and the intersection is empty, you could refer to it as a collection without FIP :-) $\endgroup$ – Stefan Hamcke Jul 25 '13 at 12:52
  • $\begingroup$ @hardmath There's 'disjoint' or 'mutually disjoint' $\endgroup$ – BCLC Dec 6 '15 at 17:48
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According to my advanced probability professor,

sets $A_1, A_2, ...$ are (mutually) disjoint if

$$\bigcap_{i=1}^{\infty} A_i = \emptyset $$

Sets $A_1, A_2, ...$ are pairwise disjoint if

$$A_i \cap A_j = \emptyset \ \forall i \ne j$$

Apparently, most texts use 'disjoint' to refer to 'pairwise disjoint'. Whenever a text uses 'pairwise disjoint', we can assume 'disjoint' refers to 'mutually disjoint'.

I think it's the same as 'pairwise distinct' and 'distinct'


Some stuff on Math SE/meta

Do Kolmogorov's axioms really need only disjointness rather than pairwise disjointness?

http://meta.math.stackexchange.com/questions/21560/should-these-be-simply-disjoint-instead-of-pairwise-disjoint

Are pairwise mutually exclusive events the same as mutually exclusive events?

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    $\begingroup$ No. In standard usage, "disjoint", "pairwise disjoint" and "mutually disjoint" (as far as that is used) are synonyms. $\endgroup$ – Daniel Fischer Dec 6 '15 at 21:00
  • $\begingroup$ @DanielFischer My advanced probability professor deducted points from people who said disjoint instead of pairwise disjoint $\endgroup$ – BCLC Dec 7 '15 at 4:45
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    $\begingroup$ Yes, and if an author decides to call the property yellow or feverish, that's valid too. A bad idea, but valid. Of course calling it "disjoint" makes more sense, but in a way it's worse, since it takes a term and deviates from its standard usage, which invites a lot of confusion. $\endgroup$ – Daniel Fischer Dec 7 '15 at 11:54
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    $\begingroup$ @DanielFischer Sigh. I just got into an argument with someone over disjoint meaning the intersection of all sets in a family is empty, which is weaker than pairwise disjointness. I was adamant that this is the correct interpretation/usage and that I was right. Then I googled it, stumbled here, and found out I'm wrong. :( $\endgroup$ – layman Dec 14 '16 at 3:51
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    $\begingroup$ If one says that a list of more than two items are distinct, than this means they are pairwise distinct, so "distinct" and "pairwise distinct" are the same thing. (If you want to say the items are not all the same, just say that.) Also "mutually" means the same as "pairwise". There are notions like "relatively prime" where adding "mutually" makes a difference, but "distinct" or "disjoint" or "disconnected" are not among them. $\endgroup$ – Marc van Leeuwen Mar 9 at 6:16

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