I've been reading different books, and I found two slightly different definitions for a family of sets $\{A_i\}_{i \in I}$ to be "Pairwise disjoint" or simply "Disjoint" mainly:

Definition 1 Let $I$ be a nonempty set, and let $\{A_i\}_{i \in I}$ be a family of nonempty sets indexed by $I$, This family is pairwise disjoint if $i,j\in I$ such that $i \neq j$ and $A_i \neq A_j$ imply that $A_i\cap A_j=\varnothing$.

Definition 2 Let $I$ be a nonempty set, and let $\{A_i\}_{i \in I}$ be a family of nonempty sets indexed by $I$, This family is pairwise disjoint if $A_i\cap A_j=\varnothing$ for all $i,j\in I$ such that $i \neq j$ .

The subtle different is that definition 1 allows that the family may have different indexes for the same set. Now I've been reading about the axiom of choice, and I found the following formulation on this question

If $A$ is a family of pairwise disjoint, non-empty sets then there is $C$ such that for all $a\in A$, $|a\cap C|=1$.

My first question: Is this formulation of the axiom of choice valid for both definition of pairwise disjoint family of sets in the sense that this formulation would remain equivalent to others statements like Zorn's lema, Existing of a choice function, etc. Or in the sense that we could find a counter example if we chose one of the definitions. Furthermore, my Second question is; if I found some others statements referring to "Pairwise disjoint family of sets" without specifying if we are using defnition one or two what should I do?.

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    $\begingroup$ I find both of these books odd; there's no reason to get indexing mixed up in the notion of pairwise disjointness. $\endgroup$ Feb 19, 2020 at 5:53
  • $\begingroup$ Definition 1) is not accepted by Mathematicians. We do not consider $\{A.A,A,...\}$ as a disjoint collection of sets if $A$ is non-empty. $\endgroup$ Feb 19, 2020 at 5:54
  • $\begingroup$ @KaviRamaMurthy - But a constant function $i\mapsto A$ for $i\in I$ is perfectly well accepted, which is what an indexed family can do. $\endgroup$ Feb 19, 2020 at 5:56
  • $\begingroup$ A not complicated definition: a (not empty) collection C of (not empty) sets is pairwise disjoint when for all distinct A,B in C, A and B are disjoint. $\endgroup$ Feb 19, 2020 at 6:13
  • $\begingroup$ if we have an equivalence relation $R$ on a set $X$, then I'm sure we all think that $X/R = \{ [x] \mid x\in X \}$is a collection of "Pairwise disjoint sets" but what would you say of ,$X/R = \{[x]\}_{x\in X}$ ? $\endgroup$
    – Chris.JL
    Feb 19, 2020 at 21:00

1 Answer 1


I have never seen "pairwise disjoint family indexed by $I$" used to mean Definition 1; it always means Definition 2 in my experience. Note though that you can also talk about a pairwise disjoint family where "family" just means "set" rather than "indexed set". In that case, Definition 1 is what it means for the set (not the indexed set) $\{A_i\}_{i\in I}$ to be pairwise disjoint. Incidentally, I strongly recommend not using the same notation for a set and an indexed set; I would use $(A_i)_{i\in I}$ for the indexed set (which is really a function with domain $I$) and $\{A_i\}_{i\in I}$ for the set (which is the image of that function). On a related note, if you are using "family" to mean indexed set rather than just set, your statement of the axiom of choice should be

If $(A_i)_{i\in I}$ is a family of pairwise disjoint, non-empty sets then there is $C$ such that for all $i\in I$, $|A_i\cap C|=1$.

The version you stated cannot be used with either of your definitions since neither definition defines what it means for a set $A$ (rather than an indexed set $(A_i)_{i\in I}$) to be a pairwise disjoint family. (Of course, you can make such a definition, and get a third version of the axiom of choice. It is trivial to see that this third version is equivalent to the axiom of choice using Definition 2, since given any non-indexed family $A$ you can make it an indexed family by the identity map $A\to A$ (that is, you take $I=A$ and $A_i=i$).)

In any case, the axiom of choice for Definition 1 is equivalent for the axiom of choice for Definition 2. The forward direction is trivial; for the reverse direction, assume the axiom of choice for Definition 2 and let $(A_i)_{i\in I}$ be a family of pairwise disjoint sets by Definition 1. Let $J=\{A_i:i\in I\}$ and consider the indexed set $(B_j)_{j\in J}$ defined by $B_j=j$.

I claim $(B_j)_{j\in J}$ is a pairwise disjoint family according to Definition 2. Indeed, suppose $j,j'\in J$ and $j\neq j'$. By definition of $J$, there exist $i,i'\in I$ such that $j=A_i$ and $j'=A_{i'}$. Then $A_i\neq A_{i'}$ and $i\neq i'$ (since $i\mapsto A_i$ is a function), so since $(A_i)_{i\in I}$ satisfies Definition 1, $A_i\cap A_{i'}=\emptyset$. But $A_i=j=B_j$ and $A_{i'}=j'=B_{j'}$, so $B_j\cap B_j=\emptyset$.

Thus by the axiom of choice for Definition 2, there exists a set $C$ such that $|B_j\cap C|=1$ for each $j\in J$. Now for any $i\in I$, let $j=A_i$, so $A_i=B_j$. We thus see that $|A_i\cap C|=|B_j\cap C|=1$. So, this set $C$ also has the required property for the family $(A_i)_{i\in I}$.

  • $\begingroup$ Marvelous this is exactly what I was waiting for an answer, really nice trick when defining $J$. I also appreciate the notation corrections. Thank you for your time and now I'm going to add some of the sources from where I extract the "odd" definition. $\endgroup$
    – Chris.JL
    Feb 19, 2020 at 15:59

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