# Distinction between the notions of pairwise disjoint

1.12 There is a slight distinction between the notions of pairwise disjoint for nonindexed collections of sets and indexed collections of sets, namely, an indexed collections of set $$\{A_i\}_{i\in I}$$ can fail to be pairwise disjoint even though the collection, $$C=\{A_i:i\in I\}$$, is pairwise disjoint. Provide an example that illustrates this fact.

Two subsets $$A$$ and $$B$$ of $$X$$ are said to be disjoint if $$A\cap B=\varnothing$$.

A collection $$C$$ of subsets of $$X$$ is said to be pairwise disjoint if each two distinct members of $$C$$ are disjoint.

An indexed collection $$\{A_i\}_{i\in I}$$ of subsets of $$X$$ is said to be pairwise disjoint if $$A_i\cap A_j=\varnothing$$ whenever $$i\neq j$$.

This is a question from a book A course in real analysis, however I cannot find any example to verify the the distinction between different notions, I always think they are the same, even though some very special case such as index set is empty.

• Why not post the two definitions in question? That would be a good place to start on a problem like this. – Carl Mummert Nov 7 '14 at 11:58
• I guess the difference is, the non indexed version, takes identical copies just once. The indexed one treats them differently. You can view the indexed one as collection of set with disjoint union with their index – Loreno Heer Nov 7 '14 at 12:03
• @Carl I have add definition – noname1014 Nov 7 '14 at 12:11

If two sets $A,B$ are taken from collection $\mathcal C$ and $A\neq B$ implies in that situation that $A\cap B=\emptyset$ then the (non-indexed) collection $\mathcal C$ can be labeled as 'pairwise disjoint'. Looked at as an indexed collection more is needed: $i\neq j$ must imply that $A_i\cap A_j=\emptyset$
Underlying is the fact that you can have two distinct indices $i\neq j$ with $A_i=A_j$.
Take $A_i$ for $i=1,2,3$ and $A_1=A_2\neq\emptyset$ and $A_1\cap A_3=\emptyset$.
Then collection $\{A_1,A_2,A_3\}=\{A_1,A_3\}$ is pairwise disjoint, but indexed collection $\{A_i\}_{i\in\{1,2,3\}}$ is not.