A description/proof of that for set of sets, dichotomically nondisjoint does not necessarily mean pair-wise nondisjoint
Topics
About: set
The table of contents of this article
Starting Context
- The reader knows a definition of dichotomically disjoint set of sets.
Target Context
- The reader will have a description and a proof of the proposition that any set of sets is not necessarily pair-wise nondisjoint if it is dichotomically nondisjoint.
Orientation
There is a list of definitions discussed so far in this site.
There is a list of propositions discussed so far in this site.
Main Body
1: Note
'dichotomically nondisjoint' is the same with 'not dichotomically disjoint'. The both mean that there is no disjoint dichotomy.
'pair-wise nondisjoint' is different from 'not pair-wise disjoint'. The former means that any pair is not disjoint, while the latter means that at least 1 pair is not disjoint while some pairs may be disjoint.
2: Description
For any set of sets, \(\{S_\alpha\vert \alpha \in A_1\}\), where \(A\) is any possibly uncountable indices set, it is not necessarily pair-wise nondisjoint if it is dichotomically nondisjoint.
3: Proof
A counter example will suffice. Let the set be \(\{S_1, S_2, S_3\}\) such that \(S_1 \cap S_2 \neq \emptyset\), \(S_2 \cap S_3 \neq \emptyset\), and \(S_3 \cap S_1 = \emptyset\). It is dichotomically nondisjoint, but is not pair-wise nondisjoint.
