384: For Set of Sets, Dichotomically Nondisjoint Does Not Necessarily Mean Pair-Wise Nondisjoint

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A description/proof of that for set of sets, dichotomically nondisjoint does not necessarily mean pair-wise nondisjoint

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Topics

About: set

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The table of contents of this article

• Starting Context
• Target Context
• Orientation
• Main Body
• 1: Note
• 2: Description
• 3: Proof

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Starting Context

• The reader knows a definition of dichotomically disjoint set of sets.

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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.

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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.

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Main Body

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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.

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2: Description

For any set of sets, $\{S_{\alpha }|\alpha \in A_1\}$, where $A$ is any possibly uncountable indices set, it is not necessarily pair-wise nondisjoint if it is dichotomically nondisjoint.

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3: Proof

A counter example will suffice. Let the set be $\{S_1,S_2,S_3\}$ such that $S_1\cap S_2\ne \mathrm{\varnothing }$, $S_2\cap S_3\ne \mathrm{\varnothing }$, and $S_3\cap S_1=\mathrm{\varnothing }$. It is dichotomically nondisjoint, but is not pair-wise nondisjoint.

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References

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