383: Dichotomically Disjoint Set of Sets

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A definition of dichotomically disjoint set of sets

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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: Definition
• 2: Note

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Starting Context

• The reader knows a definition of topological space.
• The reader knows a definition of continuous map.

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Target Context

• The reader will have a definition of dichotomically disjoint set of sets.

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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: Definition

Any set of sets, $\{S_{\alpha }|\alpha \in A_1\}$, where $A$ is any possibly uncountable indices set, such that there is a dichotomy, $A_1=A_2\cup A_3$, such that $A_2,A_3\ne \mathrm{\varnothing }$ and $A_2\cap A_3=\mathrm{\varnothing }$ and $({\cup }_{\alpha \in A_2}{S}_{\alpha })\cap ({\cup }_{\alpha \in A_3}{S}_{\alpha })=\mathrm{\varnothing }$

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2: Note

'dichotomically disjoint' is different from 'pair-wise disjoint', because if pair-wise disjoint, dichotomically disjoint, but if dichotomically disjoint, not necessarily pair-wise disjoint. For example, for $\{S_1,S_2,S_3\}$ such that $S_1\cap S_2=\mathrm{\varnothing }$, $S_2\cap S_3\ne \mathrm{\varnothing }$, and $S_3\cap S_1=\mathrm{\varnothing }$, it is dichotomically disjoint as $S_1\cap (S_2\cup S_3)=\mathrm{\varnothing }$, but is not pair-wise disjoint. If pair-wise disjoint, dichotomically disjoint, because $A_2=\{{\alpha }_0\}$ and $A_3=\{\alpha \in A_1|\alpha \ne {\alpha }_0\}$ for any ${\alpha }_0\in A_1$ will do, for example.

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References

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