A definition of dichotomically disjoint set of sets
Topics
About: set
The table of contents of this article
Starting Context
- The reader knows a definition of topological space.
- The reader knows a definition of continuous map.
Target Context
- The reader will have a definition of dichotomically disjoint set of sets.
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: Definition
Any set of sets, \(\{S_\alpha\vert \alpha \in A_1\}\), where \(A\) is any possibly uncountable indices set, such that there is a dichotomy, \(A_1 = A_2 \cup A_3\), such that \(A_2, A_3 \neq \emptyset\) and \(A_2 \cap A_3 = \emptyset\) and \((\cup_{\alpha \in A_2} S_\alpha) \cap (\cup_{\alpha \in A_3} S_\alpha) = \emptyset\)
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 \(S_1 \cap S_2 = \emptyset\), \(S_2 \cap S_3 \neq \emptyset\), and \(S_3 \cap S_1 = \emptyset\), it is dichotomically disjoint as \(S_1 \cap (S_2 \cup S_3) = \emptyset\), but is not pair-wise disjoint. If pair-wise disjoint, dichotomically disjoint, because \(A_2 = \{\alpha_0\}\) and \(A_3 = \{\alpha \in A_1\vert \alpha \neq \alpha_0\}\) for any \(\alpha_0 \in A_1\) will do, for example.
