A description/proof of that union of dichotomically nondisjoint set of real intervals is real interval
Topics
About: set
The table of contents of this article
Starting Context
- The reader knows a definition of dichotomically disjoint set of sets.
- The reader knows a definition of \(\mathbb{R}\) interval.
Target Context
- The reader will have a description and a proof of the proposition that the union of any possibly uncountable dichotomically nondisjoint set of \(\mathbb{R}\) intervals is a \(\mathbb{R}\) interval.
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: Description
For any dichotomically nondisjoint set of \(\mathbb{R}\) intervals, \(\{I_\alpha\vert \alpha \in A\}\) where \(A\) is any possibly uncountable indices set, the union, \(\cup_{\alpha \in A} I_\alpha\), is an \(\mathbb{R}\) interval.
2: Proof
For any points, \(r_1, r_2 \in \cup_{\alpha \in A} I_\alpha\), such that \(r_1 \lt r_2\), for any \(r_3 \in \mathbb{R}\) such that \(r_1 \lt r_3 \lt r_2\), \(r_3 \in \cup_{\alpha \in A} I_\alpha\)? Let us suppose that \(r_3 \notin \cup_{\alpha \in A} I_\alpha\). Then, each \(I_\alpha\) would be entirely smaller than \(r_3\) or entirely larger than \(r_3\), because if \(I_\alpha\) contained both a point smaller than \(r_3\) and a point larger than \(r_3\), \(r_3\) would be contained in \(I_\alpha\), and so, in the union. So, there would be the dichotomy such that all the intervals that are smaller than \(r_3\) are in a part and all the intervals that are larger than \(r_3\) are in the other part. The dichotomy would be disjoint, a contradiction.
