A description/proof of that for monotone ordinal numbers operation, 2 domain elements are in membership relation if corresponding images are in same relation
Topics
About: set
The table of contents of this article
Starting Context
- The reader knows a definition of ordinal number.
- The reader knows a definition of monotone operation.
- The reader admits the trichotomy of the membership relation on the ordinal numbers collection.
Target Context
- The reader will have a description and a proof of the proposition that for any monotone operation from the ordinal numbers collection into the ordinal numbers collection, any 2 domain elements are in a membership relation if the corresponding images are in the same membership relation.
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 monotone operation, \(f: O \rightarrow O\) where \(O\) is the ordinal numbers collection, and any 2 domain elements, \(o_1, o_2 \in O\), \(o_1 \in o_2\) if \(f (o_1) \in f (o_2)\); \(o_1 = o_2\) if \(f (o_1) = f (o_2)\).
2: Proof
Let us suppose that \(f (o_1) \in f (o_2)\). Let us suppose that \(\lnot o_1 \in o_2\). By the trichotomy of the membership relation on the ordinal numbers collection, \(o_1 = o_2\) or \(o_2 \in o_1\). If \(o_1 = o_2\), \(f (o_1) = f (o_2)\), a contradiction against \(f (o_1) \in f (o_2)\), by the trichotomy of the membership relation on the ordinal numbers collection. If \(o_2 \in o_1\), \(f (o_2) \in f (o_1)\), a contradiction against \(f (o_1) \in f (o_2)\), by the trichotomy of the membership relation on the ordinal numbers collection. So, \(o_1 \in o_2\).
Let us suppose that \(f (o_1) = f (o_2)\). Let us suppose that \(o_1 \neq o_2\). By the trichotomy of the membership relation on the ordinal numbers collection, \(o_1 \in o_2\) or \(o_2 \in o_1\). If \(o_1 \in o_2\), \(f (o_1) \in f (o_2)\), a contradiction against \(f (o_1) = f (o_2)\), by the trichotomy of the membership relation on the ordinal numbers collection. If \(o_2 \in o_1\), \(f (o_2) \in f (o_1)\), a contradiction against \(f (o_1) = f (o_2)\), by the trichotomy of the membership relation on the ordinal numbers collection. So, \(o_1 = o_2\).
