A description/proof of that for monotone continuous operation from ordinal numbers collection into ordinal numbers collection, image of limit ordinal number is limit ordinal number
Topics
About: set
The table of contents of this article
Starting Context
- The reader knows a definition of monotone operation.
- The reader knows a definition of continuous operation from all the ordinal numbers collection into all the ordinal numbers collection.
- The reader admits the proposition that any ordinal number is a limit ordinal number if and only if it is nonzero and is the union of its all the members.
- The reader admits the proposition that the inclusion relation satisfies the trichotomy on the ordinal numbers collection.
- The reader admits the proposition that the inclusion relation is equivalent with the membership relation for the ordinal numbers collection.
Target Context
- The reader will have a description and a proof of the proposition that for any monotone continuous operation from the ordinal numbers collection into the ordinal numbers collection, the image of any limit ordinal number is a limit ordinal number.
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 continuous operation, \(f: O \rightarrow O\) where \(O\) is the all the ordinal numbers collection, and any limit ordinal number, \(o_1\), \(o_2 := f (o_1)\) is a limit ordinal number.
2: Proof
By the definition of continuousness, \(o_2 = \cup_{o \in o_1} f (o)\). \(f (o) \in f (o_1) = o_2\) as \(f\) is monotone, and \(o_2\) is nonzero, because \(o_1\) is nonzero and there is at least 1 such an \(o\). So, \(\cup_{o \in o_1} f (o) \subseteq \cup_{o \in o_2} o\). The issue is that is there \(o' \in o_1\) such that \(o \subseteq f (o')\) for each \(o \in o_2\)? Let us suppose that there was no \(o' \in o_1\) such that \(o \subseteq f (o')\) for an \(o \in o_2\). By the proposition that the inclusion relation satisfies the trichotomy on the ordinal numbers collection, \(f (o') \subset o\) for every \(o' \in o_1\), so, \(o_2 = \cup_{o' \in o_1} f (o') \subset o \subset o_2\), because \(o \subset o_2\) if \(o \in o_2\) by the proposition that the inclusion relation is equivalent with the membership relation for the ordinal numbers collection, a contradiction. So, there is an \(o' \in o_1\) such that \(o \subseteq f (o')\) for each \(o \in o_2\), so, \(\cup_{o \in o_2} o \subseteq \cup_{o \in o_1} f (o)\). So, \(\cup_{o \in o_2} o = \cup_{o \in o_1} f (o) = o_2\). By the proposition that any ordinal number is a limit ordinal number if and only if it is nonzero and is the union of its all the members, \(o_2\) is a limit ordinal number.
