A description/proof of that for monotone operation from ordinal numbers collection into ordinal numbers collection, value equals or contains argument
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 transfinite induction principle.
Target Context
- The reader will have a description and a proof of the proposition that for any monotone operation from the all the ordinal numbers collection into the all the ordinal numbers collection and any argument, the value equals or contains the argument.
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, \(f: O \rightarrow O\), from the all the ordinal numbers collection into the all the ordinal numbers collection where \(O\) is the all the ordinal numbers collection and any argument, \(o \in O\), \(o \in= f (o)\) where \(\in=\) means being \(\in\) or \(=\).
2: Proof
\(0 \in= f (0)\).
Let us suppose that for any \(o_0 \in O\), for any \(o_1 \in o_0\), \(o_1 \in= f (o_1)\). Then, \(o_0 \in= f (o_0)\)?
Let us suppose that \(o_0\) is a successor ordinal number, \(o_0 = {o_2}^+\). \(o_2 \in o_0\), so, \(o_2 \in= f (o_2)\). If \(o_2 \in f (o_2)\), \({o_2}^+ \in= f (o_2)\), but \(f (o_2) \in f ({o_2}^+)\), so, \({o_2}^+ \in f ({o_2}^+)\). If \(o_2 = f (o_2)\), \({o_2}^+ = (f (o_2))^+\), but \((f (o_2))^+ \in= f ({o_2}^+)\), so, \({o_2}^+ \in= f ({o_2}^+)\).
Let us suppose that \(o_0\) is a limit ordinal number. \(o_0 = \cup \{o\vert o \in o_0\}\). \(f (o) \in f (o_0)\) for any \(o \in o_0\). \(sup \{f (o)\vert o \in o_0\} \in= f (o_0)\), because \(f (o_0)\) is an upper bound of \(\{f (o)\vert o \in o_0\}\). \(o_0 = sup \{o\vert o \in o_0\}\) by the way. As \(o \in= f (o)\), \(sup \{o\vert o \in o_0\} \in= sup \{f (o)\vert o \in o_0\}\). \(o_0 = sup \{o\vert o \in o_0\} \in= sup \{f (o)\vert o \in o_0\} \in= f (o_0)\).
So, yes for the question.
By the transfinite induction principle, \(o_0 \in= f (o_0)\) for any ordinal number, \(o_0 \in O\).
