A description/proof of that derived operation of monotone continuous operation from ordinal numbers collection into ordinal numbers collection is monotone continuous
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 knows a definition of derived operation of monotone continuous operation from all the ordinal numbers collection into all the ordinal numbers collection.
Target Context
- The reader will have a description and a proof of the proposition that the derived operation of any monotone continuous operation from the all the ordinal numbers collection into the all the ordinal numbers collection is monotone continuous.
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, the derived operation, \(f': O \rightarrow O\), is monotone and continuous.
2: Proof
\(f'\) is monotone, because it chooses fixed-points in the ascending order by definition.
As for the continuousness, for any limit ordinal number, \(o_0 \in O\), \(f' (o_0) = \cup_{o \in o_0} f' (o)\)? \(\cup_{o \in o_0} f' (o) = sup \{f' (o)\vert o \in o_0\} := o_1\). If \(o_1 = f (o_1)\), \(o_1 = f' (o_0)\), because \(o_1\) will be the smallest fixed-point except \(\{f' (o)\vert o \in o_0\}\). Then, \(o_1 = f (o_1)\)? \(f (f' (o)) = f' (o)\). Taking the supremum of the sets of the both sides, \(sup \{f (f' (o))\vert o \in o_0\} = sup \{f' (o)\vert o \in o_0\} = o_1\). But \(sup \{f (f' (o))\vert o \in o_0\} = f (o_1)\)? \(o_1\) is a limit ordinal number, because if it was a successor ordinal number, it would have the maximum in \(\{f' (o)\vert o \in o_0\}\), which would be impossible, because \(o_0\) is a limit ordinal number and if \(f' (o')\) was the maximum, there would be an \(o''\) such that \(o' \in o'' \in o_0\) and \(f' (o') \in f' (o'')\). So, \(f (o_1) = sup \{f (o)\vert o \in o_1\}\) as \(f\) is continuous, and \(f (o_1)\) is a limit ordinal number by the likewise reason. \(sup \{f (f' (o))\vert o \in o_0\} = sup \{f (o)\vert o \in o_1\}\)? For any \(o \in o_0\), \(f' (o) \in o_1\), so, \(f (f' (o))\) in the left hand side appears as \(f (o)\) in the right hand side, so, the right hand side is equal to or larger than the left hand side. For any \(o \in o_1\), is there an \(o' \in o_0\) such that \(f (o) \in f (f' (o'))\)? As \(sup \{f' (o)\vert o \in o_0\} = o_1\), there is an \(o' \in o_0\) such that \(o \in f' (o')\), because otherwise, \(o\) would be a smaller upper bound. So, \(f (o) \in f (f' (o'))\), so, the left hand side is equal to or larger than the right hand side. So, the left hand side equals the right hand side. So, yes, \(sup \{f (f' (o))\vert o \in o_0\} = sup \{f (o)\vert o \in o_1\}\). So, yes, \(sup \{f (f' (o))\vert o \in o_0\} = f (o_1)\). So, yes, \(o_1 = f (o_1)\). So, yes, \(f' (o_0) = \cup_{o \in o_0} f' (o)\).
