294: Derived Operation of Monotone Continuous Operation from Ordinal Numbers Collection into Ordinal Numbers Collection Is Monotone Continuous

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A description/proof of that derived operation of monotone continuous operation from ordinal numbers collection into ordinal numbers collection is monotone continuous

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Topics

About: set

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The table of contents of this article

• Starting Context
• Target Context
• Orientation
• Main Body
• 1: Description
• 2: Proof

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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.

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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.

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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.

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Main Body

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1: Description

For any monotone continuous operation, $f:O\to O$, where $O$ is the all the ordinal numbers collection, the derived operation, $f^{\prime }:O\to O$, is monotone and continuous.

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2: Proof

$f^{\prime }$ is monotone, because it chooses fixed-points in the ascending order by definition.

As for the continuousness, for any limit ordinal number, $o_0\in O$, $f^{\prime }(o_0)={\cup }_{o\in o_0}{f}^{\prime }(o)$? ${\cup }_{o\in o_0}{f}^{\prime }(o)=sup\{f^{\prime }(o)|o\in o_0\}:=o_1$. If $o_1=f(o_1)$, $o_1=f^{\prime }(o_0)$, because $o_1$ will be the smallest fixed-point except $\{f^{\prime }(o)|o\in o_0\}$. Then, $o_1=f(o_1)$? $f(f^{\prime }(o))=f^{\prime }(o)$. Taking the supremum of the sets of the both sides, $sup\{f(f^{\prime }(o))|o\in o_0\}=sup\{f^{\prime }(o)|o\in o_0\}=o_1$. But $sup\{f(f^{\prime }(o))|o\in o_0\}=f(o_1)$? $o_1$ is a limit ordinal number, because if it was a successor ordinal number, it would have the maximum in $\{f^{\prime }(o)|o\in o_0\}$, which would be impossible, because $o_0$ is a limit ordinal number and if $f^{\prime }(o^{\prime })$ was the maximum, there would be an $o^{″}$ such that $o^{\prime }\in o^{″}\in o_0$ and $f^{\prime }(o^{\prime })\in f^{\prime }(o^{″})$. So, $f(o_1)=sup\{f(o)|o\in o_1\}$ as $f$ is continuous, and $f(o_1)$ is a limit ordinal number by the likewise reason. $sup\{f(f^{\prime }(o))|o\in o_0\}=sup\{f(o)|o\in o_1\}$? For any $o\in o_0$, $f^{\prime }(o)\in o_1$, so, $f(f^{\prime }(o))$ 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 $o^{\prime }\in o_0$ such that $f(o)\in f(f^{\prime }(o^{\prime }))$? As $sup\{f^{\prime }(o)|o\in o_0\}=o_1$, there is an $o^{\prime }\in o_0$ such that $o\in f^{\prime }(o^{\prime })$, because otherwise, $o$ would be a smaller upper bound. So, $f(o)\in f(f^{\prime }(o^{\prime }))$, 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, $sup\{f(f^{\prime }(o))|o\in o_0\}=sup\{f(o)|o\in o_1\}$. So, yes, $sup\{f(f^{\prime }(o))|o\in o_0\}=f(o_1)$. So, yes, $o_1=f(o_1)$. So, yes, $f^{\prime }(o_0)={\cup }_{o\in o_0}{f}^{\prime }(o)$.

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References

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