319: For Monotone Continuous Operation from Ordinal Numbers Collection into Ordinal Numbers Collection, Image of Limit Ordinal Number Is Limit Ordinal Number

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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

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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 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.

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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.

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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, and any limit ordinal number, $o_1$, $o_2:=f(o_1)$ is a limit ordinal number.

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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 $o^{\prime }\in o_1$ such that $o\subseteq f(o^{\prime })$ for each $o\in o_2$? Let us suppose that there was no $o^{\prime }\in o_1$ such that $o\subseteq f(o^{\prime })$ for an $o\in o_2$. By the proposition that the inclusion relation satisfies the trichotomy on the ordinal numbers collection, $f(o^{\prime })\subset o$ for every $o^{\prime }\in o_1$, so, $o_2={\cup }_{o^{\prime }\in o_1}f(o^{\prime })\subset o\subset o_2$, 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 $o^{\prime }\in o_1$ such that $o\subseteq f(o^{\prime })$ 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.

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References

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