321: For Monotone Ordinal Numbers Operation, 2 Domain Elements Are in Membership Relation if Corresponding Images Are in Same Relation

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A description/proof of that for monotone ordinal numbers operation, 2 domain elements are in membership relation if corresponding images are in same relation

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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 ordinal number.
• The reader knows a definition of monotone operation.
• The reader admits the trichotomy of the membership relation on 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 operation from the ordinal numbers collection into the ordinal numbers collection, any 2 domain elements are in a membership relation if the corresponding images are in the same membership relation.

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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 operation, $f:O\to O$ where $O$ is the ordinal numbers collection, and any 2 domain elements, $o_1,o_2\in O$, $o_1\in o_2$ if $f(o_1)\in f(o_2)$; $o_1=o_2$ if $f(o_1)=f(o_2)$.

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2: Proof

Let us suppose that $f(o_1)\in f(o_2)$. Let us suppose that $\mathrm{\neg }{o}_1\in o_2$. By the trichotomy of the membership relation on the ordinal numbers collection, $o_1=o_2$ or $o_2\in o_1$. If $o_1=o_2$, $f(o_1)=f(o_2)$, a contradiction against $f(o_1)\in f(o_2)$, by the trichotomy of the membership relation on the ordinal numbers collection. If $o_2\in o_1$, $f(o_2)\in f(o_1)$, a contradiction against $f(o_1)\in f(o_2)$, by the trichotomy of the membership relation on the ordinal numbers collection. So, $o_1\in o_2$.

Let us suppose that $f(o_1)=f(o_2)$. Let us suppose that $o_1\ne o_2$. By the trichotomy of the membership relation on the ordinal numbers collection, $o_1\in o_2$ or $o_2\in o_1$. If $o_1\in o_2$, $f(o_1)\in f(o_2)$, a contradiction against $f(o_1)=f(o_2)$, by the trichotomy of the membership relation on the ordinal numbers collection. If $o_2\in o_1$, $f(o_2)\in f(o_1)$, a contradiction against $f(o_1)=f(o_2)$, by the trichotomy of the membership relation on the ordinal numbers collection. So, $o_1=o_2$.

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References

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