322: For Well-Ordered Structure and Its Sub Structure, Ordinal Number of Sub Structure Is Member of or Is Ordinal Number of Base Structure

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A description/proof of that for well-ordered structure and its sub structure, ordinal number of sub structure is member of or is ordinal number of base structure

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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 well-ordered structure.
• The reader knows a definition of ordinal number.
• The reader admits the proposition that for any 2 well-ordered structures, one is 'well-ordered structures - order-preserving map morphisms' isomorphic to the other or to an initial segment of the other.
• The reader admits the transfinite induction principle.
• 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 well-ordered structure and its any sub structure, the ordinal number of the sub structure is a member of or is the ordinal number of the base structure.

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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 well-ordered structure, $⟨S_1,<⟩$, with the ordinal number, $o_1$, and any its sub structure, $⟨S_2,{<}^{\prime }⟩$ where $S_2\in S_1$ and ${<}^{\prime }$ is the restriction of $<$ to $S_2\times S_2$, with the ordinal number, $o_2$, $o_2\in =o_1$ where $\in =$ means $\in$ or $=$.

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2: Proof

Let us prove that $\mathrm{\neg }⟨S_1,<⟩\cong ⟨seg{p}_{2,0},{<}^{\prime }⟩$ for any $p_{2,0}\in S_2$ where $\cong$ means being 'well-ordered structures - order-preserving map morphisms' isomorphic and $seg\bullet$ means the initial segment with respect to the argument. Let us suppose that there was a 'well-ordered structures - order-preserving map morphisms' isomorphism, $f:S_1\to seg{p}_{2,0}$. Let us prove that $p_1\in =f(p_1)$ for any $p_1\in S_1$, by the transfinite induction principle. Let us suppose that $p_1\in =f(p_1)$ for any $p_1\in seg{p}_{1,0}$ for a $p_{1,0}\in S_1$. $p_{1,0}$ is the smallest element of $S_1\setminus seg{p}_{1,0}$, and $f(p_{1,0})$ is the smallest element of $seg{p}_{2,0}\setminus f(seg{p}_{1,0})$. Then, $p_{1,0}\in =f(p_{1,0})$, because if $f(p_{1,0})\in p_{1,0}$, $f(p_{1,0})=p_{1,1}$ for a $p_{1,1}\in seg{p}_{1,0}$, and $p_{1,1}\in =f(p_{1,1})\in f(p_{1,0})$, a contradiction against $f(p_{1,0})=p_{1,1}$. So, $p_{1,0}\in =f(p_{1,0})$. So, by the transfinite induction principle, $p_1\in =f(p_1)$ for any $p_1\in S_1$. Then, $p_{2,0}\in =f(p_{2,0})$, which means that $f$ was not into $seg{p}_{2,0}$, a contradiction, so, there is no such a 'well-ordered structures - order-preserving map morphisms' isomorphism, which means that $\mathrm{\neg }⟨S_1,<⟩\cong ⟨seg{p}_{2,0},{<}^{\prime }⟩$ for any $p_{2,0}\in S_2$.

So, by the proposition that for any 2 well-ordered structures, one is 'well-ordered structures - order-preserving map morphisms' isomorphic to the other or to an initial segment of the other, $⟨S_1,<⟩\cong ⟨S_2,{<}^{\prime }⟩$ or $⟨S_2,{<}^{\prime }⟩\cong ⟨seg{p}_{1,0},<⟩$ for a $p_{1,0}\in S_1$. For the former, $o_1=o_2$. For the latter, $o_2\in o_1$, by the definition of $ϵ$-image, because the $ϵ$-image of $⟨S_1,<⟩$ includes all the elements of the $ϵ$-image of $⟨seg{p}_{1,0},<⟩$ plus at least the extra element corresponding to $p_{1,0}$, so, $o_2\subset o_1$. By the proposition that the inclusion relation is equivalent with the membership relation for the ordinal numbers collection, $o_2\in =o_1$.

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References

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