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
Topics
About: set
The table of contents of this article
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.
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.
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 well-ordered structure, \(\langle S_1, \lt \rangle\), with the ordinal number, \(o_1\), and any its sub structure, \(\langle S_2, \lt' \rangle\) where \(S_2 \in S_1\) and \(\lt'\) is the restriction of \(\lt\) to \(S_2 \times S_2\), with the ordinal number, \(o_2\), \(o_2 \in= o_1\) where \(\in=\) means \(\in\) or \(=\).
2: Proof
Let us prove that \(\lnot \langle S_1, \lt \rangle \cong \langle seg \text{ } p_{2, 0}, \lt' \rangle\) for any \(p_{2, 0} \in S_2\) where \(\cong\) means being 'well-ordered structures - order-preserving map morphisms' isomorphic and \(seg \text{ } \bullet\) means the initial segment with respect to the argument. Let us suppose that there was a 'well-ordered structures - order-preserving map morphisms' isomorphism, \(f: S_1 \rightarrow seg \text{ } p_{2, 0}\). 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 \(p_1 \in seg \text{ } p_{1, 0}\) for a \(p_{1, 0} \in S_1\). \(p_{1, 0}\) is the smallest element of \(S_1 \setminus seg \text{ } p_{1, 0}\), and \(f (p_{1, 0})\) is the smallest element of \(seg \text{ } p_{2, 0} \setminus f (seg \text{ } p_{1, 0})\). 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 \(p_{1, 1} \in seg \text{ } p_{1, 0}\), 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 \(seg \text{ } p_{2, 0}\), a contradiction, so, there is no such a 'well-ordered structures - order-preserving map morphisms' isomorphism, which means that \(\lnot \langle S_1, \lt \rangle \cong \langle seg \text{ } p_{2, 0}, \lt' \rangle\) 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, \(\langle S_1, \lt \rangle \cong \langle S_2, \lt' \rangle\) or \(\langle S_2, \lt' \rangle \cong \langle seg \text{ } p_{1, 0}, \lt \rangle\) 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 \(\epsilon\)-image, because the \(\epsilon\)-image of \(\langle S_1, \lt \rangle\) includes all the elements of the \(\epsilon\)-image of \(\langle seg \text{ } p_{1, 0}, \lt \rangle\) 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, \(o_2 \in= o_1\).
