A description/proof of that ordinal number is grounded and its rank is itself
Topics
About: set
The table of contents of this article
Starting Context
- The reader knows a definition of ordinal number.
- The reader knows a definition of grounded set.
- The reader knows a definition of rank of set.
- The reader admits the transfinite induction principle.
- The reader admits the proposition that any ordinal number is a member of an ordinal number.
- The reader admits the proposition that any ordinal number is a well-ordered set.
- The reader admits the proposition that any ordinal number is a transitive set.
- The reader admits the proposition that for any transitive set with the at least partial ordering by membership (supposing that the ordering by membership is really a partial ordering), any element is the initial segment up to it.
Target Context
- The reader will have a description and a proof of the proposition that any ordinal number is grounded and its rank is itself.
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
Any ordinal number, \(o\), is grounded, and its rank, \(rank \text{ }o\), is itself, which is \(rank \text{ }o = o\).
2: Proof
For any ordinal number, \(\delta\), the ordering by membership is a partial ordering, because for any \(o \in \delta\), \(\o \notin o\), and for any \(o, o', o'' \in \delta\) such that \(o' \in o\) and \(o'' \in o'\), \(o'' \in o\).
Let us prove that \(o \subseteq V_o\) where \(V_0 = \emptyset\) and \(V_o = \cup \{Pow V_{o'}\vert o' \in o\}\). As the collection of all the ordinal numbers is not any set in the ZFC set theory, we cannot use the transfinite induction principle on the collection, so, let us think of any ordinal number, \(\delta\), which is a well-ordered set, and if \(o \subseteq V_o\) for each \(o \in \delta\), \(o \subseteq V_o\) will hold for any ordinal number, \(o\), because any ordinal number is a member of an ordinal number.
Let us define the subset, \(S := \{o \in \delta\vert o \subseteq V_o\}\), of \(\delta\). For any \(o' \in \delta\), if \(seg \text{ }o' \subseteq S\), \(o' \in S\)? For any \(o'' \in o'\), \(o'' \subseteq V_{o''}\), \(o'' \in Pow V_{o''}\). By the proposition that any ordinal number is a transitive set and the proposition that for any transitive set with the at least partial ordering by membership (supposing that the ordering by membership is really a partial ordering), any element is the initial segment up to it, \(o' = seg \text{ }o'\). For any \(p \in o'\), there is an \(o'' \in o'\) such that \(p = o''\), \(V_{o'} = \cup_{o'' \in o'} Pow V_{o''}\), but \(o'' \in Pow V_{o''}\) as is shown before, so, \(p \in V_{o'}\), so, \(o' \subseteq V_{o'}\), so, \(o' \in S\). By the transfinite induction principle, \(o \subseteq V_o\) for every \(o \in \delta\).
So, any ordinal number is grounded.
Let us prove that there is no ordinal number, \(o' \in o\), such that \(o \subseteq V_{o'}\), for any ordinal number, \(o\). As before, let us think of any ordinal number, \(\delta\), and use the transfinite induction principle on \(\delta\), then, the proposition will hold for any ordinal number, after all.
Let us define the subset, \(S := \{o \in \delta\vert \text{ there is no } o' \in o \text{ such that } o \subseteq V_{o'}\}\), of \(\delta\). For any \(o' \in \delta\), if \(seg \text{ }o' \subseteq S\), \(o' \in S\)? Let us suppose that there was an \(o'' \in o'\) such that \(o' \subseteq V_{o''}\). As \(V_{o''} = \cup_{o''' \in o''} Pow V_{o'''}\), for any \(p \in o'\), \(p \in \cup_{o''' \in o''} Pow V_{o'''}\), so, \(p \in Pow V_{o'''}\) for an \(o'''\), but \(p\) can be chosen to be \(o''\) because \(o' = seg \text{ }o'\), then, \(o'' \in Pow V_{o'''}\), so, \(o'' \subseteq V_{o'''}\), a contradiction against the supposition that \(seg \text{ }o' \subseteq S\), so, there is no \(o'' \in o'\) such that \(o' \subseteq V_{o''}\), so, \(o' \in S\). By the transfinite induction principle, there is no \(o' \in o\) such that \(o \subseteq V_{o'}\) for every \(o \in \delta\).
So, \(rank \text{ }o = o\) for any ordinal number \(o\).
