- #1
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For each ordinal number i, recurisively define:
[tex]
\begin{equation*}\begin{split}
Z_0 = \{ \varnothing \} \\
Z_i = (\bigcup_{j < i} Z_j) \cup \mathcal{P}(\bigcup_{j < i} Z_j)
\end{split}\end{equation*}
[/tex]
Definition: A set S is a Z-set iff there is an ordinal i such that [itex]S \in Z_i[/itex].
Definition: For any Z-set S, define the "birthday function" B(S) to be the least ordinal i such that [itex]S \in Z_i[/itex].
The class of Z-sets is an interesting class...
First, notice that [itex]i < j \implies Z_i \subseteq Z_j[/itex].
Suppose x and y are Z-sets. Let i = B(x) and j = B(y). Let k be the maximum of i and j. Clearly x and y are both in [itex]Z_k[/itex]. Therefore, [itex]\{x, y\} \in Z_{k+1}[/itex].
Suppose S is any set of Z-sets. Let k = \bigcup_{s \in S} B(s). (Recall that ordinal numbers in ZF are sets, and you can union a collection of ordinals to obtain an ordinal larger than any in the collection) All of the elements of S are in [itex]Z_k[/itex], therefore [itex]S \in Z_{k+1}[/itex].
It is vacuously true that if S is any set in [itex]Z_0[/itex], then any s in S is an element of [itex]Z_0[/itex].
Let i be an ordinal number. Suppose that for all j < i, we have that for any [itex]S \in Z_j[/itex] we have [itex]\forall s \in S: s \in Z_j[/itex].
Let [itex]S \in Z_i[/itex]. If [itex]S \in Z_j[/itex] for some [itex]j < i[/itex], then by our hypothesis, [itex]\forall s \in S: s \in Z_j \subseteq Z_i[/itex]. The only other possibility is that [itex]S \in \mathcal{P}(\bigcup_{j < i} Z_j}[/itex]. Therefore, [itex]S \subseteq \bigcup_{j < i}Z_j[/itex], and therefore [itex]\forall s \in S, \exists j < i: s \in Z_j \subseteq Z_i[/itex].
Therefore, [itex]\forall S \in Z_i: \forall s \in S: s \in Z_i[/itex].
By the principle of transfinite induction, this statement must be true:
For any ordinal i and any set S in [itex]Z_i[/itex], every element of S is in [itex]Z_i[/itex].
In other words, if S is a Z-set, then every element of S is a Z-set.
Furthermore, if [itex]S \in Z_i[/itex], we have [itex]\forall s \in S: s \subseteq Z_i[/itex], so [itex]\bigcup S \subseteq Z_i[/itex], therefore [itex]\bigcup S \in Z_{i+1}[/itex].
So what have we proven?
If x and y are Z-sets, then {x, y} is a Z-set.
If x is a Z-set, then the sumset of x is a Z-set.
If x is a Z-set, then the powerset of x is a Z-set.
Also, it is clear that [itex]\varnothing \in Z_0[/itex] and [itex]\mathcal{N} \in Z_\mathcal{N}[/itex] (where I'm using the typical set-theoretic version of the natural numbers)
(In fact, for any ordinal i, [itex]i \in Z_i[/itex])
This looks a lot like the axioms of the pair set, the sum set, the power set, the empty set, and the axiom of infinity! It seems that the class of Z-sets coincides with the class of sets guaranteed to exist by the axioms of Zermelo set theory!
Is that cool or what?
[tex]
\begin{equation*}\begin{split}
Z_0 = \{ \varnothing \} \\
Z_i = (\bigcup_{j < i} Z_j) \cup \mathcal{P}(\bigcup_{j < i} Z_j)
\end{split}\end{equation*}
[/tex]
Definition: A set S is a Z-set iff there is an ordinal i such that [itex]S \in Z_i[/itex].
Definition: For any Z-set S, define the "birthday function" B(S) to be the least ordinal i such that [itex]S \in Z_i[/itex].
The class of Z-sets is an interesting class...
First, notice that [itex]i < j \implies Z_i \subseteq Z_j[/itex].
Suppose x and y are Z-sets. Let i = B(x) and j = B(y). Let k be the maximum of i and j. Clearly x and y are both in [itex]Z_k[/itex]. Therefore, [itex]\{x, y\} \in Z_{k+1}[/itex].
Suppose S is any set of Z-sets. Let k = \bigcup_{s \in S} B(s). (Recall that ordinal numbers in ZF are sets, and you can union a collection of ordinals to obtain an ordinal larger than any in the collection) All of the elements of S are in [itex]Z_k[/itex], therefore [itex]S \in Z_{k+1}[/itex].
It is vacuously true that if S is any set in [itex]Z_0[/itex], then any s in S is an element of [itex]Z_0[/itex].
Let i be an ordinal number. Suppose that for all j < i, we have that for any [itex]S \in Z_j[/itex] we have [itex]\forall s \in S: s \in Z_j[/itex].
Let [itex]S \in Z_i[/itex]. If [itex]S \in Z_j[/itex] for some [itex]j < i[/itex], then by our hypothesis, [itex]\forall s \in S: s \in Z_j \subseteq Z_i[/itex]. The only other possibility is that [itex]S \in \mathcal{P}(\bigcup_{j < i} Z_j}[/itex]. Therefore, [itex]S \subseteq \bigcup_{j < i}Z_j[/itex], and therefore [itex]\forall s \in S, \exists j < i: s \in Z_j \subseteq Z_i[/itex].
Therefore, [itex]\forall S \in Z_i: \forall s \in S: s \in Z_i[/itex].
By the principle of transfinite induction, this statement must be true:
For any ordinal i and any set S in [itex]Z_i[/itex], every element of S is in [itex]Z_i[/itex].
In other words, if S is a Z-set, then every element of S is a Z-set.
Furthermore, if [itex]S \in Z_i[/itex], we have [itex]\forall s \in S: s \subseteq Z_i[/itex], so [itex]\bigcup S \subseteq Z_i[/itex], therefore [itex]\bigcup S \in Z_{i+1}[/itex].
So what have we proven?
If x and y are Z-sets, then {x, y} is a Z-set.
If x is a Z-set, then the sumset of x is a Z-set.
If x is a Z-set, then the powerset of x is a Z-set.
Also, it is clear that [itex]\varnothing \in Z_0[/itex] and [itex]\mathcal{N} \in Z_\mathcal{N}[/itex] (where I'm using the typical set-theoretic version of the natural numbers)
(In fact, for any ordinal i, [itex]i \in Z_i[/itex])
This looks a lot like the axioms of the pair set, the sum set, the power set, the empty set, and the axiom of infinity! It seems that the class of Z-sets coincides with the class of sets guaranteed to exist by the axioms of Zermelo set theory!
Is that cool or what?