For each ordinal number i, recurisively define:(adsbygoogle = window.adsbygoogle || []).push({});

[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?

**Physics Forums - The Fusion of Science and Community**

Dismiss Notice

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# A curious construction

Loading...

Similar Threads for curious construction | Date |
---|---|

A How can we construct ordinals after large Veblen? | Dec 11, 2017 |

I Constructing an Index | May 26, 2017 |

B How to construct a proof? | Jul 19, 2016 |

How to construct the natural numbers with hypersets | May 13, 2015 |

Chances of getting pregnant when using 2-3 contraceptives together. Just curious! | Oct 25, 2009 |

**Physics Forums - The Fusion of Science and Community**