Some Random Questions about Set Theory

Question

Here are some random questions about set theory that I'm confused about:

1) We know that for any limit$\gamma>\omega$, $V_{\gamma}\models{ZC}$. However I believe for most such $\gamma$, $\neg{V_{\gamma}\models\text{choice}\iff\text{well ordering principle}}$. For example, set $\gamma=\omega.2$. Now $P{(\mathbb{N}})\in{V_{\gamma}}$ but in order to well order $P(\mathbb{N})$, you need an ordinal of size (cardinality) at least the continuum which clearly isn't in $V_{\gamma}$. However this isn't a contradiction since the equivalence of the two is a theorem of ZF rather than ZC. I was looking at the proof of the equivalance in Kunen and couldn't quite figure out where replacement was used; could someone tell me where it is used?

2) Let $A$ be a set and let $P\subseteq{A}$. Let $D{(A,P)}$ be the set of definable subsets of $A$ using parameters from $P$ (loosely speaking this is done in the model theoretic sense, with $L=\{E\}$ and $E$ on $A$ is interpreted as $\in$). Let $D^{+}(A)=D(A,A)$. Now $L_{\alpha}$ is defined via recursion by: $L(0)=\emptyset$, $L(\alpha+1)=D^+{(L(\alpha)})$ and $L(\alpha)=\cup_{\beta<\alpha}L(\beta)$ for limit $\alpha$. With the notation established here: Absoluteness of $\mathbb{P}$-names , is it the case that $xRy \iff x\in{y}$, $A$ is the class of ordinals and $G(x,s)=\cup_{y\in{\text{dom}(s)}}D^+(s(y))$ if $s$ is a function and is $\emptyset$ otherwise. If not, then what should $A,R,G$ be?

3) Let $M\models{ZFC}$ be transitive. I want to show that for $\alpha\in{ON^{M}}$, $(L(\alpha))^{M}=L(\alpha)$. There are several ways to do this. If we have the theorem mentioned in Absoluteness of $\mathbb{P}$-names to the effect that (under suitable conditions) things defined by induction are absolute, then we have the theorem. Another way to do so, (which still relies on the above theorem), is to assume $D^{+}$ is an absolute notion. Under these conditions is the following an acceptable proof for $(L{(\alpha)})^{M}=L{(\alpha)}$ for successor ordinals?

Let $\alpha\in{ON^{M}}$ the IH be that it holds for all $\beta<\alpha$ for $\beta\in{ON^{M}}$. However by absoluteness of ordinals, this becomes $\beta<\alpha$ for $\beta\in{ON}$. For $\alpha={\beta+1}$; $(L(\alpha))^{M}=(D^+{L(\beta)})^M$. But $(D^+{L(\beta)})^M = (D^+{(L(\beta)}^M)$ since $D^+$ is an absolute notion. But then $(D^+{(L(\beta)}^M)=D^+{(L(\beta)}$ by the induction hypothesis.

I guess my question here reduces to: Do you start from the outside and work your way to the middle (Which is what I did here; I started with $D^+$ and worked my way in. I think this is correct, since this is how we usually write proofs) or do we start from the middle and work our way out?

Thank you.

Answer 1

Let me answer the first question.

You are wrong in saying that "most $\gamma$", because in fact there is a closed and unbounded class of strong limit cardinals in which every well-ordered set is isomorphic to a von Neumann ordinal. Since we consider "club" as "almost everywhere", this shows that the intuition we have is not quite correct.

As for the other part of the first question, you might be interested to read (at least in part),

The strength of Mac Lane set theory, Mathias, A. R. D., Ann. Pure Appl. Logic 110 (2001), no. 1-3, 107–234.

Answer 2

3) Your proof is correct. But let me restate it - maybe it becomes more clear then.

Work within $ \mathrm{ZF} $. Fix $ \alpha \in \mathbf{ON}^M $. Suppose $ L^M(\beta) = L(\beta) $ for all $ \beta < \alpha $. The absoluteness of $ \mathscr{D} $ (Kunen's notation, $ D^+ $ in your notation) yields $ \mathscr{D}^M(x) = \mathscr{D}(x) $ for all $ x \in M $. More formally, the formula $ \mathscr{D}(x) = y $ is absolute for $ M $, i.e. $ \mathrm{ZF} \vdash \forall x, y \in M \ ((\mathscr{D}(x) = y)^M \iff \mathscr{D}(x) = y) $.

Now, suppose $ \alpha = \beta + 1 $ for some $ \beta \in \mathbf{ON} $.

(1) $ L^M(\alpha) = D^M(L^M(\beta)) $ (by definition inside $ M $; $ \alpha = \beta + 1 $ is absolute for $ M $)

(2) $ D^M(L^M(\beta)) = D(L^M(\beta)) $ (by $ L^M(\beta) \in M $ and $ \forall x \in M \ \mathscr{D}^M(x) = \mathscr{D}(x) $)

(3) $ D(L^M(\beta)) = D(L(\beta)) $ (by induction hypothesis)

(4) $ D(L(\beta)) = L(\alpha) $ (by definition inside $ \mathbf{V} $)

All four equalities together prove your claim. In my opinion, there is no "way out from the middle". You can read the proof downwards from 1) to 4) but also upwards from 4) to 1). Note that the identity satisfies all properties of an equivalence relation.

By the way, you should have a look at Theorem IV 5.6 in Kunen's book Set Theory - An Introduction To Independence Proofs.