# Proof that $\forall \alpha \in \mathbf{ON}(L(\alpha) \cap \mathbf{ON}) = \alpha$.

In Kunens "Set Theory An Introduction to Independence Proofs", lemma 1.9.(b), he proves the following:

$$\forall \alpha \in \mathbf{ON}(L(\alpha) \cap \mathbf{ON}) = \alpha$$.

Now, the proof proceeds by induction. When we come to the successor-case, we want to show that $$\beta$$ in $$L(\beta)$$.

Now Kunen proceeds by then saying that:

Recall (IV 5.1) that there is a $$\Delta_0$$-formula $$\phi(x)$$ such that $$\forall x(\text{x is an ordinal} \leftrightarrow \phi(x)).$$ Since $$\Delta_0$$-formulas are absolute for all transitive sets, $$\beta = L(\beta) \cap \mathbf{ON} = \{x \in L(\beta): \phi^{L(\beta)}(x)\}, \qquad (1)$$ so $$\beta \in \mathscr{D}(L(\beta)) = L(\alpha)$$ by lemma 1.2. $$\quad \square$$

Now, when we actually look at theorem 5.1 in IV, it says:

The following relations and functions were defined in $$\sf{ZF}-\sf{P}$$ by formulas provably equivalent in $$\sf{ZF}-\sf{P}$$ to $$\Delta_0$$-formulas. $$\underline{\text{They are thus absolute for transitive models of \sf{ZF}-\sf{P}}}$$ (my underlining).

1. $$x$$ is an ordinal.
2. $$\ldots$$.

But does this not mean that we need $$L(\beta)$$ to be a model for $$\sf{ZF}-\sf{P}$$ to assert that $$\phi^{L(\beta)}(x) \leftrightarrow \phi(x)$$, which seems to be what Kunen is using in $$(1)$$? If yes, as far as I can see, he has not proved this before lemma 1.9.(b) in VI. What he has proved, is that $$L(\beta)$$ is transitive (see lemma 1.6.(a) in VI).

So why is the step $$\phi(x) \rightarrow \phi^{L(\beta)}(x)$$ motivated?

Any clarification would be appreciated.

Addendum: My suspicion is that Kunen is using corollary 3.6, IV, which says:

If $$\mathbf{M}$$ is transitive, and $$\phi$$ is $$\Delta_0$$, then $$\phi$$ is absolute for $$\mathbf{M}$$.

I am however unsure how this result fits in with lemma 5.1, IV (partially quoted above).

There's no need to analyze what axioms hold in $$L(\beta).$$ You just need to show that $$\beta=\{x\in L(\beta): \phi(x)\}\in L(\beta+1),$$ where $$\phi(x)$$ says $$x$$ is transitive and linearly ordered by $$\in.$$ We know that $$\phi$$ defines the ordinals in $$V,$$ and that's enough.
Now, since $$\phi$$ is (syntactically) $$\Delta_0,$$ it is absolute for any transitive model of anything and we have $$\phi^{L(\beta)}(x) \leftrightarrow\phi(x)$$ for $$x\in L(\beta)$$ so $$\{x\in L(\beta): \phi(x)\}= \{x\in L(\beta): \phi^{L(\beta)}(x)\}\in L(\beta+1).$$
That said, I think ZF-P vastly overstates what is needed for the equivalence of "$$x$$ is transitive and well-ordered by $$\in$$" and "$$x$$ is transitive and linearly-ordered by $$\in$$". I believe that literally only requires foundation, so the equivalence should hold in any transitive model.
• So part of what you are saying in the last paragraph before your addendum, is really what I said (AFAICT) in my addendum, i.e. Kunen uses the fact that any $\Delta_0$-formula is absolute for a transitive class (cor. 3.6)? What I am however still confused about, is then why is $\mathbf{M}$ being a model for $\sf{ZF}-\sf{P}$ needed in theorem 5.1 (IV) (yes, there are also other statements there, besides "$x$ being an ordinal")? Commented Aug 1 at 4:39
• Oh, right, if we don't have foundation, we don't know that $x$ is well-ordered by $\in$. Commented Aug 1 at 4:42
• @Ben123 Per my last paragraph I think just foundation is needed in the background and then "$x$ is an ordinal" is equivalent to $\phi(x)$ in any transitive set or class, but I'm less sure of that than I am that he's only using that it's equivalent to $\phi(x)$ in $V$ in the argument. Commented Aug 1 at 4:47
• @Ben123 Yes, he is just using absoluteness of syntactically $\Delta_0$ formulas to get $\phi(x)\leftrightarrow \phi^{L(\beta)}(x).$ He is using both 5.1 and 3.6, though he doesn't explicitly reference the latter, just says "Since $\Delta_0$ -formulas are absolute for all transitive sets." Commented Aug 1 at 5:47