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

Question

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).

Answer 1

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 $$\{x\in L(\beta): \phi(x)\}= \{x\in L(\beta): \phi^{L(\beta)}(x)\}\in L(\beta+1).$$

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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.