Understanding the proof of $V=L \rightarrow \Diamond$

Question

I am trying to understand the proof that $\Diamond$ holds in the constructible universe. I am following Kunen's Set Theory, where the proof is on pages 230-231 (the more recent, 2011 edition).

I am able to follow the arguments in the proof, up to the second to last paragraph, which starts with:

Applying $M \preccurlyeq H(\omega_2)$ and absoluteness: For each ordinal $\alpha \in M$, $A_\alpha, C_\alpha \in M$ and mos$(A_\alpha) = A_{\text{mos}(\alpha)}$ and mos$(C_\alpha) = C_{\text{mos}(\alpha)}$.

I don't understand why the above statement holds true. Any help clarifying this would be appreciated.

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Edit: For context, the proof is trying to show that $\langle A_\alpha : \alpha < \omega_1 \rangle$ is a $\Diamond$-sequence. Towards a contradiction, we assume it is not a $\Diamond$-sequence.

The definition of the sequence goes as follows. For limit ordinals $\alpha$, define $$P(\alpha, A, C) \iff A, C \subseteq \alpha \land C \text{ is club in } \alpha \land \neg \exists \xi \in C (A \cap \xi = A_\xi).$$ Then let $(A_\alpha, C_\alpha)$ be the $<_L$ first pair such that $P(\alpha, A, C)$. If there is no such pair, or if $\alpha$ is not a limit, let $A_\alpha = C_\alpha = \emptyset$.

Also, mos refers to the Mostowski isomorphism from a countable $M \preccurlyeq H(\omega_2)=L(\omega_2)$ to some transitive $T$.

Answer 1

For $\alpha\in M\implies A_\alpha,C_\alpha\in M$, the key observation is that the sequence $\mathfrak{S}=(A_\alpha,C_\alpha)_{\alpha<\omega_1}$ is an element of $M$. This is just because it is definable without parameters in $H_{\omega_2}$ (check this).

Now of course this doesn't mean $\mathfrak{S}\subseteq M$ - and indeed, if $M$ is "small" (e.g. countable) it won't be - but it does mean that $M$ sees as much of $\mathfrak{S}$ as it should: $$\alpha\in Ord\cap M\implies (A_\alpha,C_\alpha)\in M.$$ We're not using anything special about $\mathfrak{S}$, just that if a sequence and an index are in $M$ then so is the term of the sequence with that index.

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For the Mostowski collapse calculations, think about what happens to the $L$-ordering and to $P$-ness when we take the Mostowski collapse. The point is that we know that $mos(A_\alpha)\subseteq mos(\alpha)$, so we just need to check that $mos(A_\alpha)$ satisfies $P$ and is the $L$-least subset of $\alpha$ to satisfy $P$; and the ideal situation would be that everything "stays the same" when we take the Mostowski collapse.

Answer 2

If $\alpha$ is countable, then $M$ thinks that it is countable, so an enumeration exists there, and so the range of the enumeration is a subset of $M$. Therefore $\alpha$ is a subset of $M$ as well, so in the Mostowski collapse, $\alpha$ is not moved. And therefore subsets of $\alpha$ are not moved either.

Moreover, when you collapse the sequence, you do not change anything, pointwise. So you actually end up with an initial segment of the entire sequence, for whatever $\omega_1$ was collapsed to.