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I am trying to understand the proof of the following lemma from the paper Can the fundamental group of a space be the rationals? by Saharon Shelah.

Let $\mathcal{E}$ be an analytic equivalence relation on $\mathcal{P}(\mathbb{N})$ such that when $n\notin B$ and $A=B\cup\{n\}$, $A$ and $B$ are not $\mathcal{E}$-equivalent. Then there is a perfect subset of $\mathcal{P}(\mathbb{N})$ of pairwise nonequivelent $A\subseteq\mathbb{N}$.

I am aware that Pawlikowski gave a neat proof of this using just basic descriptive set theory, but I want to understand the logic of this proof. The proof goes as follows (some phrases slightly altered).

Let $M$ be a countable elementary substructure of $(H(\mathfrak{c}^+),\mathcal{E})$ to which the real parameter in the definition of $\mathcal{E}$ belongs. Now if $\langle A_1,A_2\rangle$ is a pair of subsets of $\mathbb{N}$ which is Cohen generic over $M$, then they are $\mathcal{E}$-equivalent iff they are $\mathcal{E}$-equivalent in $M[A_1,A_2]$, by the absoluteness criterions. Now we show that they cannot be $\mathcal{E}$-equivalent in $M[A_1,A_2]$. Otherwise some finite information forces this, so for some $n$, if $\langle A'_1,A_2\rangle$ is Cohen generic over $M$ and $A_1\cap\{0,1,...,n\}=A_1'\cap\{0,1,...,n\}$, then $A_1',A_2$ are $\mathcal{E}$-equivalent in $M[A'_1,A_2]$. Let $A_1'$ differ from $A_1$ at $n+1$ only (if $n+1\in A_1$ then $n+1\notin A_1'$, and vice versa). Clearly $\langle A'_1,A_2\rangle$ is also Cohen generic, so they are $\mathcal{E}$-equivalent in $M[A'_1,A_2]$. By the above absoluteness we really have $A'_1\sim A_2$ and $A_1\sim A_2$, so $A'_1\sim A_1$, contradicting the property of $\mathcal{E}$.

I want to ask questions about pretty much every line, but here are the points that I feel most confused about:

  1. What's the definition of $N[G]$ when $N$ is non-transitive? From Hamkins' answer to this question it seems essentially we have to take the collapse of $N$ and then do forcing, so $N$ is not a subset of $N[G]$. Is my understanding correct?

  2. If $N$ contains the "real parameter defining $\mathcal{E}$", isn't $\mathcal{E}$ already an element of $N$, since $\mathcal{E}$ is definable in $H(\mathfrak{c}^+)$ from parameter and $N$ is an elementary substructure (at least that should be the case for the particular $\mathcal{E}$ in the paper)? What's the meaning of $(H(\mathfrak{c}^+),\mathcal{E})$ then?

  3. What does "absoluteness criterions" refer to? Is it Shoenfield's Absoluteness? Does that apply to model without power set (plus $\mathcal{P}(\mathbb{N})$ exists)?

  4. Where is it used that $\mathcal{E}$ is analytic?

Thank you in advance for your answer.

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    $\begingroup$ Cohen forcing is proper, which means the generic extension commutes with the Mostowski collapse (so to speak), so you can collapse to a transitive set, force over it, and then "uncollapse" in the obvious way. Moreover, since the generic for a Cohen real is a subset of $\omega$, it actually isn't moved by the collapse to begin with. $\endgroup$
    – Asaf Karagila
    Commented Sep 10, 2021 at 10:07
  • $\begingroup$ As for the parameter, you want to make sure that not only $\cal E$ belongs to the model, but also that the definition that you're working with is interpreted correctly. Consider, for a fixed $r$, the equivalence relation that is either trivial if $r\in L$ or the identity if $r$ is Cohen generic over $L$. Depending on the parameter, the definition you use will produce a different relation, even if the relation, as a set, belongs to your model. $\endgroup$
    – Asaf Karagila
    Commented Sep 10, 2021 at 10:09
  • $\begingroup$ @AsafKaragila But if $\mathcal{E}\in N$ and $N$ is elementary w.r.t. the language $\{\in\}$, isn't it also elementary in $\{\in,\mathcal{E}\}$? $\endgroup$
    – xXF
    Commented Sep 10, 2021 at 14:11

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