Observation 4A. There are $T_{\mathbf n}$ , $(\lambda _{\eta }, \mu _{\eta } : \eta \in T_{\mathbf n})$ and $(nor_{\eta } : \eta \in T_{\mathbf n})$ satisfying the requirements of Definition 2, where the norm is defined as in Definition 4 (hence $\mathbf n \in \mathbf N$ is special).

Proof. It’s easy to check that the following $(T_{\mathbf n}, (\mu _{\eta }, \lambda _{\eta } : \eta \in T_{\mathbf n}))$ together with the norm from Definition 4 form a special $\mathbf n \in \mathbf N$ where $T_{\mathbf n} \cap \omega ^n$ , $(\mu _{\eta }, \lambda _{\eta } : \eta \in T_{\mathbf n} \cap \omega ^n)$ are defined by induction on $n<\omega $ as follows:

1. a. $T_{\mathbf {n}} \cap \omega ^0 =\{<>\}$ .

2. b. At stage $n+1$ , for $\eta \in T_{\mathbf {n}} \cap \omega ^n$ , by induction according to $<_{lex}$ , define $\mu _{\eta }=\beth _{\lambda _{<\eta }}$ , $\lambda _{\eta }=\beth _{{\mu _{\eta }}^3}$ and the set of successors of $\eta $ in $T_{\mathbf n}$ is defined as $\{\eta {}^\frown (l) : l<\lambda _{\eta }\}$ .

For example, we shall prove the co-bigness property:

Suppose that $\eta \in T_{\mathbf n}\, (a_i : i<i(*))$ are as in Definition 2(f). Denote $k_1=|suc_{T_{\mathbf n}}(\eta )|$ and $k_2=max \{|suc_{T_{\mathbf n}}(\eta ) \setminus (a_i)| : i<i(*)\}$ . Therefore, $\frac {log_*(k_1)}{\mu _n^2}-\frac {log_*(k_2)}{\mu _n^2}\leq nor_{\eta }(a_i)$ (so necessarily $k+\frac {1}{\mu _{\eta }} \leq \frac {log_*(k_1)}{\mu _n^2}-\frac {log_*(k_2)}{\mu _n^2}$ ). Let $a=\underset {i<i(*)}{\cup }a_i$ and $k_3=|suc_{T_{\mathbf n}}(\eta )\setminus a| \leq i(*)k_2 \leq \mu _{\eta }k_2$ . Therefore $\frac {log_*(k_1)}{\mu _{\eta }^2}-\frac {log_*(\mu _{\eta }k_2)}{\mu _n^2} \leq \frac {log_*(k_1)}{\mu _{\eta }^2}-\frac {log_*(k_3)}{\mu _{\eta }^2}=nor_{\eta }(a)$ . We have to show that $k\leq nor_{\eta }(a)$ , so it’s enough to show that $k\leq \frac {log_*(k_1)}{\mu _{\eta }^2}-\frac {log_*(\mu _{\eta }k_2)}{\mu _n^2}$ . Recalling that $k+\frac {1}{\mu _{\eta }} \leq \frac {log_*(k_1)}{\mu _{\eta }^2}-\frac {log_*(k_2)}{\mu _{\eta }^2}$ , it’s enough to show that $\frac {log_*(\mu _{\eta }k_2)}{\mu _{\eta }^2}-\frac {log_*(k_2)}{\mu _{\eta }^2} \leq \frac {1}{\mu _{\eta }}$ .

Case 1: $k_2 \leq \mu _{\eta }$ . In this case, it’s enough to show that $log_*(\mu _{\eta }k_2)-log_*(k_2) \leq \mu _{\eta }$ , and indeed, $log_*(\mu _{\eta }k_2)-log_*(k_2) \leq log_*(\mu _{\eta }^2) \leq \mu _{\eta }$ .

Case 2: $\mu _{\eta }<k_2$ . By the properties of $log_*$ , $log_*(k_2) \leq log_*(\mu _{\eta }k_2) \leq log_*(k_2^2) \leq log_*(k_2)+1$ , therefore $\frac {log_*(\mu _{\eta }k_2)}{\mu _{\eta }^2}-\frac {log_*(k_2)}{\mu _{\eta }^2} \leq \frac {1}{\mu _{\eta }}$ .

Claim 6. $\mathbb {Q}_{\mathbf n}^{i} \models ccc$ for $i\in \{0,\frac {1}{2},1,2\}$ .

Proof. First we shall prove the claim for $\mathbb {Q}_{\mathbf n}^i$ where $i\in \{0,\frac {1}{2},1\}$ . Observe that if $p\in \mathbb {Q}_{\mathbf n}^i$ and $0<k<\omega $ , then there is $p\leq q \in \mathbb {Q}_{\mathbf n}^i$ such that $nor_{\eta }(Suc_q(\eta ))>k$ for every $\eta \in T_q^+$ . The statement is trivial for $i=1$ , so suppose that $i\in \{0,\frac {1}{2}\}$ . In order to prove this fact, let $Y=\{ \eta \in T_p : $ for every $\eta \leq \nu \in T_p$ , $nor_{\nu }(Suc_{T_p}(\nu ))>k \}$ , then Y is dense in $T_p$ (suppose otherwise, then we can construct a strictly increasing sequence of members $\eta _i \in T_p$ such that $nor_{\eta _i}(Suc_{T_p}(\eta _i)) \leq k$ , so $\underset {i<\omega }{\cup }\eta _i \in lim(T_p)$ contradicts the definition of $\mathbb {Q}_{\mathbf n}^i$ ). Now pick $tr(p) \leq \eta \in Y$ , then $q=p^{[\eta \leq ]}$ is as required. It also follows from this claim that $\mathbb {Q}_{\mathbf n}^{\frac {1}{2}}$ is dense in $\mathbb {Q}_{\mathbf n}^0$ (this is a private case where we let $k=2$ in the claim).

Now suppose towards contradiction that $\{p_{\alpha } : \alpha <\aleph _1\} \subseteq \mathbb {Q}_{\mathbf n}^i$ is an antichain, for every $\alpha $ , there is $p_{\alpha } \leq q_{\alpha }$ such that $nor_{\eta }(Suc_{q_{\alpha }}(\eta ))>2$ for every $\eta \in q_{\alpha }$ . For some uncountable $S\subseteq \aleph _1$ , $tr(q_{\alpha })=\eta _*$ for every $\alpha \in S$ . By the claim below, $q_{\alpha }, q_{\beta }$ are compatible for $\alpha ,\beta \in S$ , contradicting our assumption.

As for $\mathbb {Q}_{\mathbf n}^2$ , given $I=\{p_i : i<\aleph _1\}\subseteq \mathbb {Q}_{\mathbf n}^{2}\, (\mathbb {Q}_{\mathbf n}^{1})$ , the set $\{(tr(p),nor(p)) : p\in I\}$ is countable, hence there is $p_*\in I$ such that for uncountably many $p_i\in I$ we have $(tr(p_i),nor(p_i))=(tr(p_*),nor(p_*))$ . By the claim below, those $p_i$ are pairwise compatible.

Claim 7.

1. 1) $p,q\in \mathbb {Q}_{\mathbf n}^{2}$ are compatible in $\mathbb {Q}_{\mathbf n}^{2}$ iff $tr(p)\leq tr(q)\in T_p$ or $tr(q)\leq tr(p)\in T_q$ .

2. 2) Similarly, $p,q \in \mathbb {Q}_{\mathbf n}^i$ are compatible in $\mathbb {Q}_{\mathbf n}^i$ for $i\in \{0,\frac {1}{2},1\}$ iff $tr(p) \leq tr(q) \in T_p \vee tr(q) \leq tr(p) \in T_q$ .

Proof. In both clauses, the implication $\rightarrow $ is obvious, we shall prove the other direction.

1) First observe that if $p\in \mathbb {Q}_{\mathbf n}^{2}$ and $\nu \in T_p$ , then $p^{[\nu ]}\in \mathbb {Q}_{\mathbf n}^{2}$ and $p\leq p^{[\nu ]}$ (where $p^{[\nu ]}$ is the set of nodes in p comparable with $\nu $ ).

$\square _1$ If $tr(p) \leq tr(q) \in T_p$ then $T_p \cap T_q$ has arbitrarily long nodes.

Proof. Let $\eta =tr(q)$ , then by the definition of the norm and $\mathbb {Q}_{\mathbf n}^{2}$ , $\frac {1}{2}<\frac {|suc_{T_p}(\eta )|}{|suc_{T_{\mathbf n}}(\eta )|},\frac {|suc_{T_q}{\eta }|}{|suc_{T_{\mathbf n}}(\eta )|}$ . Hence there is $\nu \in suc_{T_p}(\eta )\cap suc_{T_q}(\eta )$ . Repeating the same argument, we get sequences in $T_p \cap T_q$ of length n for every n large enough.

$\square _2$ Claim: If $tr(p_1)=tr(p_2)=\eta $ , $p_1,p_2 \in \mathbb {Q}_{\mathbf n}^{2}$ , and there is $h<\omega $ such that $min \{ nor(p_1), nor(p_2) \} \leq h$ and $h<lg(\eta )$ , then $p_1$ and $p_2$ are compatible.

Proof. For every $\nu \in T_{p_1}\cap T_{p_2}$ , by the co-bigness property, $min\{nor_{\nu }(suc_{p_1}(\nu )), suc_{p_2}(\nu )\}-\frac {1}{\mu _{\nu }}\leq nor(suc_{p_1}(\nu ) \cap suc_{p_2}(\nu ))$ . By the definition of $nor(p_i)$ (recalling that $lg(\eta )^2 \leq \mu _{\eta }$ ), $1+\frac {1}{h+1} \leq (1+\frac {1}{h+1})+(\frac {1}{(h+1)^2}-\frac {1}{\mu _{\eta }}) \leq (1+\frac {1}{h+1})+(\frac {1}{h}-\frac {1}{h+1}-\frac {1}{\mu _{\nu }})=1+\frac {1}{h}-\frac {1}{\mu _{\nu }} \kern1.5pt{\leq}\kern1.5pt min\{nor(p_1),nor(p_2)\}-\frac {1}{\mu _{\nu }} \kern1.5pt{\leq}\kern1.5pt min\{nor_{\nu }(suc_{p_1}(\nu )),\ suc_{p_2}(\nu )\}-\frac {1}{\mu _{\nu }}$ . Therefore $1+\frac {1}{h+1}\leq nor(suc_{p_1}(\nu ) \cap suc_{p_2}(\nu ))$ , so $p_1 \cap p_2$ is as required. Hence,

$\square _3\, p$ and q are compatible.

Proof. Suppose WLOG that $tr(p)\leq tr(q) \in T_p$ and pick h such that $1+\frac {1}{h} \leq nor(p),nor(q)$ . By $\square _1$ , there is $\eta \in T_p \cap T_q$ such that $h<\lg (\eta )$ . Now $p\leq p^{[\eta ]}, q\leq q^{[\eta ]}$ , and $(p^{[\eta ]},q^{[\eta ]})$ satisfy the assumptions of $\square _2$ , therefore they’re compatible and so are p and q.

The proof is similar if $tr(q)\leq tr(p)\in T_q$ . The implication in the other direction is easy.

2) The proof is similar. First observe that if $\eta \in lim(T_p) \cap lim(T_q)$ , then $lim(nor_{\eta \restriction l}(suc_{T_p}(\eta \restriction l)) : l<\omega )=\infty =lim(nor_{\eta \restriction l}(suc_{T_q}(\eta \restriction l)) : l<\omega )$ , so by the co-bigness property (Definition 2(f)), $lim(nor_{\eta \restriction l}(suc_{T_{p\cap q}}(\eta \restriction l)) : l<\omega )=\infty $ . Now let $\nu =tr(q) \in T_p \cap T_q$ , as $2-\frac {1}{\mu _{tr(p)}} \leq nor(p),nor(q)$ , it follows from the co-bigness property and Definition 2(g) that $\nu \leq \eta \in T_p \cap T_q \rightarrow 2<|Suc_{p\cap q}(\eta )|$ , so $p\cap q$ is a perfect tree. It’s easy to see that there exists $\eta \in p\cap q$ such that $nor_{\nu }(Suc_{p\cap q}(\nu ))>2$ for every $\eta \leq \nu \in p\cap q$ (otherwise, we can repeat the argument in the proof of Claim 6, and get a branch through $p\cap q$ along which the norm doesn’t tend to infinity). Therefore, $p^{[\leq \eta ]} \cap q^{[\leq \eta ]} \in \mathbb {Q}_{\mathbf n}^i\, (i\in \{0,\frac {1}{2}\})$ is a common upper bound. Finally, note that if $i=1$ , then for every $n<\omega $ there exist $k^p(n+1),k^q(n+1)$ as in Definition 3(3). By the co-bigness property, for every $\eta \in T_p \cap T_q$ of length $>max\{k^p(n+1),k^q(n+1)\}$ , $n\leq nor_{\eta }(Suc_{p\cap q}(\eta ))$ . Therefore, the common upper bound is in $\mathbb {Q}_{\mathbf n}^1$ as well.

Claim 8. Let $I\subseteq \mathbb {Q}_{\mathbf n}^{2}$ be an antichain and $A=\cup \{T_q^+ : q\in I\}\subseteq T_{\mathbf n}$ . The following conditions are equivalent:

1. (a) I is a maximal antichain.

2. (b) If $\eta \in T_{\mathbf n}$ and $0<n<\omega $ then there is no $p\in \mathbb {Q}_{\mathbf n}^{2}$ such that:

   1. (α) $tr(p)=\eta $ .

   2. (β) $nor(p)=n$ .

   3. (γ) p is incompatible with every $q\in I$ .

3. (c) Like (b), but replacing $(\gamma )$ by

   1. (γ)' $T_p^+\cap A=\emptyset $ .

   2. (d) Like (b), but replacing $(\gamma )$ by

   1. (γ)'' For every $m>n\, T_p^+\cap A$ is disjoint to $\{\nu \in T_{\mathbf n} : lg(\nu )\leq m\}$ .

4. (e) If $\eta \in T_{\mathbf n}$ and $n<\omega $ then for some $m>n$ there is no set T such that:

   1. (α) $T\subseteq T_{\mathbf n}$ .

   2. (β) $\eta \in T$ .

   3. (γ) If $\nu \in T^+$ then $\eta \leq \nu $ and $lg(\nu )\leq m$ .

   4. (δ) If $\eta \leq \nu _1 \leq \nu _2$ and $\nu _2\in T$ then $\nu _1 \in T$ .

   5. (ε) $T\cap A=\emptyset $ .

   6. (ζ) If $\nu \in T$ and $lg(\nu )<m$ then $1+\frac {1}{n} \leq nor_{\nu }(suc_T(\nu ))$ .

Proof. $\neg (a) \rightarrow \neg (b):$ If p is incompatible with every $q\in I$ then $(p,tr(p),nor(p))$ is a counterexample to (b).

$\neg (b)\rightarrow \neg (c):$ If $(p,tr(p),nor(p))$ is a counterexample to (b), then it is a counterexample to (c) by the characterisation of compatibility in $\mathbb {Q}_{\mathbf n}^{2}$ in Claim 7.

$\neg (c) \rightarrow \neg (d):$ Obvious.

$\neg (d) \rightarrow \neg (e)$ : Let $T=T_p$ with p being a counter example to $(d)$ and let $\eta =tr(p), n$ witness $\neg (d)$ . We shall check that for every $m>n$ , $\{ \nu : tr(p) \leq \nu \in T \wedge lg(\nu ) \leq m\}$ satisfies $(\alpha )-(\zeta )$ if $(e)$ .

$\neg (e) \rightarrow \neg (a):$ If $(\eta , n)$ is a counterexample, then for every m there is $T_m$ satisfying $(\alpha )-(\zeta )$ of clause $(e)$ . Let D be a non-principal ultrafilter on $\omega $ and define $T:=\{\nu \in T_{\mathbf n} : \nu \leq \eta $ or $\{m : m>n, \nu \in T_m\}\in D\}$ . It remains to show that $T\in \mathbb {Q}_{\mathbf n}^{2}$ (as $T^+$ is disjoint to A, it follows that I is not a maximal antichain). The proof is similar to Claim 12.

Claim 9. Let $\mathbf n \in \mathbf N$ .

1. A) The sets $\mathbb {Q}_{\mathbf n}^1$ and $\mathbb {Q}_{\mathbf n}^2$ are Borel, the sets $\mathbb {Q}_{\mathbf n}^0$ and $\mathbb {Q}_{\mathbf n}^{\frac {1}{2}}$ are $\Pi _1^1$ .

2. B) The relation $\leq _{\mathbb {Q}_{\mathbf n}^i}$ is Borel for $i\in \{0,\frac {1}{2},1,2\}$ .

3. C) The incompatibility relation in $\mathbb {Q}_{\mathbf n}^i$ is Borel for $i\in \{0,\frac {1}{2},1,2\}$ .

Proof. A. The sets $\mathbb {Q}_{\mathbf n}^1$ and $\mathbb {Q}_{\mathbf n}^2$ are Borel: We shall first prove the claim for $\mathbb {Q}_{\mathbf n}^1$ . Consider $T_{\mathbf n}$ as a subset of $H(\aleph _0)$ . By definition, if $p\in \mathbb {Q}_{\mathbf n}^{1}$ then $T_p \subseteq T_{\mathbf n}\subseteq H(\aleph _0)$ . Hence $S:=\{p\subseteq H(\aleph _0) : p$ is a perfect subtree of $T_{\mathbf n} \}\subseteq P(H(\aleph _0))$ is a Borel subset of $P(H(\aleph _0))$ . For every $n,k<\omega $ define $S_{n,k}^{1}=\{p\in S : lg(tr(p))<k$ and if $\rho \in T_p$ and $k\leq lg(\rho )$ then $n\leq nor_{\rho }(suc_p(\rho ))\}$ . Each $S_{n,k}^{1}$ is closed (for fixed $\rho $ and n, the condition $n\leq nor_{\rho }(suc_p(\rho ))$ is closed, and so we get an intersection of closed sets), hence $S \cap (\underset {n}{\cap }\underset {k}{\cup }S_{n,k}^{1})$ is Borel, so it’s enough to show that $p\in \mathbb {Q}_{\mathbf n}^{1}$ iff $p\in S \cap (\underset {n}{\cap }\underset {k}{\cup }S_{n,k}^{1})$ and $2-\frac {1}{\mu _{tr(p)}} \leq nor(p)$ , which follows directly from the definition of $\mathbb {Q}_{\mathbf n}^1$ .

In the case of $\mathbb {Q}_{\mathbf n}^{2}$ , we replace $\underset {n}{\cap }\underset {k}{\cup }S_{n,k}^{1}$ with $\underset {n,k}{\cup }S_{n,k}^{2}$ where $S_{n,k}^{2}=\{p\in S : lg(tr(p))=n \wedge nor(p)=k\}$ . Each $S_{n,k}^{2}$ is Borel and since “being a perfect subtree” is Borel, $\mathbb {Q}_{\mathbf n}^{2}$ is Borel.

The sets $\mathbb {Q}_{\mathbf n}^0$ and $\mathbb {Q}_{\mathbf n}^{\frac {1}{2}}$ are $\Pi _1^1$ : The demand “ $\underset {n<\omega }{lim}(nor_{\eta \restriction n}(Suc_p(\eta \restriction n)))=\infty $ for every $\eta \in lim(T_p)$ ” is $\Pi _1^1$ , and it’s easy to see that $\{p\in S : tr(p) \leq \eta \in T_p \rightarrow nor_{\eta }(Suc_{T_p}(\eta ))>2\}$ is Borel.

B. The relation $\leq _{\mathbb {Q}_{\mathbf n}^i}$ is Borel for $i\in \{0,\frac {1}{2},1,2\}$ : For $i\in \{0,\frac {1}{2},1,2\}$ , the relation $\leq _{\mathbb {Q}_{\mathbf n}^{i}}$ is simply the reverse inclusion relation restricted to $\mathbb {Q}_{\mathbf n}^{i}$ , hence it is Borel.

C. The incompatibility relation in $\mathbb {Q}_{\mathbf n}^i$ is Borel for $\{0,\frac {1}{2},1,2\}$ : The incompatibility relation is Borel by Claim 7.

Claim 10.

1. A) Assume that $p_l \in \mathbb {Q}_{\mathbf n}^{i}\, (l<n)$ where $i\in \{0,1\}$ , $\underset {l<n}{\wedge }tr(p_l)=\rho $ , $n\leq lg(\rho )$ and for every $\eta \in p_l^+$ we have $2\leq k+1\leq nor_{\eta }(suc_{p_l}(\eta ))$ , then $\{p_l : l<n\}$ have a common upper bound p such that $tr(p)=\rho $ and $k\leq nor_{\eta }(suc_p(\eta ))$ for every $\eta \in T_p^+$ .

2. B) Assume that $p_l \in \mathbb {Q}_{\mathbf n}^{2}\, (l<n)$ , $\underset {l<n}{\wedge }tr(p_l)=\rho $ , $n\leq lg(\rho )$ and for every $\eta \in p_l^+\, (l<n)$ we have $1+\frac {1}{k}\leq nor_{\eta }(suc_{p_l}(\eta ))$ . In addition, assume that $k\leq lg(\rho )$ and $k(k+1) \leq \mu _{\eta }$ for every $\eta \in p_l^+\, (l<n)$ , then $\{p_l : l<n\}$ have a common upper bound p such that $tr(p)=\rho $ and $1+\frac {1}{k+1}\leq nor_{\eta }(suc_p(\eta ))$ .

Proof. A) Suppose first that $i=0$ . Let $p=\underset {l<n}{\cap }p_l$ , then $p\subseteq T_{\mathbf n}^{[\rho \leq ]}$ is a subtree containing $\rho $ . If $\nu \in p$ then $\nu \in p_l$ for every $l<n$ , hence $Suc_p(\nu )=\underset {l<n}{\cap }Suc_{p_l}(\nu )$ . As $n\leq lg(\rho ) \leq \mu _{\eta }$ for every $\rho \leq \eta \in p$ , it follows from the properties of the norm in the definition of $\mathbf n \in \mathbf N$ that $k\leq nor_{\eta }(Suc_p(\eta ))$ . Therefore, $T_p$ is a perfect tree, and similarly to the proof of Claim 7, it follows that the norm along infinite branches tends to infinity, hence $p\in \mathbb {Q}_{\mathbf n}^0$ . Suppose now that $i=1$ . The above arguments are still valid, and in addition, similarly to the argument on $\mathbb {Q}_{\mathbf n}^1$ in the proof of claim 7(2), it’s easy to see that by the co-bigness property, $p\in \mathbb {Q}_{\mathbf n}^1$ .

Claim 11. Let $\mathbf n \in \mathbf N$ . “ $\{p_n : n<\omega \}$ is a maximal antichain” is Borel for $\{p_n : n<\omega \}\subseteq \mathbb {Q}_{\mathbf n}^{2}$ .

Proof. By Claim 8.

Claim 12. Assume $\{p_n : n<\omega \} \subseteq \mathbb {Q}_{\mathbf n}^{2}$ , $\underset {n}{\wedge }tr(p_n)=\eta $ and $\underset {n}{\wedge }nor(p_n) \leq k$ . Then there is $p_* \in \mathbb {Q}_{\mathbf n}^{2}$ such that:

1. (a) $tr(p_*)=\eta $ , $nor(p_*) \leq k$ .

2. (b) $p_* \Vdash _{\mathbb {Q}_{\mathbf n}^{2}} ``(\exists ^{\infty }n)(p_n \in G_{\mathbb {Q}_{\mathbf n}^{2}})$ .”

Proof. Let D be a uniform ultrafilter on $\omega $ and define $T_{p_*}:=\{\nu \in T_{\mathbf n} : \{n : \nu \in p_n\}\in D\}$ . If $\nu \in T_{p_*}$ , then for some n, $\nu \in T_{p_n}\subseteq T_{\mathbf n}^{[\eta \leq ]}$ (recalling that $tr(p_n)=\eta $ ), hence $T_{p_*}\subseteq T_{\mathbf n}^{[\eta \leq ]}$ . Obviously, $l\leq lg(\eta ) \rightarrow \eta \restriction l \in T_{p_*}$ as $\eta =tr(p_n)\in p_n$ for every n.

$(*)_1$ If $\eta \triangleleft \nu \triangleleft \rho $ and $\rho \in T_{p_*}$ , then $\nu \in T_{p_*}$ .

Why? Define $A_{\rho }=\{n : \rho \in p_n\}$ and define $A_{\nu }$ similarly. $A_{\rho } \in D$ by the definition of $T_{p_*}$ . Obviously $A_{\rho }\subseteq A_{\nu }$ , hence $A_{\nu } \in D$ and $\nu \in T_{p_*}$ .

$(*)_2$ If $\nu \in T_{p_*}$ then $1+\frac {1}{k}\leq nor_{\nu }(suc_{p_*}(\nu ))$ .

Claim 12’. For $\iota \in \{0,\frac {1}{2},1,2\}$ , $\underset {\sim }{\eta _{\mathbf n}^{\iota }}$ is a generic for $\mathbb {Q}_{\mathbf n}^{\iota }$ , i.e., $\Vdash _{\mathbb {Q}_{\mathbf n}^{\iota }} ``V[\underset {\sim }{G_{\mathbb {Q}_{\mathbf n}^{\iota }}}]=V[\underset {\sim }{\eta _{\mathbf n}^{\iota }}]$ .”

Proof. Easy.

Claim 19. Forcing with $\mathbb {Q}_{\mathbf n}^{\iota }\, (i\in \{0,\frac {1}{2},1,2\})$ adds a Cohen real.

Proof. For every $\eta \in T_{\mathbf n}$ let $g_{\eta }: suc_{T_{\mathbf n}}(\eta )\rightarrow \{0,1\}$ be a function such that $|g_{\eta }^{-1}\{l\}|>\frac {\lambda _{\eta }}{2}-1\, (l=0,1)$ (recall that $\lambda _{\eta }=|suc_{T_{\mathbf n}}(\eta )|$ ). Define a $\mathbb {Q}_{\mathbf n}^{\iota }$ -name $\underset {\sim }{\nu }$ by $\underset {\sim }{\nu }(n)=g_{\underset {\sim }{\eta _{\mathbf n}^{\iota }} \restriction n}(\underset {\sim }{\eta _{\mathbf n}^{\iota }}\restriction (n+1))$ (recalling $\underset {\sim }{\eta _{\mathbf n}^{\iota }}$ is the generic). Clearly, $\Vdash _{\mathbb {Q}_{\mathbf n}^{\iota }}$ “ $\underset {\sim }{\nu } \in 2^{\omega }$ .” We shall prove that it’s forced to be Cohen.

$(*)$ If $p\in \mathbb {Q}_{\mathbf n}^{\iota }$ and $i=1\rightarrow 2\leq nor_{\rho }(suc_p(\rho ))$ for every $\rho \in T_p$ , then for every $\eta \in 2^{\omega }$ , for some $\rho \in T_p$ , $lg(\rho )=lg(tr(p))+m$ and if $lg(tr(p))\leq i<tr(p)+m$ then $p^{[\rho ]}\Vdash $ “ $\underset {\sim }{\nu }(i)=\eta (i)$ .”

We prove it by induction on m. For $m=1$ , as $|suc_{T_{\mathbf n}}(tr(p))\setminus suc_p(tr(p))|<\frac {|suc_{T_{\mathbf n}}(tr(p))|}{2}-1$ (by clause (g) of Definition $2$ ) and for every $i\in \{0,1\}$ we have $|g_{tr(p)}^{-1}\{i\}|>\frac {\lambda _{tr(p)}}{2}-1$ , hence there are $\rho _0,\rho _1 \in suc_p(tr(p))\setminus \{\rho \}$ such that $g_{tr(p)}(\rho _0)=0, g_{tr(p)}(\rho _1)=1$ and by the definition of $\underset {\sim }{\nu }$ , $p^{[\rho _0]}\Vdash $ “ $\underset {\sim }{\nu }(tr(p)+1)=0$ ” and $p^{[\rho _1]}\Vdash $ “ $\underset {\sim }{\nu }(tr(p)+1)=1$ .” Suppose that we proved the theorem for m, then for some $\rho \in T_p$ of length $lg(tr(p))+m$ the conclusion holds. Now repeat the argument of the first step of the induction for $p^{[\leq \rho ]}$ to obtain $\rho \leq \rho '$ of length $lg(tr(p))+m+1$ as required.

By $(*)$ , $\underset {\sim }{\nu }$ is forced to lie in every open dense set, hence it’s Cohen.

Claim 20. If

A) then B) where

1. A) (a) $p_i \in \mathbb {Q}_{\mathbf n}^{\iota }$ for $i<m$ .

   1. (b) $tr(p_i)=\rho $ for $i<m$ .

   2. (c) If $\iota \in \{0,1\}$ then $2\leq nor(p_i)$ for every $i<m$ .

   3. (d) If $\iota =2$ then $2\leq nor(p_i)$ for every $i<m$ .

   4. (e) $lg(\rho )<m_*<m$ .

   5. (f) There is $\rho <\eta \in T_{\mathbf n}$ such that $\lambda _{<\eta } \leq m_*<m \leq \mu _{\eta }$ (e.g., it follows from the assumption $m\leq \mu _{\eta } \iff m_* \leq \lambda _{\leq \eta }$ ).

2. B) There is an equivalence relation E on $\{0,1,\ldots ,m-1\}$ with $\leq m_*$ equivalence classes such that if $i<m$ then $\{p_j : j\in (i/E)\}$ has a common upper bound.

Proof. Let $\eta \in T_{\mathbf n}^{[\rho \leq ]}$ be as in clause (f). Let $k_*=lg(\eta )$ and define $\lambda _{\mathbf {n},k}:=\Pi \{\lambda _{\nu } : \nu \in T_{\mathbf n}, lg(\nu )<k\}$ , $T_{\mathbf n, \rho , k}:=\{\nu \in T_{\mathbf n} : \rho \leq \nu \in T_{\mathbf n},lg(\nu )=k\}$ . Recall that $\lambda _{\nu }$ is the size of $suc_{\mathbf n}(\nu )$ , hence $|T_{\mathbf n, \rho , k_*}|$ is the product of all $\lambda _{\nu }$ such that $\rho \leq \nu $ and $lg(\nu )<k_*$ , which is $\leq \lambda _{\mathbf {n},k_*}$ . For each $i<m$ let $\rho _i \in p_i$ be of length $k_*$ , then $\rho _i \in T_{\mathbf n, \rho , k_*}$ by the definition of $T_{\mathbf n, \rho , k_*}$ and the assumptions on $p_i$ . Define $\rho _{i}^{+}$ for $i<m$ as follows: if $\lambda _{\eta } < \lambda _{\rho _i}$ , define $\rho _{i}^{+}:= \rho _i$ . Otherwise we let $\rho _{i}^{+}\in suc_{p_i}(\rho _i)$ .

Define the equivalence relation $E:=\{(i,j) : \rho _{i}^{+}=\rho _{j}^{+}\}$ . Let $j<m$ , for every $i\in (j/E)$ define $p_i'=p_{i}^{[\rho _{j}^{+}]}$ (this is well defined, as $\rho _{i}^{+}=\rho _{j}^{+}$ ), then $tr(p_i')=\rho _{j}^{+}$ for every $i \in (j/E)$ . By the choice of $\eta $ , for $j<m$ , $|j/E|\leq m\leq \mu _{\eta } \leq \mu _{\rho _j^+}$ (by the choice of $\rho _j^+$ and Definition $2$ ).

By Claim 10, the set $\{p_i' : i \in (j/E)\}$ has a common upper bound, hence $\{p_i : i \in (j/E)\}$ has a common upper bound.

By the choice of $p_i^+$ , the number of E-equivalence classes is bounded by $\lambda _{<\eta }$ . As $\lambda _{<\eta } \leq m_*$ , we’re done.

Claim 21. We have $p_* \Vdash _{\mathbb {P}}$ “ $\underset {\sim }{\rho }$ is not $(\mathbb {Q}_{\mathbf n}^{\iota }, \underset {\sim }{\eta _{\mathbf n}^{\iota }})$ -generic over V” when:

1. a) $\iota \in \{1,2\}$ and $\alpha _* <\omega $ .

2. b) $(\mathbb {P}_{\alpha }, \underset {\sim }{\mathbb {Q}_{\alpha }} : \alpha <\alpha _*)$ is an FS iteration with limit $\mathbb {P}=\mathbb {P}_{\alpha _*}$ .

3. c) $\mathbf {n}_{\alpha } \in \mathbf {N}$ is special (note: $\mathbf {n}_{\alpha }$ is not a $\mathbb {P}_{\alpha }$ -name).

4. d) $\Vdash _{\mathbb {P}_{\alpha }} $ “ $\underset {\sim }{\mathbb {Q}_{\alpha }}={(\mathbb {Q}_{\mathbf {n}_{\alpha }}^{\iota })}^{V^{\mathbb {P}_{\alpha }}}$ .”

5. e) $\mathbf {n} \in \mathbf {N}$ is special.

6. f) For every $\alpha $ , $\mathbf {n}$ and $\mathbf {n}_{\alpha }$ are far (i.e., $\eta _1 \in T_{\mathbf n} \wedge \eta _2 \in T_{\mathbf {n}_{\alpha }}\rightarrow \lambda _{\eta _1}^{\mathbf n} \ll \mu _{\eta _2}^{\mathbf {n}_{\alpha }}$ or $\lambda _{\eta _2}^{\mathbf {n}_{\alpha }} \ll \mu _{\eta _1}^{\mathbf n})$ . Moreover, for every $\alpha <\alpha _*$ for every l large enough, for some $m\in \{l,l+1\}$ we have:

If $\rho \in T_{\mathbf n}, lg(\rho )=l$ , $\nu _1 , \nu _2 \in T_{\mathbf {n}_{\alpha (l)}}$ and $lg(\nu _1) < m \leq lg(\nu _2))$ then $\lambda _{\mathbf {n}_{\alpha (l)},\nu _1} \ll \mu _{\mathbf {n}, \rho }$ and $\lambda _{\mathbf {n}, \rho } \ll \mu _{\mathbf {n}_{\alpha (l)}, \nu _2}$ .

1. g) $p_* \Vdash _{\mathbb {P}} $ “ $\underset {\sim }{\rho } \in lim(T_{\mathbf n})$ .”

Proof. For $\eta \in T_{\mathbf n}$ define $W_{\mathbf n,\eta }:=\{w : w\subseteq suc_{T_{\mathbf n}}(\eta )$ and $i=1\rightarrow lg(\eta ) \leq nor_{\eta }(w)$ and $i=2 \rightarrow 2\leq nor_{\eta }(w)\}$ . For $n<\omega $ define $\Lambda _n=\{\eta \in T_{\mathbf n} : lg(\eta )<n\}$ , so $T_{\mathbf n}=\underset {n<\omega }{\cup }\Lambda _n$ . Define $S_n:=\{\bar {w} : \bar {w}=(w_{\eta } : \eta \in \Lambda _n \wedge w_{\eta } \in W_{\mathbf n,\eta })\}$ and $S=\underset {n<\omega }{\cup }S_n$ . $(S, \leq )$ is a tree with $\omega $ levels such that each level is finite and $lim(S)=\{\bar {w} : \bar {w}=(w_{\eta } : \eta \in T_{\mathbf n})$ and $\bar {w} \restriction \Lambda _n \in S_n$ for every $n\}$ . For $\bar {w} \in lim(S)$ let $\mathbf {B}_{\bar {w}}:=\{ \rho \in lim(T_{\mathbf n}):$ for every n large enough, $\rho \restriction (n+1) \in w_{\rho \restriction n}\}$ , so $\mathbf {B}_{\bar {w}}=\underset {m<\omega }{\cup }\mathbf {B}_{\bar {w},m}$ where $\mathbf {B}_{\bar {w},m}=\{\rho \in lim(T_{\mathbf n}) :$ if $m\leq n$ then $\rho \restriction (n+1) \in w_{\rho \restriction n}\}$ . We shall prove that

$(*) \Vdash _{\mathbb {Q}_{\mathbf n}^{\iota }} $ “ $\underset {\sim }{\eta _{\mathbf n}^{\iota }} \in \mathbf {B}_{\bar {w}}$ ” for every $\bar {w} \in lim(S)$ . In fact, for every $p\in \mathbb {Q}_{\mathbf n}^{\iota }$ there is a stronger q and $m<\omega $ such that $lim(q) \subseteq \mathbf {B}_{\bar w, m}$ .

Let $p\in \mathbb {Q}_{\mathbf n}^{\iota }$ , we shall prove that for some $p\leq q$ and $m<\omega $ , $q\Vdash \underset {\sim }{\eta _{\mathbf n}^{i}} \in \mathbf {B}_{\bar {w},m}$ . Let $\nu \in T_p$ such that $lg(\nu )$ is large enough and let $m=lg(\nu )$ . Now q will be defined by taking the subtree obtained from the intersection of $T_p^{[\leq \nu ]}$ with $(\underset {\nu \leq \rho }{\cup }w_{\rho })$ . By the co-bigness property, q is a well defined condition, and obviously $q\Vdash \underset {\sim }{\eta _{\mathbf n}^{i}} \in \mathbf {B}_{\bar {w},m}$ .

By $(*)$ it suffices to prove that for some $\bar {w} \in lim(S)$ , $p_* \nVdash _{\mathbb {P}} $ “ $\underset {\sim }{\rho } \in \mathbf {B}_{\bar {w}}$ .”

Proof. Assume towards contradiction that $p\Vdash $ “ $\underset {\sim }{\rho } \in \mathbf {B}_{\bar {w}}$ for every $\bar {w} \in lim(S)$ ,” so there is a sequence $(p_{\bar {w}} : \bar {w} \in lim(S))$ and a sequence $(m(\bar {w}) : \bar {w} \in lim(S))$ such that:

1. a) $p_* \leq p_{\bar {w}}$ .

2. b) $p_{\bar {w}} \Vdash \underset {\sim }{\rho } \in \mathbf {B}_{\bar {w},m(\bar {w})}$ .

By increasing the conditions $p_{\bar {w}}$ if necessary, we may assume WLOG that:

1. 1. $tr(p_{\bar {w}}(\alpha ))$ is an object for every $\bar {w}$ and every $\alpha \in Dom(p_{\bar {w}})$ .

2. 2. If $\iota =1$ and $\alpha \in Dom(p_{\bar {w}})$ , then $p_{\bar w} \restriction \alpha \Vdash _{\mathbb {P}_{\alpha }} $ “ $\nu \in p_{\bar w}(\alpha ) \rightarrow nor_{\nu }(Suc_{p_{\bar w}(\alpha )}(\nu )) >2$ .”

If $\iota =2$ and $\alpha \in Dom(p_{\bar {w}})$ , then for some $m \ll lg(tr(p_{\bar {w}}(\alpha )))$ , $p_{\bar {w}} \restriction \alpha \Vdash _{\mathbb {P}_{\alpha }} \nu \in p_{\bar {w}}(\alpha ) \rightarrow 1+\frac {1}{m} \leq nor(suc_{p_{\bar {w}}(\alpha )}(\nu ))$ .

In order to prove (1)+(2), we shall prove by induction on $\beta \leq \alpha _*$ that for every $p \in \mathbb {P}_{\beta }$ there is $p\leq q \in \mathbb {P}_{\beta }$ satisfying (2) and forcing a value to the relevant trunks.

The induction step: assume that $\beta =\gamma +1$ . As $p(\gamma )$ is a $\mathbb {P}_{\gamma }$ -name of a condition in $\mathbb {Q}_{\mathbf n}^2$ , there are $p \restriction \gamma \leq p' \in \mathbb {P}_{\gamma }$ and $\rho $ such that $p' \Vdash _{\mathbb {P}_{\gamma }} tr(p(\gamma ))=\rho $ . As $p' \Vdash _{\mathbb {P}_{\gamma }} p(\gamma ) \in \mathbb {Q}_{\mathbf n}^2$ and by the definition of $\mathbb {Q}_{\mathbf n}^2$ , there is $p' \leq p''$ and $m\leq \mu _{lg(\rho )}$ such that $p'' \Vdash _{\mathbb {P}_{\gamma }} \nu \in p(\gamma ) \rightarrow 1+\frac {1}{m} \leq nor(suc_{p(\gamma )}(\nu ))$ . Now choose $m \ll m_1$ , so $p'' \Vdash _{\mathbb {P}_{\gamma }} $ “there is $\nu \in p(\gamma )$ such that $lg(\nu )=m_1$ .” Therefore there are $p''\leq p^*$ and $\nu $ of length $m_1$ such that $p^* \Vdash _{\mathbb {P}_{\gamma }} $ “ $\nu \in p(\gamma ) \wedge (\nu \leq \eta \in p(\gamma ) \rightarrow 1+\frac {1}{m} \leq nor(suc_{p(\gamma )}(\eta )))$ .” By the induction hypothesis, there is $p^* \leq q' \in \mathbb {P}_{\gamma }$ satisfying (1)+(2). Now define $q:=q' \cup (\gamma ,p(\gamma )^{[\nu \leq ]})$ , obviously q is as required. The proof for $\mathbb {Q}_{\mathbf n}^1$ is similar.

Now we shall define a partition of $lim(S)$ to $\aleph _0$ sets as follows:

Let $W_{m,u,\bar {\rho }}=\{\bar {w} \in lim(S) : m(\bar {w})=m, Dom(p_{\bar {w}})=u\in [\alpha _*]^{<\aleph _0}, \bar {\rho }=(tr(p_{\bar {w}}(\alpha )) : \alpha \in u)\}$ . Choose $(m_*, u_*, \bar {\rho }_*)$ such that $W=W_{m_*, u_*, \bar {\rho }_*}\subseteq lim(S)$ is not meagre. Let $\bar {u}_* \in S$ such that W is comeager above $\bar {u}_*$ . Let $l(*)$ be such that $\bar {u}_* \in S_{l(*)}$ .

Denote $\bar {\rho }^*=(\rho _{\alpha }^{*} : \alpha \in u_*)$ , let $(\alpha _n : n<n(*))$ list $u_*$ in increasing order and let $\alpha _{n(*)}=\alpha _*$ . Therefore, if $\bar {u}_* \leq \bar {w} \in W$ then $Dom(p_{\bar {w}})=\{\alpha _0,\ldots ,\alpha _{n(*)-1}\}$ and $tr(p_{\bar {w}}(\alpha _n))=\rho _{\alpha _n}^*$ for every $n<n(*)$ .

By our assumption, $\mathbf n$ is far from $\mathbf {n}_{\alpha }$ . As increasing $\bar {u}_*$ is not going to change the argument, we may assume that $l(*)$ is large enough so $\underset {\alpha \in u_*}{\wedge }lg(\rho _{\alpha }^*)<l(*)$ and if $l<n(*)$ , $\nu \in T_{\mathbf n}$ , $\rho \in T_{\mathbf {n}_{\alpha _l}}$ and $lg(\bar {u}_*) \leq lg(\nu )$ , then $\lambda _{\mathbf n,\nu } \ll \mu _{\mathbf {n}_{\alpha _l},\rho }$ or $\lambda _{\mathbf {n}_{\alpha _l},\rho } \ll \mu _{\mathbf n,\nu }$ . Note that we don’t have to assume that $lg(\bar {u}_*)\leq lg(\rho )$ : For every $n<n(*)$ , there is $m_n$ as guaranteed by $(f)$ , with $(\mathbf {n}_{\alpha _n}, lg(\nu ),m_n)$ here standing for $(\mathbf {n}_{\alpha },l,m)$ there. If $lg(\rho ) \leq m_n$ , then by taking an arbitrary $\nu _2$ of length $>m_n$ , it follows from $(f)$ that $\lambda _{\mathbf {n}_{\alpha _n},\rho } \ll \mu _{\mathbf n,\nu }$ . If $m_n< lg(\rho )$ , then by taking an arbitrary $\nu _2$ of length $\leq m_n$ , we get $\lambda _{\mathbf {n},\nu } \ll \mu _{\mathbf {n}_{\alpha _n},\rho }$ .

Recalling $(f)$ (and by increasing $\bar {u}_*$ if necessary), let $(m_n : n<n(*))$ be a series of natural numbers such that $(\mathbf n, \mathbf {n}_{\alpha _n}, l(*), m_n)$ satisfy that assumptions of $(f)$ (with $(\mathbf n, \mathbf {n}_{\alpha _n}, l(*), m_n)$ here standing for $(\mathbf n,\mathbf {n}_{\alpha },l,m)$ there).

Let $\Lambda _m^0=\Lambda _{m+1} \setminus \Lambda _m=\{ \rho \in T_{\mathbf n} : lg(\rho )=m\}$ and let $S_m^0=\{\bar {w} : \bar {w}=(w_{\eta } : \eta \in \Lambda _m^0),$ for every $\eta \in \Lambda _m^0,\, w_{\eta } \in W_{\mathbf n,\eta } \}$ .

Recalling that above $\bar {u}_*$ , W is nowhere meagre, for every $\bar {v} \in S_{l(*)}^0$ there is $\bar {w}_{\bar {v}} \in W \subseteq lim(\mathbb {S})$ such that $\bar {u}_*{}^\frown \bar {v} \leq \bar {w}_{\bar {v}}$ .

Choose $p_n, U_n$ by induction on $n\leq n(*)$ such that:

1. 1. $p_n \in \mathbb {P}_{\alpha _n}$ .

2. 2. If $m<n$ then $p_m \leq p_n \restriction \alpha _m$ .

3. 3. $U_n \subseteq S_{l(*)}^0$ .

4. 4. If $m<n$ then $U_n \subseteq U_m$ .

5. 5. If E is an equivalence relation on $U_n$ with $\leq \Pi \{|T_{\mathbf {n}_{\alpha _l},m_l}| : n\leq l<n(*)\}$ equivalence classes, then for some $\bar {v}_* \in U_n$ , $\cap \{ \underset {\rho \in T_{\mathbf n,l(*)}}{\cup } w_{\bar {v}_\rho } : \bar {v} \in \bar {v}_*/E\}=\emptyset $ .

6. 6. If $\bar {v} \in U_n$ then $p_{\bar {w}_{\bar {v}}} \restriction \alpha _n \leq p_n$ .

Suppose we’ve carried the induction, then for every $\bar {v} \in U_{n(*)}$ , $p_{\bar {w}_{\bar {v}}}=p_{\bar {w}_{\bar {v} \restriction \alpha _{n(*)}}} \leq p_{n(*)}$ , hence by the choice of $p_{\bar {w}_{\bar {v}}}$ , $p_{n(*)} \Vdash \underset {\sim }{\rho } \in \cap \{\mathbf {B}_{\bar {w}_{\bar {v}}, m_*} : \bar {v} \in U_{n(*)}\}$ . Therefore it’s enough to show that $\cap \{\mathbf {B}_{\bar {w}_{\bar {v}}, m_*} : \bar {v} \in U_{n(*)}\}=\emptyset $ . By its definition, $\mathbf {B}_{\bar {w}_{\bar {v}},m_*}=lim(T_{\bar {v}})$ where $T_{\bar {v}}=\{ \eta \in T_{\mathbf n} : $ if $m_*<lg(\eta )$ then $\eta (m+1) \in w_{\eta \restriction m}$ for every $m_* \leq m \}$ . Therefore, if we show that $\cap \{T_{\bar {v}} \cap T_{\mathbf n,l(*)+1} : \bar {v} \in U_{n(*)}\}=\emptyset $ , then it will follow that $\cap \{lim(T_{\bar {v}}) : \bar {v} \in U_{n(*)}\}=\emptyset $ . This follows from part (5) of the induction hypothesis, as $\cap \{ \underset {\rho \in T_{\mathbf n,l(*)}}{\cup } w_{\bar {v}_\rho } : \bar {v} \in U_{n(*)} \}=\emptyset $ . This contradiction proves the claim.

Carrying the induction: For $n=0$ , choose any $p_0 \in \mathbb {P}_{\alpha _0}$ and let $U_0=S_{l(*)}^0$ . It’s enough to show that $U_0$ satisfies $(5)$ . Let E be an equivalence relation on $U_0$ with $m_{**} \leq \Pi \{|T_{\mathbf {n}_{\alpha (l)},m_l}| : l<n(*)\}$ equivalence classes and denote $\Pi \{|T_{\mathbf {n}_{\alpha (l)},m_l}| : l<n(*)\}$ by $m'$ . For every $m<m_{**}$ , denote by $U_{0,m}$ the mth equivalence class of E. Suppose towards contradiction that for every $m<m_{**}$ there is some $\eta _m$ in $\cap \{\underset {\rho }{\cup }w_{\rho } : \bar {w} \in U_{0,m}\}$ . For every m there is $\rho _m$ such that $\eta _m \in suc_{T_{\mathbf n}}(\rho _m)$ . Choose $\bar {w}=(w_{\rho } : \rho \in T_{\mathbf {n},l(*)})$ by letting $w_{\rho }=suc_{T_{\mathbf n}}(\rho ) \setminus \{ \eta _m : m<m_{**} \wedge \rho _m=\rho \}$ . We shall prove that $\bar {w} \in U_0$ . It will then follow that $\bar {w} \in U_{0,m}$ for some m, therefore $\eta _m \in \underset {\rho }{\cup }w_{\rho }$ , contradicting the definition of $w_{\rho }$ . This proves that $U_0$ is as required. In order to prove that $\bar {w} \in U_0$ , note that for every $\rho $ , $|suc_{T_{\mathbf n}}(\rho ) \setminus w_{\rho }| \leq |\{m : \rho _m=\rho \}| \leq m_{**} \leq m'=\Pi \{|T_{\mathbf {n}_{\alpha (l)},m_l}| : l<n(*)\} \ll \mu _{\mathbf n, \rho }$ (the last inequality follows by $(f)$ and the choice of $m_l$ , recalling that the $m_l$ were chosen to satisfy the assumptions of (f) and recalling the definition of the $\lambda _{\mathbf n, \nu }$ ). Therefore, $\bar {w} \in U_0$ .

Suppose now that $n=k+1\leq n(*)$ . Choose $q_k \in \mathbb {P}_{\alpha _k}$ such that $p_k \leq q_k$ and $q_k$ forces a value $\Lambda _{\bar {v}}^{k}$ to $\{\rho \in p_{\bar {w}_{\bar {v}}}(\alpha _k) : lg(\rho )=m_k\}$ for every $\bar {v} \in U_k$ . For every $\rho \in T_{\mathbf {n}_{\alpha _k},m_k}$ let $U_{k,\rho }=\{\bar {v} \in U_k : \rho \in \Lambda _{\bar {v}}^{k}\}$ . If $\bar {v} \in U_k$ , then $q_k$ forces the value $\Lambda _{\bar {v}}^{k}$ to $\{\rho \in p_{\bar {w}_{\bar {v}}}(\alpha _k) : lg(\rho )=m_k\}$ , hence $U_k=\cup \{U_{k,\rho } : \rho \in T_{\mathbf {n}_{\alpha },m_k}\}$ . WLOG $U_{k,\rho }$ are pairwise disjoint. Now suppose towards contradiction that none of them satisfies requirement (5) of the induction for $k+1$ , then each $U_{k,\rho }$ has a counterexample $E_{\rho }$ , and the union $\underset {\rho }{\cup }E_{\rho }$ is therefore an equivalence relation which is a counterexample to $U_k$ satisfying (5). Therefore, for some $\rho $ , $U_{k,\rho }$ satisfies (5), so choose $U_n=U_{k,\rho }$ .

Define $p_n \in \mathbb {P}_{\alpha _k +1} \subseteq \mathbb {P}_{\alpha _n}$ as follows:

1. 1. $p_n \restriction \alpha _k =q_k$ .

2. 2. $p_n(\alpha _k)=\cap \{p_{\bar {w}_{\bar {v}}}(\alpha _k)^{[\rho \leq ]} : \bar {v} \in U_n\}$ .

Now for every $\bar {v} \in U_k$ , $p_{\bar {w}_{\bar {v}}} \restriction \alpha _k \leq p_k \leq q_k$ , hence $q_k \Vdash _{\mathbb {P}_{\alpha _k}} \nu \in p_{\bar {w}_{\bar {v}}}(\alpha _k) \rightarrow 1+\frac {1}{m} \leq nor(suc_{p_{\bar {w}_{\bar {v}}}(\alpha _k)}(\nu ))$ . We shall prove that $q_k \Vdash _{\mathbb {P}_{\alpha _k}} p_n(\alpha _k) \in \mathbb {Q}_{\mathbf {n}_{\alpha }}^2$ . As, $|U_n| \leq |S_{l(*)}^0| \leq 2^{\Sigma \{\lambda _{\mathbf n,\rho '} : \rho ' \in \Lambda _{l(*)}^0\}}<\mu _{\mathbf {n}_{\alpha _k},\rho }$ (with the last inequality following from (f) and the choice of the $m_k$ ’s), the assumptions of Claim 10 hold, the conclusion follows by the proof of Claim 10. A similar argument (using the first part of Claim 10) proves the claim for the case of $\mathbb {Q}_{\mathbf n}^1$ .

So $p_n$ obviously satisfies requirements 1, 2, and 6.

Claim 22. For $\iota \in \{0,\frac {1}{2},1,2\}$ we have $\Vdash _{Cohen} $ “there is a Borel set $\mathbf {B} \subseteq lim(T_{\mathbf {n}_*})$ such that $lim(T_{\mathbf {n}_*})^V \subseteq \mathbf {B}$ and $\mathbf {B}$ is $(\mathbb {Q}_{\mathbf {n}_*}^{\iota }, \underset {\sim }{\eta _{\mathbf {n}_*}^{\iota }})$ -null” (where by “ $(\mathbb {Q}_{\mathbf {n}_*}^{\iota }, \underset {\sim }{\eta _{\mathbf {n}_*}^{\iota }})$ -null” we mean that for every p there is a stronger q with $lim(q) \cap \mathbf {B}=\emptyset $ ).

Proof. Let $\mathbb {Q}$ be the set of finite functions with domain $\{\eta \in T_{\mathbf {n}_*} : lg(\eta )<k\}$ for some $k<\omega $ such that $f(\rho ) \in suc_{T_{\mathbf {n}_*}}(\rho )$ . $(\mathbb {Q}, \subseteq )$ is countable and for every $q\in \mathbb Q$ there are $q\leq q_1,q_2 \in \mathbb Q$ which are incompatible, hence is equivalent to Cohen forcing. Let $\underset {\sim }{f}:=\underset {g\in \underset {\sim }{G}}{\cup }g$ . For $f\in S=\Pi \{suc_{T_{\mathbf {n}_*}}(\rho ) : \rho \in T_{\mathbf {n}_*}\}$ define $\mathbf {B}_f:= \{\eta \in lim(T_{\mathbf {n}_*}) :$ for infinitely many n we have $\eta \restriction (n+1)=f(\eta _n)\}$ . For every $n<\omega $ let $\mathbf {B}_{f,n}=\{\eta \in lim(T_{\mathbf {n}_*}) : \eta \restriction (m+1)\neq f(\rho )$ if $n\leq m$ and $n\leq lg(\rho )\}$ . Clearly, $\Vdash $ “ $\underset {\sim }{f} \in S$ ,” $\mathbf {B}_f^c=\underset {n<\omega }{\cup }\mathbf {B}_{f,n}$ , and obviously each $\mathbf {B}_{f,n}$ is Borel, hence $\mathbf {B}_f$ is Borel. For every $\eta \in T_{\mathbf {n}_*}$ let $w_{\eta }=suc_{T_{\mathbf {n}_*}}(\eta )\setminus \{f(\eta )\}$ . As in Claim 21, $\Vdash _{\mathbb {Q}_{\mathbf {n}_*}^{\iota }} $ “ $\underset {\sim }{\eta _{\mathbf {n}_*}^{\iota }} \in \mathbf {B}_{\bar {w}}$ ” for $\bar {w}$ and $\mathbf {B}_{\bar {w}}$ as in that proof. Hence $\Vdash _{\mathbb {Q}_{\mathbf {n}_*}^{\iota }} $ “ $\underset {\sim }{\eta _{\mathbf {n}_*}^{\iota }} \notin \mathbf {B}_f$ ,” so $\mathbf {B}_f$ is $(\mathbb {Q}_{\mathbf {n}_*}^{\iota }, \underset {\sim }{\eta _{\mathbf {n}_*}^{\iota }})$ -null. Let $G\subseteq \mathbb Q$ be generic and let $g=\underset {\sim }{f}[G]$ , so $\mathbf {B}_g$ is a $(\mathbb {Q}_{\mathbf {n}_*}^{\iota }, \underset {\sim }{\eta _{\mathbf {n}_*}^{\iota }})$ -null Borel set in $V[G]$ . We shall prove that $V[G] \models lim(T_{\mathbf {n}_*})^V \subseteq \mathbf {B}_g$ . Let $\eta \in lim(T_{\mathbf {n}_*})^V$ and $m<\omega $ , it’s enough to show that in V, $\Vdash _{\mathbb Q} $ “for some $m\leq k$ and $\rho \in T_{\mathbf {n}_*}, \underset {\sim }{f}(\rho )=\eta \restriction (k+1)$ .” Let $p\in \mathbb {Q}$ , we can extend p to a function $p\leq q$ with domain $\{\eta \in T_{\mathbf {n}_*} : lg(\eta )<k\}$ for some $m\leq k$ . Now let $q\leq s$ be an extension of q with domain $\{\eta \in T_{\mathbf {n}_*} : lg(\eta ) \leq k\}$ such that $s(\eta \restriction k)=\eta \restriction (k+1)$ . Obviously, s forces the required conclusion, so we’re done.

Proof. Clause a) By the definition of $\mathbb {P}_{\mathbf q}$ and claim 17, so $|\mathbb {P}_{\mathbf q}| \leq \Sigma \{ |\mathbb {P}_{\mathbf q,J}| : J\subseteq L$ is finite $\} \leq 2^{\aleph _0}+|L|^{<\aleph _0}=2^{\aleph _0}+\mu =\mu $ .

Clause b) By a) we have $\Vdash _{\mathbb {P}_{\mathbf q}} ``2^{\aleph _0} \leq \mu $ ,” and as $|L|=\mu $ we have $\Vdash _{\mathbb {P}_{\mathbf q}} ``\mu =|L|\leq |\{\underset {\sim }{\eta _t} : t\in L\}| \leq 2^{\aleph _0}$ .” Together we’re done.

Clause c) $(\alpha )$ By the definitions of $V[X^+]$ and $\mathbb {P}_{\mathbf q}$ . In particular, $\neg AC_{\aleph _0}$ , as we can use $(A_n : n<\omega )$ where $A_n:=\{\{\eta _{t_l} : l<n\} : t_0<_L\dots <_L t_{n-1}\}$ . As $V[X^+]$ is really just $HOD(V\cup X^{<\omega })$ in $V[G]$ , $V[X^+] \models ZF$ follows by the standard arguments in the literature.

Clause c) $(\beta )$ Let $A\in V[X^+]$ be a subset of $lim(T_{\mathbf {m}_*})$ . A is definable in $V[G]$ by a first order formula $\phi (x, \bar {a},{c})$ such that $c\in V$ and $\bar {a}=(\eta _{t_0},\dots ,\eta _{t_{n-1}})$ is a finite sequence from X. Let $J=\{s\in L_{\mathbf q} : s\leq t_l$ for some $l\}$ . For $s\in L\setminus J$ let $L_s=\{t_l : l<n\} \cup \{s\}$ , then $L_s \in \mathbf {L}_{\mathbf q}$ hence by 14 we have $\mathbb {P}_{L_s}\lessdot \mathbb {P}_{L_{\mathbf q}}$ . Let $\underset {\sim }{T_s}=TV(\phi (\underset {\sim }{\eta _s},\bar {a},c))$ , so $\underset {\sim }{T_s}$ is a $\mathbb {P}_{L_{\mathbf q}}$ -name and actually a $\mathbb {P}_{L_s}$ -name.

Let $(p_{s,i} : i<\omega )$ be a maximal antichain in $\mathbb {P}_{L_s}$ and let $W_s \subseteq \omega $ such that $p_{s,i}\Vdash \underset {\sim }{T_s}=true$ if and only if $i\in W_s$ . Define the $\mathbb {P}_{\{t_l : l<n\}}$ -name $\underset {\sim }{U}:= \{i<\omega : p_{s,i} \restriction \{t_l : l<n\} \in \underset {\sim }{G_{\mathbb {P}_{\{t_l : l<n\}}}}\}$ .

If $G_0 \subseteq \mathbb {P}_{\{t_l : l<n\}}$ is generic over V and $U=\underset {\sim }{U}[G_0]$ , then in $V[G_0]$ , $(lim(p_{s,i}(s)[G_0]) : i \in U)$ are pairwise disjoint: by Claim $7$ , if $p,q\in \mathbb {Q}_{\mathbf n}^{\iota }$ are incompatible and $\eta \in lim(p)$ , then $\eta \notin lim(q)$ (otherwise, WLOG $lg(tr(p))\leq lg(tr(q))$ , and both $tr(p)$ and $tr(q)$ are initial segments of $\eta $ , hence $tr(p)\leq tr(q) \in T_p$ which is a contradiction by Claim $7$ ). Hence it’s enough to show that $((p_{s,i}(s)[G_0]) : i \in U)$ is an antichain in $V[G_0]$ . Assume towards contradiction that for some $i \neq j\in U$ there is a common upper bound q for $p_{s,i}(s)[G_0]$ and $p_{s,j}(s)[G_0]$ . Therefore there is a $\mathbb {P}_{\{t_l : l<n\}}$ -name $\underset {\sim }{q}$ and $r\in G_0$ such that $r\Vdash _{\mathbb {P}_{\{t_l : l<n\}}} ``p_{s,i}(s), p_{s,j}(s) \leq \underset {\sim }{q}$ .” Since $i,j\in U$ , we have $p_{s,i}\restriction \{t_l : l<n\}, p_{s,j}\restriction \{t_l : l<n\} \in G_0$ , and as $G_0$ is directed, there is a common upper bound $r_1 \in G_0$ for $p_{s,i}\restriction \{t_l : l<n\}, p_{s,j}\restriction \{t_l : l<n\}$ and r. Now let $r^+:=r_1 \cup \{(s,\underset {\sim }{q})\} \in \mathbb {P}_{L_s}$ , then obviously $r^+$ is a common upper bound (in $\mathbb {P}_{L_s}$ ) for $p_{s,i}$ and $p_{s,j}$ , which contradicts our assumption.

Moreover, $(p_{s,i}(s)[G_0] : i\in U)$ is a maximal antichain: If $q\in {\mathbb {Q}_{\mathbf n}^{\iota }}^{V[G_0]}$ is incompatible with $p_{s,i}(s)[G_0]$ for every $i\in U$ , then as before, there are $r\in G_0$ and a $\mathbb {P}_{\{t_l : l<n\}}$ -name $\underset {\sim }{q}$ such that r forces that $\underset {\sim }{q}$ is incompatible with $p_{s,i}(s)$ for every $i\in U$ . As before we can get a member of $\mathbb {P}_{L_s}$ that is incompatible with $(p_{s,i} : i<\omega )$ , contradicting its maximality. Hence $(p_{s,i}(s)[G_0] : i\in U)$ is a maximal antichain in $V[G_0]$ .

If $s_1,s_2 \in L_{\mathbf q}\setminus J$ , by the homogeneity assumption, there is an automorphism f of $L_{\mathbf q}$ over J such that $f(s_1)=s_2$ . Therefore the natural map induced by f is mapping $\bar {a}$ to itself and $\underset {\sim }{\eta _{s_1}}$ to $\underset {\sim }{\eta _{s_2}}$ . Hence $\underset {\sim }{T_{s_1}}$ is mapped to $\underset {\sim }{T_{s_2}}$ . As $(\hat {f}(p_{s_1,i}) : i<\omega )$ and $W_{s_1}$ have the same properties (with respect to $\underset {\sim }{T_{s_2}}$ ) as $(p_{s_2,i} : i<\omega )$ and $W_{s_2}$ , we may assume WLOG that $W_{s_1}=W_{s_2}$ (denote it by W) and $\hat {f}(p_{s_1,i})=p_{s_2,i}$ .

Therefore, if $G_0 \subseteq \mathbb {P}_{\{t_l : l<n\}}$ is generic and $i\in \underset {\sim }{U}[G_0]$ , then there is $p_i \in {(\mathbb {Q}_{\mathbf n}^{\iota }})^{V[G_0]}$ and W such that for every $s\in L\setminus J$ , $p_{s,i}(s)[G_0]=p_i$ and $W_s=W$ (and W can be found in the ground model).

Work now in $V[G_0]$ : Let $B:=\cup \{lim(p_i) : i\in W \cap U\}$ , so B is a Borel set and we shall prove that $A=B$ modulo the ideal: by clauses $(c)(\gamma )+(c)(\delta )$ proved below, it’s enough to show that if $s\in L_{\mathbf q}\setminus J$ , then ${\eta _s}\notin A\Delta B$ (note that, by its definition, $J\in V$ and hence $J\in V[X^+]$ ).

Let $s\in L_{\mathbf q}\setminus J$ and $i\in U$ , then $p_{s,i} \in \mathbb {P}_{L_s}/G_0$ and by the choice of $p_{s,i}$ , $p_{s,i} \Vdash _{\mathbb {P}_{L_s}/G_0} ``\phi (\underset {\sim }{\eta _s}, \bar {a},c) $ iff $\underset {\sim }{T_s}=true$ iff $i\in W$ .” In other words, in $V[G_0]$ we have $p_i \Vdash _{\mathbb {Q}_{\mathbf n}^{\iota }} ``\phi (\underset {\sim }{\eta _s}, \bar {a},c)$ iff $i\in W$ .” Since $(p_i : i\in U)$ is a maximal antichain, every $G\subseteq \mathbb {Q}_{\mathbf n}^{2}$ generic over $V[G_0]$ must contain exactly one of the $p_i$ , hence in $V[G_0]:\, \Vdash _{\mathbb {Q}_{\mathbf n}^{\iota }} ``\phi (\underset {\sim }{\eta _s}, \bar {a},c)$ iff $i\in W$ for the $p_i$ such that $p_i \in \underset {\sim }{G}$ .” Now $p_{s,i}(s)=p_i \in \underset {\sim }{G}$ iff $\underset {\sim }{\eta _s} \in lim(T_{p_{s,i}(s)})=lim(T_{p_i})$ , hence we got $\Vdash _{\mathbb {Q}_{\mathbf n}^{2}} ``\phi (\underset {\sim }{\eta _s}, \bar {a},c)$ iff $i\in W$ where i is such that $\underset {\sim }{\eta _s} \in lim(T_{p_i})$ .” Therefore $\Vdash _{\mathbb {Q}_{\mathbf n}^{\iota }} ``\underset {\sim }{\eta _s} \in A$ iff $\underset {\sim }{\eta _s}\in B$ .”