Claim 2.7.

1. (a) $Y_{\epsilon } \neq \emptyset $ for every $\epsilon <\omega _1$ .

2. (b) For every $\epsilon <\zeta <\omega _1$ and $(\Lambda _1,h_1) \in Y_{\epsilon }$ , there exists $(\Lambda ,h) \in Y_{\zeta }$ such that $\Lambda _1 \subseteq \Lambda $ .

Proof. Proof of (a). We prove the claim by induction on $\epsilon <\omega _1$ . This is clear for $\epsilon =0.$ Now suppose that $\epsilon =\zeta +1$ and the claim holds for $\zeta .$ Let $(\Lambda ,h) \in Y_{\zeta }$ and define $(\Lambda ', h')$ as follows:

• • $\Lambda ' = \bigcup _{k<\omega } \langle k \rangle ^{\frown } \Lambda $ , where $\langle k \rangle ^{\frown } \Lambda = \{ \langle \rangle \} \cup \{ \langle k \rangle ^{\frown } t: t \in \Lambda \}$ ;

• • $\mathrm {dom}(h')=\Lambda '$ ;

• • $h'(\langle \rangle )=\epsilon $ ;

• • for $k<\omega $ and $t \in \Lambda $ , $h'(\langle k \rangle ^{\frown } t)=h(t).$

It is easily seen that $(\Lambda ', h') \in Y_{\epsilon }.$ Now suppose that $\epsilon $ is a limit ordinal, and the claim holds for all $\zeta < \epsilon .$ Let $(\zeta _{k} : k<\omega )$ be an increasing sequence of ordinals cofinal in $\epsilon $ . By the induction hypothesis, choose an increasing sequence $((\Lambda _{k},h_{k}) : k<\omega )$ such that $(\Lambda _{k},h_{k}) \in Y_{\zeta _{k}}$ . Define $(\Lambda ', h')$ as follows:

• • $\Lambda ' = \bigcup _{k<\omega } \langle k \rangle ^{\frown } \Lambda _k$ ;

• • $\mathrm {dom}(h')=\Lambda '$ ;

• • $h'(\langle \rangle )=\epsilon $ ;

• • for $k<\omega $ and $t \in \Lambda _k$ , $h'(\langle k \rangle ^{\frown } t)=h_k(t).$

Again, it is easy to show that $(\Lambda ', h') \in Y_{\epsilon }.$

Proof of (b). Assume $\epsilon <\zeta <\omega _1$ and $(\Lambda _1,h_1) \in Y_{\epsilon }$ . By clause (a), pick some $(\Lambda _2,h_2) \in Y_{\zeta }$ and define $(\Lambda , h)=(\Lambda _1,h_1)+(\Lambda _2,h_2)$ as follows:

• • $\Lambda =\Lambda _1 \cup \Lambda _2$ .

• • $h: \Lambda \rightarrow \zeta +1$ is defined as

  $$ \begin{align*} h(\eta)= \left\{ \begin{array}{@{}ll} h_1(\eta) \hspace{3.8cm} &\text{ if } \eta \in \Lambda_1 \setminus \Lambda_{2};\\ h_2(\eta) \hspace{3.8cm} &\text{ if } \eta \in \Lambda_2 \setminus \Lambda_{1};\\ \max\{h_1(\eta),h_2(\eta)\} \hspace{1.5cm} &\text{ if } \eta \in \Lambda_1 \cap \Lambda_2. \end{array} \right. \end{align*} $$

It is easy to see that $(\Lambda ,h) \in Y_{\zeta }$ and $\Lambda _1 \subseteq \Lambda $ .

Claim 2.8. $B_{\epsilon } \in I_{\mathbb D,\aleph _0}$ .

Proof. We have

Thus we have to show that $\eta _{dom} \notin B_{\epsilon }$ . For this, it is enough to show that $I_{\Lambda }$ is a maximal antichain, as then, it will follow that some $t \in \Omega _{\Lambda }^+$ is an initial segment of $\eta _{dom}$ , and therefore, by the definition of $B_\epsilon $ we have $\eta _{dom} \notin B_{\epsilon }$ , as requested.

First, we show that $I_{\Lambda }$ is an antichain. Suppose that $t \neq s$ are in $\Omega _{\Lambda }^+$ . We have to show that $(t, f_t)$ and $(s, f_s)$ are incompatible in $\mathbb D.$ Suppose towards contradiction that $(t, f_t)$ and $(s, f_s)$ are compatible, in particular t and s are $\lhd $ -comparable. Let us assume that $t \lhd s$ . Let us write t and s as

$$\begin{align*}t= t_0 ^{\frown} \langle 2n_0+1\rangle ^{\frown} t_1 ^{\frown} \langle 2n_1+1 \rangle ^{\frown} \dots ^{\frown} t_k ^{\frown} \langle 2n \rangle, \end{align*}$$

and

$$\begin{align*}s=s_0 ^{\frown} \langle 2m_0+1\rangle ^{\frown} s_1 ^{\frown} \langle 2m_1+1 \rangle ^{\frown} \dots ^{\frown} s_l ^{\frown} \langle 2m \rangle, \end{align*}$$

where $n_i, m_j, n, m < \omega $ , and $t_i, s_j \in \max (\Lambda )$ . The assumptions $t \lhd s$ and $t_i, s_j \in \max (\Lambda )$ imply that $k<l$ , $s_i=t_i$ for $i \leq k$ and $m_i=n_i$ for $i<k$ , hence we can write s as

$$\begin{align*}s=t_0 ^{\frown} \langle 2n_0+1\rangle ^{\frown} t_1 ^{\frown} \langle 2n_1+1 \rangle ^{\frown} \dots ^{\frown} t_k ^{\frown} \langle 2m_k+1 \rangle ^{\frown} \dots ^{\frown} s_l ^{\frown} \langle 2m \rangle. \end{align*}$$

Thus we must have $2n=2m_k+1$ which is impossible. Therefore, $I_{\Lambda }$ is an antichain.

Next we show that $I_{\Lambda }$ is a maximal antichain. Let $p=(t_p, f_p) \in \mathbb D$ . If there exists $t \in \Omega _{\Lambda }^+$ such that $t \unlhd t_p$ , then $(t, f_t) \leq p$ and we are done. Therefore, we may assume that there is no such t. Let

$$\begin{align*}\Omega^* &=\{\langle \rangle\} \cup \{t: (\exists t_0,\ldots, t_{k-1} \in \max(\Lambda)) t \\ &=t_0 ^{\frown} \langle 2n_0+1\rangle ^{\frown} \dots ^{\frown} t_{k-1}~ ^{\frown} \langle 2n_{k-1}+1\rangle \text{~and~} t \unlhd t_p\}. \end{align*}$$

Then $\Omega ^* \neq \emptyset $ , and since $t_p$ has finite length, there is an element t of $\Omega ^*$ of maximal length. Let $s_1 \in \Lambda _{\epsilon }$ be such that $t ^{\frown } s_1 \unlhd t_p$ and $s_1$ is maximal. There are two cases.

Case I. $t ^{\frown } s_1 =t_p$ . Let k be maximal such that

$$ \begin{align*}s_2 := s_1^{\frown} \langle f_p( \ell g(t)+\ell g(s_1)+i) : i<k \rangle \in \Lambda_{\epsilon}.\end{align*} $$

Note that by the construction, $\Lambda _{\epsilon }$ has no infinite branches, and since $s_1 \in \Lambda _\epsilon ,$ it follows that there is such maximal k. It then follows from the choice of $s_2$ that $s_2 \in \max (\Lambda _\epsilon )$ . Let

$$ \begin{align*}s= t ^{\frown} s_1 ^{\frown} \langle f_p( \ell g(t)+\ell g(s_1)+i) : i<k \rangle=t_p ^{\frown} \langle f_p( \ell g(t)+\ell g(s_1)+i) : i<k \rangle\end{align*} $$

then $t_p \unlhd s \in \Omega _{\Lambda }^+$ and $(s,f_s) \in I_\Lambda $ is compatible with p.

Case II. $t ^{\frown } s_1 \lhd t_p$ . First note that $s_1 \in \max (\Lambda _{\epsilon })$ , as otherwise, we will have $\mathrm {Suc}_{\Lambda _{\epsilon }}(t)=\{t^{\frown } \langle k \rangle : k<\omega \}$ , and then for some $k<\omega $ , $t ^{\frown } s_1^{\frown } \langle k \rangle \unlhd t_p$ , and $s_1^{\frown } \langle k \rangle \in \Lambda _\epsilon ,$ which contradicts the choice of $s_1$ as a maximal element of $\Lambda _\epsilon $ with $t ^{\frown } s_1 \unlhd t_p$ . Now there are two possibilities:

• • $t_p( \ell g(t)+\ell g(s_1))$ is odd: Then $t ^{\frown } s_1 ^{\frown } \langle t_p( \ell g(t)+\ell g(s_1)) \rangle \in \Omega ^*$ , which contradicts the maximality of t.

• • $t_p( \ell g(t)+\ell g(s_1))$ is even: Then we have $t ^{\frown } s_1 ^{\frown } \langle t_p( \ell g(t)+\ell g(s_1)) \rangle \in \Omega _\Lambda ^+$ and $t ^{\frown } s_1 ^{\frown } \langle t_p( \ell g(t)+\ell g(s_1)) \rangle \unlhd t_p$ , which contradicts our assumption that there is no element in $\Omega _\Lambda ^+$ which is $\unlhd t_p$ .

Thus case II cannot happen. The claim follows.

Claim 2.9. $\bigcup _{\epsilon <\omega _1}B_{\epsilon } \notin I_{\mathbb D,\aleph _0}$ .

Proof. Suppose towards contradiction that $\bigcup _{\epsilon <\omega _1}B_{\epsilon } \in I_{\mathbb D,\aleph _0}$ . It then follows that there exists a Borel set B such that $\bigcup _{\epsilon <\omega _1}B_{\epsilon } \subseteq B$ and $\Vdash _{\mathbb D}$ “ .”

For each condition $p \in \mathbb D$ set

$$\begin{align*}X(p)=\{g \in \omega^\omega: t_p \lhd g \text{~and~} f_p \leq g \}. \end{align*}$$

By the definition of the ideal $I_{\mathbb D,\aleph _0}$ , there exists a sequence $\bar I=\langle I_n: n<\omega \rangle $ of countable pre-dense subsets of $\mathbb D$ such that letting $I_n=\{p_{n,l} : l<\omega \}$ , we have $B \subseteq \omega ^\omega \setminus \big (\bigcap _{n<\omega } \bigcup _{l<\omega }X(p_{n,l})\big )$ .

Let $\chi $ be a large enough regular cardinal and let N be a countable elementary submodel of $L_{\chi }[\bar I, \langle \Lambda _{\epsilon } : \epsilon <\omega _1 \rangle ]$ such that $\bar I,\langle \Lambda _{\epsilon } : \epsilon <\omega _1 \rangle \in N$ . Let $\delta (*)=N\cap \omega _1^{L_{\chi }[\bar I, \langle \Lambda _{\epsilon } : \epsilon <\omega _1 \rangle ]}$ . Then $\delta (*) < \omega _1$ , and hence $\Lambda =\Lambda _{\delta (*)}$ is well-defined. Let $\bar I^*=\langle I^*_m : m<\omega \rangle $ enumerate all countable pre-dense subsets of $\mathbb D$ in N, and for every $m<\omega $ , set $I^*_m=\{ p_{m,l}^* : l<\omega \}$ . Note that for every $n<\omega $ , $I_n \in N$ , hence there exists some $j(n)<\omega $ such that $I_n=I^*_{j(n)}$ . We now prove the following:

1. (*) : There exists a sequence $\langle q_n: n<\omega \rangle $ of conditions such that:

   1. (1) $q_n=(t_n,f_n) \in \mathbb D \cap N$ .

   2. (2) $n=m+1 \Rightarrow q_m \leq q_n$ .

   3. (3) If $n=m+1$ , then there exists l such that $p_{m,l}^* \leq q_n$ .

   4. (4) $t_0=\langle \rangle $ .

   5. (5) If $n>0$ then $t_n=s_n ^{\frown } \langle 2m_n \rangle $ for some $s_n \in \Omega _{\Lambda }$ and $m_n<\omega $ .

We construct the sequence $\langle q_n: n<\omega \rangle $ by induction on n. For $n=0$ set $q_0=(\langle \rangle , \mathrm {id}_\omega )$ , where $\mathrm {id}_\omega $ is the identity function on $\omega .$ Now assume that $n=m+1$ and the condition $q_m=(t_m,f_m)$ is constructed, such that it satisfies the induction hypothesis. As $I^*_m=\{p_{m,l}^* : l<\omega \}$ is pre-dense, the set $E=\{p \in \mathbb D: (\exists l)(p_{m,l}^* \leq p)\} \in N$ is open dense in $\mathbb {D}$ . Let $I=\{r_l : l<\omega \} \subseteq E$ be a maximal antichain in $\mathbb D$ above $q_m$ . By elementarity, we can assume that $I \in N$ . Let $A=\{\mathrm {tr}(r_l) : l<\omega \}$ . Then by Lemma 2.4(b) and the fact that $q_m, A \in N$ , we have

• • $\mathrm {rk}_{q_m,A}(t_{m})<\omega _1$ ;

• • $\mathrm {rk}_{q_m,A}(t_{m}) \in N$ .

Thus $\mathrm {rk}_{q_m,A}(t_{m}) < \delta (*).$ Let $h_{\delta (*)} : \Lambda _{\delta (*)} \rightarrow \delta (*)+1$ witness $( \Lambda _{\delta (*)}, h_{\delta (*)}) \in Y_{\delta (*)}$ , and let $\Lambda ^*$ be the set of sequences $t \in \Lambda _{\delta (*)}$ satisfying the following conditions:

• • $q_m \leq (t_{m} ^{\frown } t, t_{m} ^{\frown } t ^{\frown } f_m \restriction [\ell g(t_{m} ^{\frown } t),\omega ))$ .

• • $\mathrm {rk}_{q_m, A}(t_{m} ^{\frown } t)< h_{\delta (*)}(t)$ .

As $\delta (*)$ is a limit ordinal, $\langle \rangle \in \Lambda ^*$ and hence $\Lambda ^* \neq \emptyset $ . Let

$$ \begin{align*}\alpha_*=\min \{\mathrm{rk}_{q_m, A}(t_{m} ^{\frown} t) : t \in \Lambda^*\},\end{align*} $$

and choose $t_* \in \Lambda ^*$ such that $\alpha _*=\mathrm {rk}_{q_m, A}(t_{m} ^{\frown } t_*)$ .

We will show that $\alpha _*=0$ . Assume towards a contradiction that $\alpha _*>0$ . As $t_* \in \Lambda ^*$ , we have $\alpha _*=\mathrm {rk}_{q_m, A}(t_m^{\frown } t_*) < h_{\delta (*)}(t_*)$ , therefore by the definition of $h_{\delta (*)}$ , for every k large enough, $\alpha _* \leq h_{\delta (*)}(t_* ^{\frown } \langle k \rangle )$ . By the definition of the rank, the set

$$ \begin{align*}U_1=\{k < \omega : \alpha_* \leq \mathrm{rk}_{q_m, A}(t_m ^{\frown}t_* ^{\frown} \langle k \rangle)\}\end{align*} $$

is finite. It is also clear that the set

$$ \begin{align*}U_2=\{k < \omega : k\leq f_m(\ell g(t_m^{\frown} t_*)) \}\end{align*} $$

is finite. It follows that the set $U=U_1 \cup U_2$ is finite, and hence, for every large enough k we have $k \notin U$ , and $\alpha _* <h_{\delta (*)}(t_* ^{\frown } \langle k \rangle )$ . For such k,

• • $\mathrm {rk}_{q_m, A}(t_m ^{\frown }t_* ^{\frown } \langle k \rangle ) < \alpha _*$ (as $k \notin U_1$ );

• • $\mathrm {rk}_{q_m, A}(t_m ^{\frown }t_* ^{\frown } \langle k \rangle ) < \mathrm {rk}_{q_m, A}(t_m ^{\frown }t_*)$ (by the definition of the rank).

Therefore, for every such k, $t_* ^{\frown } \langle k \rangle \in \Lambda ^*$ and $\mathrm {rk}_{q_m, A}(t_m ^{\frown }t_* ^{\frown } \langle k \rangle ) < \alpha _*$ , which contradicts the minimality of $\alpha _*$ . Thus $\alpha _*=0$ .

By the way we defined the rank, we can find some $t' \in A$ such that:

• • $q_m \leq (t',f_m \restriction [\ell g(t'),\omega ))$ ;

• • $\ell g(t') \leq \ell g(t_{m} ^{\frown } t_*)$ ;

• • $(t_{m} ^{\frown } t_*)(l) \leq t'(l)$ , for every $\ell g(t_m) \leq l <\ell g(t')$ .

As $A=\{\mathrm {tr}(r_l) : l<\omega \}$ and $t' \in A,$ we have $t'=\mathrm {tr}(r_{l_*})$ , for some $l_* < \omega $ . By induction on $i \geq \ell g(t')$ we choose some $k_i$ such that:

• • $f_m(i) \leq k_i$ ;

• • $f_{r_{l_*}}(i) \leq k_i$ ;

• • $s_{m,i}=t' \restriction [\ell g(t_m), \ell g(t')) ^{\frown } \langle k_j : \ell g(t') \leq j<i \rangle \in \Lambda _{\delta (*)}$ .

Note that $t' \restriction [\ell g(t_m), \ell g(t')) \in \Lambda _{\delta (*)},$ and since $\Lambda _{\delta (*)}$ has no infinite branches, it follows that there exists a maximal i for which we can choose $k_i$ as requested. Set

$$ \begin{align*}t"=t^{\prime\frown} \langle k_j : j<i+1 \rangle ^{\frown} \langle 2\big(f_{r_{l_*}}(\ell g(t')+i)+f_m(\ell g(t')+i)\big) \rangle.\end{align*} $$

Let $q_{m+1}=(t", t^{\prime \prime \frown } f \restriction [\ell g(t"),\omega )),$ where $f(i)=max \{f_m(i),f_{r_{l_*}}(i)\}$ for every $i\in [\ell g(t"),\omega ))$ . It’s easy to see that $r_{l_*},q_m \leq q_{m+1}$ and that the condition $q_{m+1}$ satisfies the requirements (1)–(5) of $(*)$ . This completes the construction of the sequence $\langle q_n: n<\omega \rangle $ .

As the sequence $\langle q_n=(t_n, f_n): n<\omega \rangle $ is increasing, the sequence $\langle t_n: n<\omega \rangle $ is increasing too, and hence $f=\bigcup _{n<\omega }t_n$ is a well-defined function.

We first show that $f \in \omega ^\omega $ . It suffices to show that $\mathrm {dom}(f)=\omega $ . Thus let $k<\omega .$ As N is an elementary submodel of $L_{\chi }[\bar I, \langle \Lambda _{\epsilon } : \epsilon <\omega _1 \rangle ]$ , there is a pre-dense set $I\in N$ such that for every $p \in I, k <\ell g(t_p)$ . Let m be such that $I=\{p_{m,l}^* : l<\omega \}$ . By the way we defined $q_{m+1},$ for some $l, p_{m,l}^* \leq q_{m+1}$ , hence $k< \ell g(t_{m+1})$ . Thus $k \in \mathrm {dom}(t_{m+1}) \subseteq \mathrm {dom}(f),$ as requested.

Now we show that $f \notin B.$ It suffices to show that f is $(N,\mathbb D)$ -generic. As the sequence $\langle q_n: n<\omega \rangle $ is increasing, it follows that $f \in X(q_n)$ for every $n<\omega $ . Let $I \in N$ be a countable pre-dense subset of $\mathbb D$ . Then for some m, $I=I^*_m=\{p_{m,l}^* : l<\omega \}$ . Let l be such that $p_{m,l}^* \leq q_{m+1}$ . It follows that $f \in X(p_{m,l}^*)$ , and hence $t_{p_{m,l}^*} \lhd f$ and $f_{p_{m,l}^*} \leq f.$ This implies f is $(N,\mathbb D)$ -generic, as wanted.

We now prove that $f \in B_{\delta (*)}$ . Let us recall that $B_{\delta (*)}=\{f \in \omega ^{\omega } : t \in \Omega _{\Lambda _{\delta (*)}}^+ \Rightarrow \neg (t \lhd f) \}.$ Suppose towards contradiction that $f \notin B_{\delta (*)}$ . Then for some $t \in \Omega _{\Lambda _{\delta (*)}}^+$ , we have $t \lhd f.$ Let n be large enough such that $t \lhd t_n \lhd f.$ Then $(t, f_t), (t_n, f_{t_n}) \in I_{\Lambda _{\delta (*)}}$ are compatible, as witnessed by $(t_n, f)$ , which contradicts the fact that $I_{\Lambda _{\delta (*)}}$ is an antichain.

But then $f \in B_{\delta (*)}$ and $B_{\delta (*)} \subseteq B$ , and hence $f \in B$ which is a contradiction. The claim follows.

Claim 2.11. For every $\alpha < \omega _1, f^{-1}(B_{\alpha }) \in I_{\mathbb Q,\aleph _0}.$

Proof. Let $\alpha < \omega _1.$ Then $\Vdash _{\mathbb Q}$ “ ,” and therefore $\Vdash _{\mathbb Q}$ “ .” Thus $f^{-1}(B_{\alpha }) \in I_{\mathbb Q,\aleph _0}$ , as required.

Claim 2.12. $\bigcup _{\alpha < \omega _1} f^{-1}(B_{\alpha }) \notin I_{\mathbb Q,\aleph _0}.$

Proof. Let $\chi $ be large enough regular and let $N \prec (H(\chi ), \in )$ be a countable elementary submodel of $H(\chi )$ containing all the relevant objects. It suffices to find $K \subseteq \mathbb Q \cap N$ such that K is $(N,\mathbb Q)$ -generic and .

Let $RO(\mathbb D)$ be the Boolean completion of $\mathbb D$ and let $\pi : \mathbb Q \rightarrow RO(\mathbb D)$ be a projection, which exists by our assumption on $\mathbb Q.$ By Lemma 2.5, there exists $G\subseteq RO(\mathbb D) \cap N$ which is $(N,RO(\mathbb D))$ -generic, such that . Let $H \subseteq \mathbb Q / G$ be $\mathbb Q / G$ -generic over $N[G],$ where $\mathbb Q / G=\{q \in \mathbb Q: \pi (q) \in G \}$ . Then $K=G \ast H$ is $(N,\mathbb Q)$ -generic and clearly .

Claim 3.4. Let $\kappa =\aleph _1^V.$ Then $L[J] \models $ “ $J \cap L[J]$ is a $\kappa $ -complete $\kappa $ -saturated ideal on $|U|$ .”

Proof. We first show that J is $\kappa $ -complete in V. Suppose towards a contradiction that $\langle X_n : n<\omega \rangle \in V$ is such that $X_n \in J$ for $n<\omega $ , but $X=\bigcup _{n<\omega }X_n \notin J$ . For $n<\omega $ set $A_n=\bigcup _{\alpha \in X_n} B_{\alpha }"$ . Then $\langle A_n : n<\omega \rangle \in V$ , and by the definition of the ideal J,

• • $A_n \in I$ , for $n<\omega $ ;

• • $A=\bigcup _{n<\omega }A_n \notin I$ .

This contradicts clause (2). It immediately follows that $L[J] \models $ “ $J \cap L[J]$ is a $\kappa $ -complete ideal on $|U|$ .”

We now show that $L[J]\models $ “ $\mathcal {P}(|U|)/(J \cap L[J])$ is $\kappa $ -cc.” Suppose not. Thus there exists a sequence $\langle A_{\alpha } : \alpha <\kappa \rangle \in L[J]$ of J-positive sets such that for every $\alpha <\beta < \kappa $ , $A_{\alpha } \cap A_{\beta } \in J$ . As J is $\kappa $ -complete in $L[J]$ , we may assume without loss of generality that $A_{\alpha } \cap A_{\beta }=\emptyset ,$ for every $\alpha <\beta < \kappa $ .

Work in V. For each $\alpha < \kappa $ set $B^*_\alpha =\bigcup _{\xi \in A_\alpha }B^{\prime \prime }_\xi .$ Then $\langle B^*_\alpha : \alpha < \kappa \rangle \in V$ satisfies the following conditions:

• • $B^*_\alpha \notin I$ for every $\alpha < \kappa $ (as $A_\alpha \notin J$ );

• • $B^*_\alpha \cap B^*_\beta =\emptyset $ , for $\alpha < \beta < \kappa .$

This contradicts clause (4). The claim follows.

Claim 4.10. Let t and k be as above.

1. (a) If $u\subseteq \mathrm {Suc}_{T_{\mathbf {n}}}(t)$ and $k+2 \leq \mathrm {nor}_{t}(u)$ , then $u\cap w_{t, k} \neq \emptyset $ .

2. (b) If $u\subseteq \mathrm {Suc}_{T_{\mathbf {n}}}(t)$ , $l<k$ and $l+1 \leq \mathrm {nor}_{t}(u)$ , then letting $v=u\setminus w_{t, k}$ , we have:

   1. (1) $v\subseteq u$ and $v\cap w_{t, k}=\emptyset $ .

   2. (2) $l\leq \mathrm {nor}_{t}(v)$ and $v\neq \emptyset $ .

   3. (3) If $\mathrm {nor}_{t}(u)>2$ , then $\min \{k, \mathrm {nor}_{t}(u)-1\} \leq \mathrm {nor}_{t}(v)$ .

Proof. (a). Otherwise, $u\subseteq Suc_{T_{\mathbf {n}}} \setminus w_{t, k}$ , and hence $\mathrm {nor}_{t}(\mathrm {Suc}_{T_{\mathbf {n}}} \setminus w_{t, k}) \geq \mathrm {nor}_{t}(u) \geq k+2$ , which is impossible.

(b). Clause (1) is clear. For clause (2), note that $v=u \cap (\mathrm {Suc}_{T_{\mathbf {n}}}(t) \setminus w_{t, k})$ , hence by the co-bigness property Definition 4.3(4)(c), $\mathrm {nor}_{t}(v)=\mathrm {nor}_{t}(u \cap (\mathrm {Suc}_{T_{\mathbf {n}}}(t) \setminus w_{t, k})) \geq l.$ For clause (3), let $j=\min \{k-1, \mathrm {nor}_{t}(u)-2\}=\min \{\mathrm {nor}_{t}(\mathrm {Suc}_{T_{\mathbf {n}}}(t) \setminus w_{t, k})-2, \mathrm {nor}_{t}(u)-2\}$ . By the co-bigness property, $\mathrm {nor}_{t}(v)=\mathrm {nor}_{t}(u \cap (\mathrm {Suc}_{T_{\mathbf {n}}}(t) \setminus w_{t, k})) \geq j+1$ , from which the result follows.

Claim 4.11. $\Vdash _{\mathbb {Q}_{\mathbf {n}}^1}$ “ , for every $n<\omega $ .

Proof. We prove the claim by induction on n. For $n=0$ the claim is obvious by the choice of . Now suppose that $n=m+1$ and the claim holds for m. Let $p\in \mathbb {Q}_{\mathbf {n}}^1$ be an arbitrary condition. We find an extension q of p, forcing . By extending p if necessary, we may assume that:

• • p decides , say it forces “ ” for every $i<n$ ;

• • $j_m+m+1 < \ell g(\mathrm {tr}(p))$ ;

• • $n+2 < \mathrm {nor}_{t}(\mathrm {Suc}_{T_p}(t))$ for every $\mathrm {tr}(p) \unlhd t \in T_p$ .

By Claim 4.10, $w_{\mathrm {tr}(p), n} \cap \mathrm {Suc}_{T_p}(\mathrm {tr}(p)) \neq \emptyset .$ Let $t \in w_{\mathrm {tr}(p), n} \cap \mathrm {Suc}_{T_p}(\mathrm {tr}(p))$ and let $q=(t, T_p^{[t]}),$ where $T_p^{[t]}=\{s \in T_p: s \unlhd t \text {~or~} t \unlhd s \}$ . Then

and hence $q \Vdash $ “ ,” as required.

Claim 4.12. Suppose $h\in \omega ^{\omega }$ and $p\in \mathbb {Q}_{\mathbf {n}}^1$ . Then there exists a condition $q \geq p$ such that $q\Vdash $ “ for every large enough n.”

Proof. Without loss of generality, we may assume that the function h is increasing. By extending p if necessary, we may also assume that $\ell g(\mathrm {tr}(p))> 2$ and for every $t \in T_p$ with $\mathrm {tr}(p) \unlhd t$ we have $2 < \mathrm {nor}_{t}(\mathrm {Suc}_{T_p}(t))$ .

Let m be maximal such that p decides , and let $\langle j_i: i < m+1 \rangle $ be such that $p \Vdash $ “ .” Let also $\langle n_i: i<\omega \rangle $ be an increasing sequence of natural numbers such that:

• • $n_0=\ell g(\mathrm {tr}(p))$ ;

• • $h(n_i) < n_{i+1},$ for every $i<\omega .$

Let

$$\begin{align*}T=\{t \in T_p: (\forall i<\omega)(\forall l\in [n_i,n_{i+1})) \big[ l < \ell g(t) \Rightarrow t \restriction (l+1) \notin w_{t \restriction l,i}\big]\}. \end{align*}$$

T is obviously downwards closed and by the co-bigness property, T is a perfect tree. Let $t \in T$ be such that $2< \mathrm {nor}_{s}(\mathrm {Suc}_{T}(s))$ for every $t \unlhd s \in T$ and set $q=(t, T^{[t]})$ . Then $q \in \mathbb {Q}_{\mathbf {n}}^1$ and $q \geq p$ . We show that

Suppose not. Thus we can find $r \geq q$ and $n> m$ such that

By extending r, we may assume that:

• • r decides , say it forces “ ”;

• • $j_n < \ell g(\mathrm {tr}(r))$ .

Let i be such that $n_i \leq j_n < n_{i+1}$ . Set $s_1= \mathrm {tr}(r) \restriction j_n$ and $s_2=\mathrm {tr}(r) \restriction (j_n+1).$ Then $s_2 \in \mathrm {Suc}_{T_q}(s_1)$ , and hence by the definition of the tree T we have $s_2 \notin w_{s_1, i}$ . On the other hand, $r \Vdash $ “ ,” and hence $s_2 \in w_{s_1, n}$ . As the sequence $\langle w_{s_1, k}: k \leq j_n \rangle $ is decreasing, we must have $n<i.$ It then follows that $ h(n)< h(i)< n_i \leq j_n,$ and hence

By $(*)_1$ and $(*)_2$ we get a contradiction.

Proof. We have to show that the set

$$\begin{align*}G=\{ (t, f) \in \mathbb{D}: t\lhd \rho \text{~and~} (\forall \ell g(t) \leq n<\omega) f(n) \leq \rho(n) \} \end{align*}$$

is a $\mathbb {D}$ -generic filter over V. Thus suppose that $I\subseteq \mathbb D$ is a maximal antichain in V. By our assumption, there exists $(t, f) \in I$ such that $t \lhd f$ and $f(n) \leq \rho (n)$ for every $n<\omega $ . It then follows that $(t, f) \in G \cap I$ , and hence $G \cap I \neq \emptyset .$ The result follows.

Proof. Set $p_1=p.$ Let $h'\in \omega ^{\omega }$ be a function such that for every $n<\omega $ ,

• • $f_n \leq ^* h'$ ;

• • $n<h'(n)<h'(n+1)$ .

Let also $h\in \omega ^{\omega }$ be defined as $h(n)=h'(n)+1$ .

By the proof of Lemma 4.9, there are $p_2$ and $n_1^*$ such that:

• • $p_1 \leq p_2$ ;

• • $2< n_1^* \leq \ell g(\mathrm {tr}(p_2))$ ;

• • $p_2 \Vdash $ “ .”

Proof. By extending p if necessary, assume that $\mathrm {nor}_{t}(\mathrm {Suc}_{T_p}(t))>2$ for every $\mathrm {tr}(p) \unlhd t \in T_p$ . Set $n_0=\ell g(\mathrm {tr}(p))$ . Now suppose that $i<\omega $ and $n_{i+1}$ is defined. We define $n_{i+2}$ .

Let $l_1 \in [n_i,n_{i+1})$ and $t_1 \in T_p \cap \omega ^{l_1}$ . We define a sequence $\langle s^{l_1, t_1}_{l} : l_1 \leq l<\omega \rangle $ by induction on $l \geq l_1$ such that:

• • $s^{l_1, t_1}_{l_1}=t_1$ ;

• • $s^{l_1, t_1}_{l+1} \in \mathrm {Suc}_{T_p}(s^{l_1, t_1}_l) \setminus w_{s^{l_1, t_1}_l,0}$ .

We can easily define such a sequence, as for each l, $\mathrm {Suc}_{T_p}(s^{l_1, t_1}_l) \setminus w_{s^{l_1, t_1}_l,0} \neq \emptyset .$ Let $\eta ^{l_1, t_1}=\bigcup _{l_1 \leq l<\omega }s^{l_1, t_1}_{l}$ . Then

$$\begin{align*}\lim_{n<\omega}(\mathrm{nor}_{\eta^{l_1, t_1} \restriction n}(\mathrm{Suc}_{T_p}(\eta^{l_1, t_1} \restriction n)))=\infty, \end{align*}$$

and therefore there exists $n^{l_1, t_1}_{i+2} \geq n_{i+1}$ such that $\mathrm {nor}_{\eta ^{l_1, t_1} \restriction m}(\mathrm {Suc}_{T_p}(\eta ^{l_1, t_1} \restriction m))>\beth _{i+1}(0)$ for every $n^{l_1, t_1}_{i+2} \leq m$ . Let

$$\begin{align*}n_{i+2} =\max\{ n^{l_1, t_1}_{i+2}: l_1 \in [n_i,n_{i+1}) \text{~and~} t_1 \in T_p \cap \omega^{l} \}+1. \end{align*}$$

Note that $n_{i+2}$ is well-defined as for each $l_1 \in [n_i,n_{i+1})$ , the set $T_p \cap \omega ^{l}$ is finite. It is easy to see that $n_{i+2}$ is as required.

Let $\langle n_i : i<\omega \rangle $ be a sequence as in Claim 4.16 for the condition $p_2.$ Let $j_*$ and $j_{**}$ be such that:

• • $j_*$ is maximal such that $p_2$ decides ;

• • $j_{**}$ is maximal such that $p_2$ decides .

Let also the sequences $\langle m_i: i < j_* \rangle $ and $\langle k_i: i < j_{**} \rangle $ be such that

and

Let $s_1=\langle m_i+l_{k_i} : i<j_{**} \rangle $ . Then

Define $h_1 \in \omega ^{\omega }$ by

$$ \begin{align*}h_1(i)=\beth_{h(i+1)+n_{i+2}}(0)+\max \{m_j : j<j_*\}+h(i+2).\end{align*} $$

Then $h_1 \geq h$ and $(s_1, s_1 \cup h_1 \restriction [\ell g(s_1),\omega )) \in \mathbb D.$ As I is a maximal antichain, there exists a condition $(s_2, h_2) \in \mathbb D$ which extends $(s_1, s_1 \cup h_1 \restriction [\ell g(s_1),\omega ))$ and a member of I. We show that there is an extension of $p_2$ which forces “ satisfies $(s_2, h_2)$ .”

Proof. We define a sequence $\langle t^*_i: i \in [\ell g(s_1), \ell g(s_2)] \rangle $ , by induction on i as follows. Let $t^*_{\ell g(s_1)}=\mathrm {tr}(p_2).$ Now suppose that $t^*_i$ is defined and $\ell g(t^*_i) \in [n_{i-\ell g(s_1)}, n_{i+1-\ell g(s_1)}]$ . By Claim 4.16, applied to $\ell g(t^*_i)$ and $t^*_i$ , there exist $l^* \in [n_{i+1-\ell g(s_1)}, n_{i+2-\ell g(s_1)}]$ and $t^* \in T_{p_2} \cap \omega ^{l^*}$ such that:

• • $t^*_i \unlhd t^*$ ;

• • for every $l \in [\ell g(t^*_i), l^*), t^* \restriction (l+1) \notin w_{t^* \upharpoonright l, 0}$ ;

• • $\beth _{i-\ell g(s_1)+1}(0)<\mathrm {nor}_{t^*}(\mathrm {Suc}_{T_p}(t^*))$ .

Let $u=\mathrm {Suc}_{T_{\mathbf {n}}}(t^*) \setminus \mathrm {Suc}_{T_{p_2}}(t^*)$ and $j=j_*+i - \ell g(s_1)$ . We show that $w_{t^*,j} \setminus w_{t^*,j+1} \nsubseteq u$ . Suppose not. It then follows that

$$ \begin{align*}|w_{t^*,j+1}| \leq \frac{|w_{t^*,j}|}{2} \leq |w_{t^*,j} \setminus w_{t^*,j+1}| \leq |u|,\end{align*} $$

and therefore, by the construction of condition $p_2$ and the choice of the w’s,

$$ \begin{align*}\beth_{i-\ell g(s_1)+1}(0) < \mathrm{nor}_{t^*}(\mathrm{Suc}_{T_{p_2}}(t^*))=\mathrm{nor}_{t^*}(\mathrm{Suc}_{T_{\mathbf {n}}}(t^*) \setminus u) \leq i-\ell g(s_1)+2,\end{align*} $$

which is not possible. Therefore $w_{t^*,j} \setminus w_{t^*,j+1} \nsubseteq u$ , and hence we can find some $t^*_{i+1} \in \mathrm {Suc}_{T_{p_2}}(t^*) \cap (w_{t^*, j} \setminus w_{t*,j+1}) \neq \emptyset .$ This defines $t^*_{i+1}.$

It is now easy to see that $p_{3, i}= (t^*_i, T_{p_2}^{[t^*_i]})$ ’s are as required.

Note that each $p_{3, i}$ from the above claim also decides say

Note also that for every $i\in (\ell g(s_1), \ell g(s_2))$ ,

$$ \begin{align*}m_{i-1} \leq \ell g(\mathrm{tr}(p_{3,i}))=\ell g(t^*_i) \leq n_{i+1} \leq \beth_{h(i)+n_{i+1}}(0) < h_1(i-1) \leq h_2(i-1).\end{align*} $$

Now choose $p_4$ such that:

• • $p_{3, \ell g(s_2)} \leq p_4$ ;

• • $\max (\mathrm {range}(s_2))<\ell g(\mathrm {tr}(p_4))$ ;

• • $p_4$ does not force a value for Footnote ⁶ .

Proof. We define $p_{5, i}$ ’s, by induction on i. Set $p_{5,\ell g(s_1)}=p_4$ . Now suppose that $i, i+1 \in [\ell g(s_1), \ell g(s_2)]$ and $p_{5, i}$ is defined.

Set $s_i^*=\mathrm {tr}(p_{5,i})$ and $i^*=s_2(i)- m_{i}$ . By the choice of the sequence $\langle n_i: i<\omega \rangle $ , we have $1 <i^*.$ Also, for notational simplicity set

$$ \begin{align*}j_i=j_*+(\ell g(s_2)-\ell g(s_1))+(i-\ell g(s_1)).\end{align*} $$

By an argument similar to the proof of Claim 4.16, we can find $s^*$ such that:

• • $s_i^* \unlhd s^*$ ;

• • for every $l\in (\ell g(s_i^*), \ell g(s^*))$ , $s^* \restriction (l+1) \notin w_{s^* \restriction l,0}$ ;

• • $j_i+i^*+2< \mathrm {nor}_{s^*}(\mathrm {Suc}_{T_{p_{5, i}}}(s^*))$ .

Choose

$$ \begin{align*}s^*_{i+1} \in \mathrm{Suc}_{T_{p_{5,i}}}(s^*) \cap \big(w_{s^*, j_i+i^*} \setminus w_{s^*, j_i+i^*+1}\big)\end{align*} $$

and define $p_{5,i+1}:=(s^*_{i+1}, T_{p_{5, i}}^{[s^*_{i+1}]})$ . Then $p_{5, i+1} \geq p_{5, i}.$ We show that it satisfies items (2)–(4) of the claim.

$p_{5, i+1}$ satisfies clause (2). To show this, we need to have:

• • $\mathrm {tr}(p_{5,i+1})=s^*_{i+1} \in w_{s^*,j_i}$ ;

• • every initial segment t of $s^*_{i+1}$ avoids $w_{t \restriction (\ell g(t)-1),j_i}$ .

These are clearly true by the way we defined $s^*_{i+1}$ .

$p_{5, i+1}$ satisfies clause (3). By the choice of the conditions $p_{3, j}$ , $l_j \leq 1$ for every $j<j_*+(\ell g(s_2)-\ell g(s_1))$ . Since $p_{5, i+1}$ extends $p_{5, i}$ , it follows from the induction hypothesis that $p_{5, i+1}\Vdash $ “ ” for all $j<i$ . Thus to guarantee clause (3), we only have to show that $p_{5, i+1}\Vdash $ “ .” By the choice of the sequence $\langle n_i: i<\omega \rangle $ and the conditions $p_{3, i}$ we have $i^* \geq 2$ . Furthermore ” and

In particular, it follows that .” Thus by its definition,

which gives the result.

$p_{5, i+1}$ satisfies clause (4). This is clear, as $s^*_{i+1} \in w_{s^*, j_i+i^*} \setminus w_{s^*, j_i+i^*+1}$ .

Let $p_5=p_{5,\ell g(s_2)}$ . It’s easy to see that $p_5 \Vdash $ “ .” Indeed, as $p_2 \leq p_5,$ we have $p_5 \Vdash $ “ .” On the other hand, for every $\ell g(s_1) \leq j < \ell g(s_2)$ we have, using Claim 4.18,

Finally let $q=p_6.$ Then q extends p and it forces “ satisfies I.” This completes the proof of Lemma 4.15.