Proof The reader can find a proof in Laflamme’s article [Reference Laflamme10]. Another good reference is found in Wohofsky master’s thesis [Reference Wohofsky16, pp. 53–59].

Proof It is enough to prove that given any non-empty family of summable ideals of cardinality less than $\mathfrak {d}$ , there is an ultrafilter meeting each ideal in the family, but still there is a tall summable ideal having empty intersection with the ultrafilter. Fix $\mathcal {D}=\{\mathscr {I}_{\alpha }:\alpha <\lambda \}$ an arbitrary family of tall summable ideals such that $\lambda <\mathfrak {d}$ , and for $\alpha \in \lambda $ , let $g_{\alpha }:\omega \to \mathbb R$ be a function such that $\mathscr {I}_{\alpha }=\{A\in \mathcal {P}(\omega ):\sum _{n\in A}g_{\alpha }(n)<\infty \}$ . Let E be the set of even natural numbers. For each $F\in [\lambda ]^{<\omega }\setminus \{\emptyset \}$ , define a function $\psi _F$ on E as follows:

$$ \begin{align*} &\qquad\qquad\qquad\qquad\qquad\qquad \psi_F(0)=0,\\ &\psi_{F}(n+2)=\min\{k\in\omega:k>\psi_{F}(n)\ \text{and}\ (\forall\alpha\in F)(\forall i\ge k)\\ &\qquad\qquad\qquad\qquad\quad(g_{\alpha}(i)\leq1/2^{n+2}(n+2)^{2})\}. \end{align*} $$

Since $\{\psi _{F}:F\in [\lambda ]^{<\omega }\setminus \{\emptyset \}\}$ has cardinality less than $\mathfrak {d}$ , there is a function $f:E\to \omega $ which is not dominated by $\{\psi _{F}:F\in [\lambda ]^{<\omega }\setminus \{\emptyset \}\}$ . We can assume that for all $n\in E$ , $f(n+2)-f(n)>n^{2}$ , and we can define the following function:

$$ \begin{align*} \tilde{f}(n)=\begin{cases} f(n),\text{ if }n\in E,\\ f(n-1)+(n-1)^{2}, \quad \text{if }n\notin E. \end{cases} \end{align*} $$

For each $F\in [\lambda ]^{<\omega }\setminus \{\emptyset \}$ , let $A_{F}$ be the set where $\psi _{F}$ is dominated by f, that is $A_{F}=\{n\in E:\psi _{F}(n)\leq f(n)\}$ . Note that the family $\{A_{F}:F\in [\lambda ]^{<\omega }\setminus \{\emptyset \}\}$ is a centered family. Indeed, pick $F_{0},\ldots ,F_{n}\in [\lambda ]^{<\omega }\setminus \{\emptyset \}$ . By the definition of the functions $\psi _{F}$ we have that if $F\subseteq G$ , then $\psi _{F}\leq \psi _{G}$ , so in particular, if $\psi _{F_{0}\cup \cdots \cup F_{n}}(k)\leq f(k)$ , then we have $\psi _{F_{i}}(k)\leq f(k)$ for $i=0,\ldots ,n$ , and so $k\in A_{0}\cap \cdots \cap A_{n}$ .

Now, for $F\in [\lambda ]^{<\omega }\setminus \{\emptyset \}$ let $B_{F}$ be the set $\bigcup _{k\in A_{F}}[\tilde {f}(k),\tilde{f}(k+1))$ . Finally, take the family $\mathcal {G}=\{B_{F}:F\in [\lambda ]^{<\omega }\setminus \{\emptyset \}\}$ . It follows that $\mathcal {G}$ is a centered family, by the previous paragraph.

Let us see that $\mathcal {G}$ has non-empty intersection with each ideal in $\mathcal {D}$ . For $\alpha \in \lambda $ , we claim that $B_{\{\alpha \}}\in \mathscr {I}_{\alpha }$ . This follows easily from the definition of ${\psi _{\{\alpha \}}}$ and ${A_{\{\alpha \}}}$ . We have that for all ${n\in A_{\{\alpha \}}}$ , ${\psi _{\{\alpha \}}(n)\leq f(n)=\tilde {f}(n)}$ , and by the definition of ${\psi _{\{\alpha \}}}$ , for every ${k\ge \psi _{\{\alpha \}}(n)}$ , ${g_{\alpha }(k)\leq \frac {1}{2^{n}n^{2}}}$ , so in particular, for every ${n\in A_{\{\alpha \}}}$ and every ${k\ge \tilde {f}(n)},{\ g_{\alpha }(k)\leq \frac {1}{2^{n}n^{2}}}$ . We have the following:

$$ \begin{align*} \sum_{n\in B_{\{\alpha\}}} g_{\alpha}(n)&=\sum_{n\in A_{\{\alpha\}}} \left[\sum_{m\in[\tilde{f}(n),\tilde{f}(n+1))}g_{\alpha}(m)\right]\\ &\leq \sum_{n\in A_{\{\alpha\}}}\left[\sum_{m\in[\tilde{f}(n),\tilde{f}(n+1))} \frac{1}{2^n n^2}\right]\\ & = \sum_{n\in A_{\{\alpha\}}}\frac{n^2}{2^n n^2}\leq \sum_{n\in A_{\{\alpha\}}}\frac{1}{2^n}\leq 1. \end{align*} $$

What remains is to find a tall summable ideal $\mathscr {I}$ which has empty intersection with some ultrafilter extending $\mathcal G$ . Consider the following function from $\omega $ to $\mathbb {R}$ :

$$ \begin{align*} h(n)=\begin{cases} 1, \quad \text{if } n\in [0,\tilde{f}(0)),\\ \frac{1}{k+1}, \quad \text{if } n\in [\tilde{f}(k),\tilde{f}(k+1)). \end{cases} \end{align*} $$

Let $\mathscr {I}_{h}$ be the corresponding summable ideal. We claim that $\mathcal {G}\subseteq \mathscr {I}_{h}^{+}$ . Pick any $B_{F}\in \mathcal {G}$ . Then,

$$ \begin{align*} \sum_{n\in B_F} h(n) &= \sum_{n\in A_{F}}\left[\sum_{m\in[\tilde{f}(n),\tilde{f}(n+1))}h(m)\right]\\ & = \sum_{n\in A_{F}}\left[\sum_{m\in[\tilde{f}(n),\tilde{f}(n+1))}\frac{1}{n+1}\right]\\ & = \sum_{n\in A_{F}}\frac{n^2}{n+1}\longrightarrow\infty. \end{align*} $$

Then we have $\mathcal G\subseteq \mathscr {I}_h^+$ , so $\mathcal {G}\cup \mathscr {I}_h^*$ can be extended to an ultrafilter $\mathcal U$ . Obviously $\mathscr {I}_h$ and $\mathcal U$ satisfy our requirements.

Proof Let $\varphi $ be a lscsm such that $\mathscr {I}=Exh(\varphi )$ . Note that for any $n\in \omega $ there is $k\in \omega $ such that for all $i\ge k$ , $\varphi (\{i\})\leq \frac {1}{2^n}$ (recall that $\mathscr {I}$ is a tall ideal). Recursively construct a sequence $\langle k_n:n\in \omega \rangle $ such that for all $i\ge k_n$ , $\varphi (\{i\})\leq \frac {1}{2^n},$ and $k_n<k_{n+1}$ . Then define $g_{\varphi }(n)=k_n$ . Let us see that $g_{\varphi }$ is a witness for the lemma. It is clear that if $\mathcal {U}$ is a weak $\mathscr {I}$ -ultrafilter, then for all finite to one $f:\omega \to \omega $ such that $f\leq g_{\varphi }$ there is $A\in \mathcal {U}$ such that $f[A]\in \mathscr {I}$ . So let us assume that for all finite to one $f\leq g_{\varphi }$ there is $A\in \mathcal {U}$ such that $f[A]\in \mathscr {I}$ . Let $f\in \omega ^{\omega }$ be an arbitrary finite to one function. If $D=\{i\in \omega :{g}_{\varphi }(i)\leq f(i)\}\in \mathcal {U}$ , note that for any $i\in D$ , $\varphi (\{f(i)\})\leq 1/2^i$ , so for any $n\in \omega $ ,

$$ \begin{align*} \varphi(f[D]\cap n)\leq \sum_{j\in f[D]\cap n}\varphi(\{j\})&\leq \\ &\quad \sum_{i\in D,f(i)<n}\varphi(\{f(i)\})\leq\sum_{i\in D,f(i)<n}\frac{1}{2^i}\leq \sum_{i\in D}\frac{1}{2^i}. \end{align*} $$

Then, by the lower semicontinuity of $\varphi $ , it follows that $\varphi (f[D])\leq \sum _{i\in D}\frac {1}{2^i}$ . This same argument can be used to proved that $\varphi (f[D]\setminus n)\leq \sum _{i\in D,f(i)\ge n}\frac {1}{2^i}$ , which implies $\lim _{n\to \infty }\varphi (f[D]\setminus n)=0$ , so $f[D]\in \mathscr {I}$ . On the other hand, if $\omega \setminus D=\{n\in \omega :f(n)<g_{\varphi }(n)\}\in \mathcal {U}$ , then there is a finite to one function $h\in \omega ^{\omega }$ such that $h\leq g$ and $\omega \setminus D=\{n\in \omega :f(n)=h(n)\}\in \mathcal {U}$ (just define $h\upharpoonright (\omega \setminus D)=f\upharpoonright (\omega \setminus D)$ , and extend it to an arbitrary finite to one function h such that $h\leq g_{\varphi }$ ). By hypothesis, there is $A\in \mathcal {U}$ such that $h[A]\in \mathscr {I}$ . Define $B=A\cap (\omega \setminus D)$ , then we have $f[B]=h[B]\in \mathscr {I}$ .

Proof Fix $\mathcal {U}$ an $(\mathscr {I},p)$ -point. Let $\dot {f}$ be a $\mathbf {PT}$ -name and $p\in \mathbf {PT}$ a condition such that $p\Vdash ` `\dot {f}\in \omega ^{\omega }"$ . By Lemma 4.7, we can assume that $p\Vdash ` `\dot {f}\leq g_{\varphi }"$ (note that by absoluteness the same $g_{\varphi }$ from the ground model works in any generic extension), so we are going to prove that in the generic extension, for any $\dot {f}\leq g_{\varphi }$ there is $A\in \mathcal {U}$ such that $\dot {f}[A]\in \mathscr {I}$ . By Theorem 4.6, $\mathcal U$ remains as a p-point in the generic extension, so we can assume that $p\Vdash ` `\dot {f} \textit {is finite to one}"$ (the case when $\dot {f}$ has a restriction on which it is constant follows immediately). By an application of Lemma 4.4, we can assume that for any $s\in split(p)$ there is a function $h_s\in \omega ^{\omega }$ such that for all $n\in \omega $ and for all but finitely many $k\in succ_{p}(s)$ , $p\upharpoonright s^{\frown } k\Vdash ` `\dot {f}\upharpoonright (\lvert s\rvert +n)=h_s\upharpoonright (\lvert s\rvert +n)"$ . Since $\mathcal U$ is an $\mathscr {I}$ -ultrafilter, for every $s\in split(p)$ we find $A_s\in \mathcal U$ such that $h_s[A_s]\in \mathscr {I}$ . Note that $h_s$ is not necessarily a finite to one function, but since $\mathcal U$ is a p-point, we can assume that $h_s\upharpoonright A_s$ is either constant or finite to one. Now, by Lemma 4.5, we can as well assume that the condition p has one of the following properties:

1. a) For all $s\in split(p)$ the restriction $h_s\upharpoonright A_s$ is finite to one.

2. b) For all $s\in split(p)$ the restriction $h_s\upharpoonright A_s$ is a constant $a_s$ .

Let us consider first the case a). Since $\mathscr {I}$ is a p-ideal and for all $s\in split(p)$ we have $h_s[A_s]\in \mathscr {I}$ , then there is $Z\in \mathscr {I}$ such that for all $s\in split(p)$ , $h_s[A_s]\subseteq ^*Z$ . Removing finitely many elements of each $A_s$ , we can assume that for all $s\in split(p)$ we have $h_s[A_s]\subseteq Z$ (recall that we are dealing with case a)). Making use of the p-point game we will construct two sequences $\{F_n:n\in \omega \}$ and $\{T_n:n\in \omega \}$ such that:

1. (1) $T_0\leq _0 p$ .

2. (2) For all $n\in \omega $ , $T_{n+1}\Vdash ` `\dot {f}[F_n]\subseteq Z"$ , and $T_{n+1}\leq _{n+1}T_n$ .

3. (3) $\bigcup _{n\in \omega } F_n\in \mathcal {U}$ .

4. (4) $T_{\omega }=\bigcap _{n\in \omega } T_n$ is a condition.

The construction is as follows:

1. (i) Player I starts playing $A_0=A_{st(p)}$ , and define $T_0=p$ .

2. (ii) Suppose Player II has answered with a set $F_0\subseteq A_0$ . Let $k_0\in \omega $ be big enough so that for all $i\in succ_{T_0}(st(T_0))\setminus k_0$ it holds that $T\upharpoonright st(T_0)^{\frown } i\Vdash ` `\dot {f}\upharpoonright (\max (F_0)+1)=h_{st(T_0)}\upharpoonright (\max (F_0)+1)$ ”. Note that this implies that for all $i\ge k_0$ , $T_0\upharpoonright st(T_0)^{\frown } i\Vdash ` `\dot {f}[F_0]\subseteq Z"$ . Then define

   $$ \begin{align*} T_1=\bigcup_{\substack{i\ge k_0,\\ i\in succ_{T_0}(st(T_0))}}T_0\upharpoonright st(T_0)^{\frown} i. \end{align*} $$

   Then Player I plays the set $A_1=\bigcap _{s\in {1^{\leq 1}}}A_{\varphi _{T_1}(s)}\setminus (\max (F_0)+1)$ .

3. (iii) Suppose at round number n Player I has played the set

   $$ \begin{align*} A_n=\bigcap_{s\in n^{\leq n}} A_{\varphi_{T_n}(s)}\setminus \left(\max\left(\bigcup_{k<n}F_k\right)+1\right). \end{align*} $$

   Suppose Player II responds with a finite set $F_n\subseteq A_n$ . Then, let $k_n\in \omega $ be big enough such that for all $r\in \varphi _{T_n}(n^{\leq n})$ and all $i\in succ_{T_n}(r)\setminus k_n$ , it holds that $T_n\upharpoonright r^{\frown } i\Vdash ` `\dot {f}\upharpoonright (\max \left (\bigcup _{k\leq n}F_n\right )+1)=h_{r}\upharpoonright (\max \left (\bigcup _{k\leq n}F_n\right )+1)"$ . Note that this implies that for each $r\in \varphi _{T_n}(n^{\leq n})$ and $i\in succ_{T_n}(r)\setminus k_n$ , $T_n\upharpoonright r^{\frown } i\Vdash ` `\dot {f}[F_n]\subseteq Z"$ . Then define $T_{n+1}$ as follows:

   $$ \begin{align*} T_{n+1}=\bigcup_{s\in n^{\leq n}}\bigcup_{\substack{i\ge k_n,\\ i\in succ_{T_n}(\varphi_{T_n}(s))}}T\upharpoonright \varphi_{T_n}(s)^{\frown} i. \end{align*} $$

   By construction it follows that $T_{n+1}\Vdash ` `\dot {f}[F_n]\subseteq Z"$ . Let Player I play the set

   $$ \begin{align*} A_{n+1}=\bigcap_{s\in(n+1)^{\leq n+1}} A_{\varphi_{T_{n+1}}(s)}\setminus\left(\max\left(\bigcup_{k\leq n}F_k\right)+1 \right). \end{align*} $$

By Lemma 2.2, this is not a winning strategy for Player I, so there is a play where Player II wins the game. Let $\{F_n:n\in \omega \}$ and $\{T_n:n\in \omega \}$ be the sequences constructed by Player I in this play, and define $T_{\omega }=\bigcap _{n\in \omega }T_n$ .

Now let us deal with case b). Fist note that there is no condition $q\leq p$ such that the set $\{a_s:s\in splitt(q)\}$ is finite. To prove this assume it is false and apply Lemma 4.5 finitely many times, then recall that the function $\dot {f}$ is a finite to one function to arrive at a contradiction. By using this remark it is possible to find a stronger condition $q\leq p$ such that $Z=\{a_s:s\in split(q)\}\in \mathscr {I}$ . The rest follows the same lines of the previous case. Using the p-point game again we construct two sequences $\{F_n:n\in \omega \}$ and $\{T_n:n\in \omega \}$ such that conditions (2)–(5) of the construction for case a) hold, and condition (1) is changed to (1’) $T_0\leq q$ . The strategy for player I is defined in the same way as it was in the previous case. Then the condition $T_{\omega }$ and the set $\bigcup _{n\in \omega }F_n$ constructed in this way satisfy $T\Vdash ` `\dot {f}[\bigcup _{n\in \omega }F_n]\subseteq Z"$ .

Proof We follow the same notation as in Definition 5.3. Let $G_{\mathbb {P}}\subseteq \mathbb {P}$ be a generic filter over V, define $r_{-1}=(p_{-1},\dot {q}_{-1})=(1_{\mathbb {P}},1_{\dot {\mathbb {Q}}})$ and $n^*=\sup (\{n\in \{-1\}\cup \omega :p_n\in G_{\mathbb {P}}\})$ . Define $\tilde {\delta }$ as a $\mathbb {P}$ -name for the sequence $\langle t_n:n\in \omega \rangle $ of conditions in $\dot {\mathbb {Q}}[G_{\mathbb {P}}]$ defined as follows:

• For all $n\in \omega $ , if $n\leq n^*$ , then $t_n=\dot {q}_n[G_{\mathbb {P}}]$ , and for all natural number $n> n^*$ , define $t_n$ to be $\dot {s}_n[G_{\mathbb {P}}]$ , where $\dot {s}_n$ is such that there is $u_n\in \mathbb {P}$ such that $(u_n,\dot {s}_n)\in \mathbb {P}*\dot {\mathbb {Q}}$ is the $\sqsubseteq $ -first extension of $(u_{n-1},\dot {s}_{n-1})$ such that $u_n\in G_{\mathbb {P}}$ , and $\dot {s}_n[G_{\mathbb {P}}]$ decides the value of $\dot {f}^{-1}(n)$ and $\dot {f}\upharpoonright n$ . Let $g\in \omega ^{\omega }$ be the function such that $t_n\Vdash ` `\dot {f}\upharpoonright n=g\upharpoonright n"$ . Define $\textrm{int}(\tilde {\delta },\dot {f})$ as a $\mathbb {P}$ -name for g.

By definition of $\tilde {\delta }$ and the definition of $\textrm{int}(\tilde {\delta },\dot {f})$ , it follows that $\tilde {\delta }$ is a fine derived sequence and the pair $(\tilde {\delta },\textrm{int}(\tilde {\delta },\dot {f}))$ is forced to a fine interpretation of $\dot {f}$ in $V[\dot {G}_{\mathbb {P}}]$ , since $\textrm{int}(\tilde \delta ,\dot {f})$ is actually a $\mathbb {P}$ -name of the function g. It is easy to see that $(\langle p_n:n\in \omega \rangle ,h)$ is a fine interpretation of $\textrm{int}(\tilde {\delta },\dot {f})$ .

Proof Fix a natural number $n\in \omega $ . We need to find the objects in clauses 1)–4). First note that by Lemma 5.4, we can find inside $\mathcal {M}$ a $\mathbb {P}$ -name for a fine derived sequence $\dot {\delta }=\dot {\delta }(\langle r_n:n\in \omega \rangle ,\dot {f})$ such that $\textrm{int}(\dot {\delta },\dot {f})$ is interpreted by $\langle p_n:n\in \omega \rangle $ as $h_0$ . For short, denote by $\dot {h}$ the name $\textrm{int}(\dot {\delta },\dot {f})$ .

Let $G\subseteq \mathbb {P}$ be a generic filter over V. By hypothesis, $\mathcal {U}$ remains as an $(\mathscr {I},p)$ -point in $V[G]$ , and by elementarity, $\mathcal {M}[G]\vDash \mathcal {U}\text { is an } (\mathscr {I},p)\text {-point}$ . So for each $k\in \omega $ , there are $u_k\leq p_k$ and $X_k\in \mathcal {U}$ such that $u_k\Vdash ` `\varphi (\dot {h}[X_k])<1"$ , since $\mathscr {I}=Exh(\varphi )$ . Clearly this can by done inside $\mathcal {M}$ , so we assume these two sequences are in $\mathcal {M}$ . Let $X\in \mathcal {M}\cap \mathcal {U}$ be a pseudointersection of $\langle X_k:k\in \omega \rangle $ . Now, working in $\mathcal {M}$ , define by recursion a sequence of natural numbers $\langle k_l:l\in \omega \rangle $ as follows:

1. (1) $k_0=0.$

2. (2) For all $i\leq k_l$ , $X\setminus k_{l+1}\subseteq X_i$ .

3. (3) For all $i\ge k_{l+1}$ , $p_i\Vdash ` `\dot {h}\upharpoonright k_l= h_0\upharpoonright k_l".$

4. (4) For all $i\leq k_l$ , there is $u_i^{l+1}\leq u_i\leq p_i$ such that $u_i^{l+1}\Vdash ` `\varphi (\dot {h}[X_i\setminus k_{l+1}])<m-\max \{\varphi (h_0[F]):F\subseteq k_l\land \varphi (h_0[F])<m\}"$ .

Note that 4) can be achieved since $\dot {h}$ is forced to be finite to one, and $u_i$ forces $\dot {h}[X_i]\in \mathscr {I}=Exh(\varphi )$ . Conditions 1)–3) are easy to arrange.

Assume the sequence $\langle k_j:j\in \omega \rangle $ has been constructed. Without loss of generality, we can assume that $Y=\bigcup _{j\in \omega }[k_{3j+1},k_{3j+2})\in \mathcal U$ . Let $j_0$ be such that $Z\setminus k_{3j_0}\subseteq X\cap Y$ (recall that Z is a pseudointersection of $\mathcal M\cap \mathcal U$ , and $X\cap Y\in \mathcal U\cap \mathcal M$ ). Note that for any $l> j_0$ we have $[k_{3l-1},k_{3l+1})\cap Z=\emptyset $ . Fix one of such l. We claim that $u_{k_{3l}}^{3l+1}\Vdash ` `\varphi (\dot {h}[Z])< m"$ . Since $u_{k_{3l}}^{3l+1}\leq p_{k_{3l}}$ , by 3) above we have

$$ \begin{align*} u_{k_{3l}}^{3l+1}\Vdash ` `\dot{h}\upharpoonright k_{3l-1}= h_0\upharpoonright k_{3l-1}", \end{align*} $$

which in turn implies

$$ \begin{align*} u_{k_{3l}}^{3l+1}\Vdash ` `\dot{h}[Z\cap k_{3l-1}]= h_0[Z\cap k_{3l-1}]" \end{align*} $$

and so

(1) $$ \begin{align} u_{k_{3l}}^{3l+1}\Vdash ` `\varphi(\dot{h}[Z\cap k_{3l-1}])\leq\max\{\varphi(h_0[F]):F\subseteq k_{3l}\land\varphi(h_0[F])<m\}". \end{align} $$

Also, by 2) and the choice of l we have

(2) $$ \begin{align} Z\setminus k_{3l+1}\subseteq X\cap Y\setminus k_{3l+1}\subseteq X_{k_{3l}}\setminus k_{3l+1.} \end{align} $$

Note that since $Z\cap [k_{3l-1},k_{3l+1})=\emptyset $ , the following holds:

(3) $$ \begin{align} u_{k_{3l}}^{3l+1}\Vdash ` `\varphi(\dot{h}[Z])\leq\varphi(\dot{h}[Z\setminus k_{3l}])+\varphi &(\dot{h}[Z\cap k_{3l}])= \\ &\qquad \varphi(\dot{h}[Z\setminus k_{3l+1}])+\varphi(\dot{h}[Z\cap k_{3l-1}])". \nonumber \end{align} $$

Note that (1) implies the following:

$$ \begin{align*} u_{k_{3l}}^{3l+1}\Vdash` `m-\max\{\varphi(h_0[F]):F\subseteq k_{3l}\land\varphi(h_0[F])<m\}\leq m-\varphi(\dot{h}[Z\cap k_{3l-1}])". \end{align*} $$

This inequality, joint with 4), gives the following:

$$ \begin{align*} u_{k_{3l}}^{3l+1}\Vdash` `\varphi(\dot{h}[X_{k_{3l}}\setminus k_{3l+1}])< m-\varphi(\dot{h}[Z\cap k_{3l-1}])" \end{align*} $$

or equivalently

$$ \begin{align*} u_{k_{3l}}^{3l+1}\Vdash ` `\varphi(\dot{h}[X_{k_{3l}}\setminus k_{3l+1}])+\varphi(\dot{h}[Z\cap k_{3l-1}])<m". \end{align*} $$

By (2) and the monotonicity of $\varphi $ , we have that

$$ \begin{align*} u_{k_{3l}}^{3l+1}\Vdash ` `\varphi(\dot{h}[Z\setminus k_{3l+1}])+\varphi(\dot{h}[Z\cap k_{3l-1}])<m". \end{align*} $$

Finally by (3) we get

$$ \begin{align*} u_{k_{3l}}^{3l+1}\Vdash ` `\varphi(\dot{h}[Z])<m". \end{align*} $$

Now let $l>j_0$ be a natural number big enough so that $n<k_{3l-1}$ , and let $q\leq u_{k_{3l}}^{3l+1}$ be an $(\mathcal {M},{\mathbb {P}})$ -generic condition. Define the following:

1. (a) In $\mathcal {M}$ , let $\tilde {s}_n$ be the $\mathbb {P}$ -name $\dot {q}_{k_{3l}}$ .

2. (b) In $\mathcal {M}$ , let be a $\mathbb {P}$ -name for the fine derived sequence $\tilde {\delta }(\langle r^{\prime }_i:i\in \omega \rangle ,\dot {f})$ , where $r^{\prime }_i=r_{i+k_{3l}}$ , for all $i\in \omega $ .

3. (c) In $\mathcal {M}$ , let $\tilde {h}_{n}$ be the $\mathbb {P}$ -name $\dot {h}$ .

4. (d) Define $p_{n,\mathcal {M}}=q$ .

We claim that $p_{n,\mathcal {M}}$ , $\tilde {s}_n$ , $\tilde {h}_n$ and are as required.

Conditions (1)–(4) are easily seen to be hold by the construction. Condition $i)$ , follows from the fact that $p_{n,\mathcal {M}}\leq p_{k_{3l}}\leq p_0$ . The first part of $ii)$ follows from a) and $(p_{k_{3l}},\dot {q}_{k_{3l}})\leq (p,\dot {q})$ and the choice of $\tilde {s}_n$ . The second part of condition $ii)$ follows from the fact that $\langle (p_k,\dot {q}_k):k\in \omega \rangle \in \mathcal {M}$ , $p_{n,\mathcal {M}}\leq p_{k_{3l}}$ and $\tilde {s}_n=\dot {q}_{k_{3l}}$ .

Let us see condition $(iii)$ . By definition $\tilde {\delta }(\langle r_i':i\in \omega \rangle ,\dot {f})$ , is a fine interpretation of $\dot {f}$ . To see that $\tilde {\delta }(\langle r_i':i\in \omega \rangle ,\dot {f})$ is forced to be below $\tilde {s}_n$ by $p_{n,\mathcal {M}}$ , let $G\subseteq \mathbb {P}$ be any generic filter such that $p_{n,\mathcal {M}}\in G$ . By definition $r_0'=(p_{k_{3l}},\dot {q}_{k_{3l}})$ . Since $p_{n,\mathcal {M}}\leq p_{k_{3l}}$ and $p_{n,\mathcal {M}}\in G$ , it follows that the first term of $\tilde {\delta }(\langle r_i':i\in \omega \rangle )$ is $\dot {q}_{k_{3l}}[G]=\tilde {s}_n[G]$ , and all the remaining terms are below the first one since $\tilde {\delta }(\langle r_i':i\in \omega \rangle ,\dot {f})$ is a decreasing sequence. To see that $p_{n,\mathcal {M}}\Vdash ` `\varphi (\tilde {h}_n[Z])<m"$ , recall that $l>j_0$ , $p_{n,\mathcal {M}}\leq u_{k_{3l}}^{3l+1}$ and $u_{k_{kl}}^{3l+1}\Vdash ` `\varphi (\dot {h}[Z])<m"$ .

To see that condition $(iv)$ holds, just note that $n < k_{3l-1}$ , $(p_{n,\mathcal {M}},\tilde {s}_n)\leq (p_{k_{3l}},\dot {q}_{k_{3l}})$ and

$$ \begin{align*} (p_{k_{3l}},\dot{q}_{k_{3l}})\Vdash` `\dot{f}\upharpoonright k_{3l}=h_0\upharpoonright k_{3l}", \end{align*} $$

which implies

$$ \begin{align*} p_{n,\mathcal{M}}\Vdash` `\tilde{s}_n\Vdash` `\dot{f}\upharpoonright k_{3l}=h_0\upharpoonright k_{3l}"" \end{align*} $$

and also

$$ \begin{align*} p_{k_{3l}}\Vdash` `\tilde{h}_n\upharpoonright k_{3l}=\dot{h}\upharpoonright k_{3l}=h_0\upharpoonright k_{3l}", \end{align*} $$

so it follows that

$$ \begin{align*} p_{n,\mathcal{M}}\Vdash` `\tilde{s}_n\Vdash` `\dot{f}\upharpoonright k_{3l}=\tilde{h}_n\upharpoonright k_{3l}"". \end{align*} $$

Since $n<k_{3l-1}$ , the required property follows.

Proof of Theorem 5.1

Let us first note that the successor step is easy to handle: let $\mathbb {P}*\dot {\mathbb {Q}}$ be a two step iteration of proper forcings such that $\mathbb {P}$ preserves $(\mathscr {I},p)$ -points and $\mathbb {P}\Vdash ` `\dot {\mathbb {Q}}\text { preserves } (\mathscr {I},p)\text {-points}"$ . Let $\mathcal {U}$ be an $(\mathscr {I},p)$ -point in the ground model. Let $G\subseteq \mathbb {P}$ be V-generic. Then, in $V[G]$ , $\mathcal {U}$ generates a $(\mathscr {I},p)$ -point, and $\dot {\mathbb {Q}}[G]$ preserves $(\mathscr {I},p)$ -points. If $H\subseteq \dot {\mathbb {Q}}$ is $V[G]$ -generic, then $\mathcal {U}$ generates an $(\mathscr {I},p)$ -point in $V[G][H]=V[G*H]$ . So $\mathbb {P}*\dot {\mathbb {Q}}$ preserves $\mathcal {U}$ as an $(\mathscr {I},p)$ -point.

We are left with the limit step. Let us assume all the hypothesis in the statement of Theorem 5.1. Note that we can assume that the iteration has length $\omega $ , since every real appears in a successor stage or in a countable cofinality stage, so we assume $\mathbb {P}_{\omega }=\langle \mathbb {P}_n,\dot {\mathbb {Q}}_n:n\in \omega \rangle $ . Let $\mathcal U$ be an $(\mathscr {I},p)$ -point, $\dot f$ a $\mathbb P_{\omega }$ -name for a function from $\omega $ to $\omega $ and $p\in \mathbb P_{\omega }$ a condition. Note that by Theorem 2.8, we have that $\mathcal {U}$ generates a p-point after forcing with $\mathbb {P}_{\omega }$ , so we can assume that $p\Vdash ` `\mathcal {U}\text { generates a } p\text {-point}"$ . If there is a condition $q\leq p$ and $X\in \mathcal U$ such that $q\Vdash ` `\dot {f}\upharpoonright X \textit {is constant}"$ , then we are done. Otherwise, there is $q\leq p$ and there exists $A\in \mathcal U$ such that $\dot f\upharpoonright A$ is forced to be finite to one by q, so we can assume that $\dot f$ is finite to one. Let $\mathcal M\prec H(\theta )$ be a countable elementary submodel such that $\dot {f},\mathcal U,p,\mathbb P_{\omega }\in \mathcal M$ . Fix $Z\in \mathcal U$ a pseudointersection of $\mathcal U\cap \mathcal M$ . In $\mathcal M$ , let $(\langle r_n:n\in \omega \rangle ,h_0)$ be a fine interpretation for $\dot {f}$ below p, and assume $\varphi (h_0[Z])< m$ for some fixed $m\in \omega $ . For each $n\in \omega $ , we assume $r_n$ has the form $\langle r_n^k:k\in \omega \rangle $ . We will construct by recursion four sequences $\langle q_n:n\in \omega \rangle $ , $\langle \dot {s}_n:n\in \omega \rangle $ , $\langle \dot {h}_n:n\in \omega \rangle $ and such that the following holds:

1. (a) For $n=0$ , $q_0$ is a $(\mathcal {M},\mathbb {P}_0)$ -generic condition.

2. (b) For all $n\in \omega $ , $q_n$ is $(\mathcal {M},\mathbb {P}_n)$ -generic and $q_{n}=(q_0,\ldots ,\dot {q}_n)$ , where $\dot {q}_n$ is a $\mathbb {P}_{n-1}$ -name for $n>0$ .

3. (c) For all positive $n\in \omega $ , $\dot {h}_n$ is a $\mathbb {P}_{n-1}$ -name and $q_{n-1}\Vdash ` `\dot {h}_n\in \mathcal {M}[\dot {G}_{\mathbb {P}_{n-1}}]"$ .

4. (d) For all positive $n\in \omega $ , is a $\mathbb {P}_{n-1}$ -name for a decreasing sequence of conditions in $\tilde {\mathbb {P}}_{[n,\omega )}[\dot {G}_{\mathbb {P}_{n-1}}]$ .

5. (e) For all positive $n\in \omega $ , $\dot {s}_n$ is a $\mathbb {P}_{n-1}$ -name and $q_{n-1}$ forces that $\dot {s}_n$ is a condition in $\tilde {\mathbb {P}}_{[n,\omega )}[\dot {G}_{\mathbb {P}_{n-1}}]\cap \mathcal {M}[\dot {G}_{\mathbb {P}_{n-1}}]$ .

6. (f) For all positive $n\in \omega $ , $q_{n-1}$ forces that is a fine interpretation of $\dot {f}$ below $\dot {s}_n$ .

7. (g) For all positive $n\in \omega $ , $q_{n-1}$ forces that $\varphi (\dot {h}_n[Z])<m$ .

8. (h) For all positive $n\in \omega $ , $q_{n-1}\Vdash ` `\dot {s}_n\Vdash ` `\dot {f}\upharpoonright n=\dot {h}_n\upharpoonright n""$ .

9. (i) For all positive $n\in \omega $ , $q_{n-1}\Vdash ` `\dot {q}_n\leq \dot {s}_{0}^n"$ and $q_0\leq r_0^0$ .

10. (j) For all $n\in \omega $ , $q_n\Vdash ` `\dot {s}_{n+1}\leq \dot {s}_{[1,\omega )}^{n}"$ .

Suppose first we have succeeded in the above construction and consider the condition $q=\bigcup _{n\in \omega }q_n$ . Note that by construction of q we have $q\leq q_0$ . We claim that $q\Vdash ` `\varphi (\dot {f}[Z])\leq m"$ . It is enough to prove $q\Vdash "(\forall n\in \omega )(\dot {f}\upharpoonright n=\dot {h}_n\upharpoonright n)"$ , since by g), this would imply that q forces the measure of the initial segments of $\dot {f}[Z]$ to be smaller than m, and by the lower semicontinuity of $\varphi $ , this implies that $\varphi (\dot {f}[Z])$ is at most m. First note that for any positive $n\in \omega $ , it holds that $q_{n}^{\frown } \dot {s}_{n+1}\leq q_{n-1}^{\frown } \dot {s}_{n}$ . To see this fix a positive $n\in \omega $ . Clause j) is equivalent to write $q_{n-1}^{\frown } \dot {q}_n\Vdash ` `\dot {s}_{n+1}\leq \dot {s}_{[1,\omega )}^n"$ . This, together with $q_{n-1}\Vdash ` `\dot {q}_n\leq \dot {s}_0^n"$ (see clause i)) and the factorization $\dot {s}_{n}=(\dot {s}_0^n,\dot {s}_{[1,\omega )}^n)$ gives the conclusion that $q_n^{\frown }\dot {s}_{n+1}\leq q_{n-1}^{\frown } \dot {s}_n$ . Then, note that this implies that for all $k\ge n$ , it holds that $q_k^{\frown } \dot {s}_{k+1}\leq q_{n-1}^{\frown } \dot {s}_{n}$ . This implies that for all $k\in \omega $ , $q\upharpoonright k=q_k\leq q_{n-1}^{\frown } \dot {s}_{n}\upharpoonright k$ , which means that $q\leq q_{n-1}^{\frown } \dot {s}_{n}$ . Since clause h) is equivalent to $q_{n-1}^{\frown } \dot {s}_n\Vdash ` `\dot {f}\upharpoonright n=\dot {h}_n\upharpoonright n"$ , we conclude that $q\Vdash ` `\dot {f}\upharpoonright n=\dot {h}_n\upharpoonright n"$ . Finally, since n was an arbitrary positive natural number, we conclude that $q\Vdash "(\forall n\in \omega )(\dot {f}\upharpoonright n=\dot {h}_n\upharpoonright n)"$ .

Let us construct the sequences. We start defining $\dot {s}_0=p_0$ , $\dot {h}_0=h_0$ and . Consider the factorization $\mathbb {P}=\mathbb {P}_0*\tilde {\mathbb {P}}_{[1,\omega )}$ . Relative to this factorization, any condition r in $\mathbb {P}_{\omega }$ is factored as $r=(r_0,\dot {r}_{[1,\omega )})$ . For the sequence $\langle r_n:n\in \omega \rangle $ we will write $r_n=(r_0^n,\dot {r}_{[1,\omega )}^{n})$ . Then note that all conditions from Lemma 5.5 are satisfied for $\mathscr {I}$ , $\varphi $ , $\mathbb {P}_0*\tilde {\mathbb {P}}_{[1,\omega )}$ , $\mathcal {U}$ , $\dot {f}$ , $p=(p_0,\dot {p}_{[1,\omega )})$ , $(\langle r_n:n\in \omega \rangle ,h_0),\ Z$ , $m,$ and $\mathcal {M}$ . So for $n=1$ , by an application of Lemma 5.5 we get $q_{1,\mathcal {M}}$ , $\tilde {s}_1$ , $\tilde {h}_1$ , that satisfy conditions 1) $-$ 4) from Lemma 5.5, and such that conditions $i)$ – $iv)$ are adapted as follows:

1. (1) $q_{1,\mathcal {M}}\leq r_0^0$ is a $(\mathcal {M},\mathbb {P}_0)$ -generic condition.

2. (2) $q_{1,\mathcal {M}}\Vdash ` `\tilde {s}_{1}\leq \dot {p}_{[1,\omega )}"$ and $q_{\mathcal {M}}\Vdash ` `\tilde {s}_{1}\in \tilde {\mathbb {P}}_{[1,\omega )}[\dot {G}_{\mathbb {P}_0}]\cap \mathcal {M}[\dot {G}_{\mathbb {P}_0}]"$ .

3. (3) $q_{1,\mathcal {M}}$ forces that is a fine interpretation of $\dot {f}$ below $\tilde {s}_1$ , as well as $\varphi (\tilde {h}_1[Z])<m$ .

4. (4) $q_{1,\mathcal {M}}\Vdash ` `\tilde {s}_1\Vdash ` `\dot {f}\upharpoonright 1=\tilde {h}_1\upharpoonright 1""$ .

Now define $\dot {q}_0=q_{1,\mathcal {M}}$ , $\dot {s}_1=\tilde {s}_1$ , $\dot {h}_1=\tilde {h}_1$ , and just as was obtained. Let us see that conditions a)–j) are satisfied. Conditions a) and b) follow trivially from 1) above. Condition c) follows from 3) above. Condition d) follows from 3) above. Condition e) follows from 2) above. Conditions f) and g) follow from 3) above. Condition h) is condition 4) above. Condition i) follows from 1) and 2). Condition j) follows from 2).

Now assume that we have constructed $q_{n-1}$ , $\dot {s}_n$ , $\dot {h}_n,$ and with the required properties. Consider the factorization $\mathbb {P}_{\omega }=\mathbb {P}_{n-1}*\dot {\mathbb {Q}}_{n}*\tilde {\mathbb {P}}_{[n+1,\omega )}$ . Let $G_{n-1}\subseteq \mathbb {P}_{n-1}$ be a generic filter over V such that $q_{n-1}\in G_{n-1}$ . Then let $s_n=\dot {s}_n[G_{n-1}]$ , $h_{n}=\dot {h}_{n}[G_{n-1}],$ and $\vec {t}_n$ be the evaluations of $\dot {s}_{n}$ , $\dot {h}_n,$ and , respectively, by the filter $G_{n-1}$ . We work in $\mathcal {M}[G_{n-1}]$ . Then we have that $\dot {\mathbb {Q}}_{n}[G_{n-1}]*\tilde {\mathbb {P}}_{[n+1,\omega )}[G_{n-1}]\in \mathcal {M}[G_{n-1}]$ . Moreover,

$$ \begin{align*} \mathcal{M}[G_{n-1}]\vDash (\vec{t}_n,h_{n})\text{ is a fine interpretation of } \dot{f} \text{ below } s_{n} \end{align*} $$

and

$$ \begin{align*} V[G_{n-1}]\vDash\varphi(h_{n}[Z])< m. \end{align*} $$

Summarizing, we have the following:

1. (1) $s_n=(s_0^n,\dot {s}_{[1,\omega )}^n),\dot {f},\mathscr {I},\varphi ,\mathcal {U},\dot {\mathbb {Q}}_{n}[G_{n-1}]*\tilde {\mathbb {P}}_{[n+1,\omega )}[G_{n-1}]\in \mathcal {M}[G_{n-1}]$ .

2. (2) $(\vec {t}_n,h_{n})\in \mathcal {M}[G_{n-1}]$ is a fine interpretation of $\dot {f}$ below $s_{n}$ .

3. (3) Z is a pseudointersection of $\mathcal {M}[G_{n-1}]\cap \mathcal {U}$ and $\varphi (h_n[Z])<m.$

Then we are in conditions to apply Lemma 5.5 for $n+1$ and $s_n$ taking the place of the condition p in the lemma. We get the following:

1. (1) A $(\mathcal {M}[G_{n-1}],\dot {\mathbb {Q}}_{n}[G_{n-1}])$ -generic condition $q_{\mathcal {M}[G_{n-1}]}$ .

2. (2) A $\dot {\mathbb {Q}}_{n}[G_{n-1}]$ -name $\tilde {s}_{n+1}$ for a condition in

   $\tilde {\mathbb {P}}_{[n+1,\omega )}[G_{n-1}][\dot {G}_{\dot {\mathbb {Q}}_n[G_{n-1}]}]\cap \mathcal {M}[G_{n-1}][\dot {G}_{\dot {\mathbb {Q}}_n[G_{n-1}]}]$ .

3. (3) A $\dot {\mathbb {Q}}_{n}[G_{n-1}]$ -name $\tilde {h}_{n+1}$ for a finite to one function from $\omega $ to $\omega $ .

4. (4) A $\dot {\mathbb {Q}}_{n}[G_{n-1}]$ -name for a sequence of conditions in $\tilde {\mathbb {P}}_{[n+1,\omega )}[G_{n-1}][\dot {G}_{\dot {\mathbb {Q}}_n[G_{n-1}]}]$ .

And such that the following hold:

1. (i) $q_{\mathcal {M}[G_{n-1}]}\leq s_n^0$ is a $(\mathcal {M}[G_{n-1}],\dot {\mathbb {Q}}_n[G_{n-1}])$ -generic condition.

2. (ii) $q_{\mathcal {M}[G_{n-1}]}\Vdash ` `\tilde {s}_{n+1}\leq \dot {s}_{[1,\omega )}^{n}"$ and $q_{\mathcal {M}}\Vdash ` `\tilde {s}_{n+1}\in \tilde {\mathbb {P}}_{[n+1,\omega )}[G_{n-1}] [\dot {G}_{\mathbb {Q}_n[G_{n-1}]}]\cap \mathcal {M}[\dot {G}_{n-1}][\dot {G}_{\mathbb {Q}_n[G_{n-1}]}]"$ .

3. (iii) $q_{\mathcal {M}[G_{n-1}]}$ forces that is a fine interpretation of $\dot {f}$ below $\tilde {s}_{n+1}$ , as well as $\varphi (\tilde {h}_{n+1}[Z])<m$ .

4. (iv) $q_{\mathcal {M}[G_{n-1}]}\Vdash ` `\tilde {s}_{n+1}\Vdash ` `\dot {f}\upharpoonright (n+1)=\tilde {h}_{n+1}\upharpoonright (n+1)""$ .

All of these happen regarding $V[G_{n-1}]$ and $\mathcal {M}[G_{n-1}]$ , and the construction was achieved by only assuming that $q_{n-1}\in G_{n-1}$ , and so everything above is forced by $q_{n-1}$ . Now, going back to V, we define the corresponding $\mathbb {P}_{n}$ -names:

1. $\alpha $ ) $\dot {q}_n$ as a $\mathbb {P}_{n-1}$ -name for $\dot {q}_{\mathcal {M}[G_{n-1}]}$ .

2. $\beta $ ) $\dot {s}_{n+1}$ as a $\mathbb {P}_{n-1}*\dot {\mathbb {Q}}_{n}$ -name for $\tilde {s}_{n+1}$ .

3. $\gamma $ ) $\dot {h}_{n+1}$ as a $\mathbb {P}_{n-1}*\dot {\mathbb {Q}}_{n}$ -name for $\tilde {h}_{n+1}$ .

4. $\delta $ ) as a $\mathbb {P}_{n-1}*\dot {\mathbb {Q}}_{n}$ -name for .

Finally, define $q_n=q_{n-1}\ ^{\frown } \dot {q}_n$ . Checking that these names are as required is routing as for case $n=1$ . This finishes the construction of the sequences, and therefore, the proof of Theorem 5.1.

Proof Let $\varphi $ be a lscsm which defines $\mathscr {I}$ . Define $Z=\{n\in \omega :\varphi (\{n\})=0\}$ . Note that since $\varphi $ is a lscsm, the set Z is coinfinite. Now, for $n\notin Z$ , define $g(n)=\varphi (\{n\})$ , and for $n\in Z$ , define $g(n)=1/2^n$ . The lower semicontinuity of $\varphi $ implies that $\mathscr {I}_g\subseteq \mathscr {I}$ .

Proof First note that for all summable ideal $\mathscr {I}_g$ , it holds that $\mathscr {I}_g=Fin(\varphi _g)=Exh(\varphi _g)$ , where $\varphi _g$ is the lscsm induced by $g:\omega \to \mathbb {R}^+$ . The forcing $\mathbb {P}_{\omega _2}$ is an $\omega _2$ -length countable support iteration of the Rational Perfect set forcing over a model of $\mathsf {ZFC}+\mathsf {CH}$ . Theorem 4.8 states that Rational Perfect forcing preserves $(\mathscr {I},p)$ -points whenever $\mathscr {I}$ in an analytic p-ideal, and Theorem 5.1 makes sure that $(\mathscr {I},p)$ -points are preserved along the iteration whenever $\mathscr {I}$ is an $F_{\sigma } \ p$ -ideal. Recall that in the Rational Perfect set model the dominating number is equal to $\omega _2$ . So let $\mathcal {D}$ be a family of analytic p-ideals such that $\vert \mathcal {D}\vert =\omega _1$ . Since every real appears before a stage of cofinality $\omega _1$ , it can be assumed that each ideal in $\mathcal {D}$ belongs to the ground model. By Lemma 6.1, for every $\mathscr {I}\in \mathcal {D}$ there is a summable ideal $\mathscr {J}(\mathscr {I})$ such that $\mathscr {J}(\mathscr {I})\subseteq \mathscr {I}$ . Define $\mathcal {D}'=\{\mathscr {J}(\mathscr {I}):\mathscr {I}\in \mathcal {D}\}$ . Since these ideals are from the ground model, there is an ultrafilter $\mathcal {U}$ that is an $(\mathscr {I},p)$ -point for every ideal $\mathscr {I}\in \mathcal {D}'$ (for example, $\mathcal {U}$ may be a Ramsey ultrafilter). This ultrafilter remains as an $(\mathscr {I},p)$ -point for every $\mathscr {I}\in \mathcal {D}'$ in the forcing extension by $\mathbb {P}_{\omega _2}$ , which implies that $\mathcal {U}$ is an $(\mathscr {I},p)$ -point for any $\mathscr {I}\in \mathcal {D}$ .

Question 7.7. Does there exist a Borel or analytic tall ideal other than $\mathcal {ED}_{fin}$ which is non-trivially Laflamme–Zhu?

Question 7.8. Is there a critical ideal $\mathscr {I}$ for the Borel (analytic) Laflamme–Zhu ideals, i.e., such that any ideal $\mathscr {J}$ is Laflamme–Zhu if and only if $\mathscr {J}\leq _{KB}\mathscr {I}$ ?

Question 7.9. In case there is an analytic ideal $\mathscr {I}\nleq _{KB}\mathcal {ED}_{fin}$ which is Laflamme–Zhu, let $\mathbf {LZ}_{Borel}$ and $\mathbf {LZ}_{Analytic}$ be the families of all Borel and analytic ideals which are Laflamme–Zhu, respectively. What is the structure of the Katětov–Blass order restricted to such classes?