Proof. Let $f\in {{}^{\omega }\omega }$ be a finite-to-one function such that

$$ \begin{align*} \limsup\nolimits_{n\in P}\mathrm{card\,}(f^{-1}(n))=\infty \end{align*} $$

holds for all $P\in p$ . We define a family $\mathcal {A}$ as follows: $A\in \mathcal {A}$ if and only if there exist $i<\omega $ and $P\in p$ such that $\mathrm { card\,}(f^{-1}(n)\setminus A)<i$ for each $n\in P$ . Then, Theorem 2.7 ensures that the ultrafilters we are building are strictly RK-greater than $p $ .

We claim that $\left \{ f^{-1}(p)\right \} \cup \mathcal {A}$ is a $P^+$ -family. Suppose not, and take a witness $\int \mathcal { W}$ . From Remark 2.4, without loss of generality, we may assume that $\int {\mathcal {W}}\subset \langle f^{-1}(p)\cup \mathcal {A}\cup \widetilde {\mathcal {W}}\rangle $ . Consider two cases:

Case 1: There exists a sequence $(B_{n})_{n<\omega }$ and a strictly increasing $k\in {{}^{\omega }\omega }$ such that $B_{n}\subset W_{k(n)}$ , $ f(B_{n})\not \in p$ , and B is compatible with $f^{-1}(p)\cup \mathcal {A} \cup \widetilde { \mathcal {W}}$ , where $B=\bigcup _{n<\omega }B_{n}$ . Take a sequence $ (f(\bigcup _{i\leq n}B_{n}))_{n<\omega }$ . This is an increasing sequence, and it is clear that $\bigcup _{n<\omega }f(\bigcup _{i\leq n}B_{n})=f(B)\in p$ . Make a partition of $f(B)$ by taking $f(\bigcup _{n\leq i+1}B_{n})\setminus f(\bigcup _{n\leq i}B_{n})$ for $i<\omega $ . As p is a P-point, there exists some $P\in p$ such that $P\cap (f(\bigcup _{n\leq i+1}B_{n})\setminus f(\bigcup _{n\leq i}B_{n}))$ is finite for all $i<\omega $ , and thus $ f^{-1}(P)\cap B_{n}$ is finite for all $i<\omega $ . Therefore, $ f^{-1}(P)\cap W_{i}\cap B$ is finite, and thus $(f^{-1}(P)\cap B)^{c}\in \int \mathcal {W}$ , which means that $\int {\mathcal {W}}$ is not compatible with $\langle f^{-1}(p)\cup \{B\}\rangle $ .

Case 2: Otherwise without loss of generality $f(W_{i})\in p$ for each $ i<\omega $ , since we are not in Case 1. Define sets $V_{1}=W_{1}$ , $ V_{i}=W_{j}\cap f^{-1}(\bigcap _{k<i}f(W_{k}))$ , and note that $ \bigcup _{i<\omega }V_{i}\in \langle \mathcal {A}\cup f^{-1}(p)\rangle $ since we are not in Case 1. Then, $(f(V_{i}))_{i<\omega }$ is a decreasing sequence, and because f is finite-to-one, $(f(V_{i})\setminus f(V_{i+1}))_{i<\omega }$ is a partition of $f(V_{1})\in p$ . As p is a P-point, there exists $P\in p$ such that $P_{i}=(f(V_{i})\setminus f(V_{i+1}))\cap P$ is finite for each $i<\omega $ . Let $g:\omega \rightarrow \omega $ be defined by $g(i)=E\left ( \frac {i+1}{2}\right ) $ , where $E\left ( x\right ) $ stands for the integer part of x. Let

$$ \begin{align*} R=\bigcup\nolimits_{i<\omega }(f^{-1}(P_{i})\cap \bigcup\nolimits_{j\in \lbrack g(i),...,i]}V_{i}). \end{align*} $$

Note that $R\cap V_{i}$ is finite for each $i<\omega $ , and that

$$ \begin{align*} \limsup\nolimits_{n\in \tilde{P}}\mathrm{card\,}(f^{-1}(n)\cap R)=\infty \end{align*} $$

for all $\tilde {P}\in p$ and $\tilde {P}\subset P$ . Thus, $R^{c}\not \in \langle \mathcal {A}\cup f^{-1}(p)\cup \widetilde {\mathcal {W}}\rangle $ , although $R^{c}\in \int \mathcal {W}$ , which completes Case 2.

To complete the proof of the theorem use Corollary 2.6.⊣

Fact 3.2. Let $\mathcal {A}$ be a centered family of subsets of $\omega $ such that $\mathcal {A} \cup \{F\}$ is not an ultrafilter subbase for any F compatible with $ \mathcal {A}$ . Let $\mathcal {F}$ be a countable family compatible with $\mathcal {A}$ . Then $\mathcal {A} \cup \mathcal {F}$ is not an ultrafilter subbase.

Proof. Without loss of generality, we may assume that $(F_{n})_{n<\omega }$ is a decreasing sequence of sets, such that $F_{n+1}\not \in \langle \mathcal {A} \cup \{F_{n}\}\rangle $ . Put $B_{n}=F_{n}\setminus F_{n+1}$ and define $ B^1=\bigcup _{n<\omega }B_{2n}$ , $B^2=\bigcup _{n<\omega }B_{2n+1}$ . Clearly at least one of sets $B^1$ , $B^2$ interact $\mathcal {A}$ —say $B_1$ does. If $B^1\not \in \langle \mathcal {A}\cup \mathcal {F}\rangle $ then we are done. Suppose that $B^1\in \langle \mathcal {A}\cup \mathcal {F}\rangle $ , and denote by $n_{0}$ , the minimal $n<\omega $ that $B^1\in \langle \mathcal {A} \cup \{F_{n}\}\rangle $ . But $F_{n_{0}+1}\cap B^1=F_{n_{0}+2}\cap B^1$ and so $F_{n_{0}+2}\in \langle \mathcal {A}\cup \mathcal {F}_{n_{0}+1}\rangle $ , which is a contradiction.⊣

Fact 3.3. Let $\mathcal {Y},\mathcal {Z}$ be centered families of subsets of $ \omega $ , which are not ultrafilter subbases. If $h\in {{}^{\omega }\omega }$ , then there exist sets Y and Z such that Y is compatible with $\mathcal {Y}$ , $ Z$ is compatible with $\mathcal {Z}$ , and $h(Z)$ is not compatible with Y.

Proof. Take any O such that O and $O^c$ are compatible with $ \mathcal {Y}$ .

If $h^{-1}(O)$ is not compatible with $ \mathcal {Z}$ then $Y=O$ , $Z=(h^{-1}(O))^c$ ;

if $h^{-1}(O^c)$ is not compatible with $ \mathcal {Z}$ then $Y=O^c$ , $Z=h^{-1}(O)$ ;

if $h^{-1}(O) $ is compatible with $ \mathcal {Z}$ and $h^{-1}(O^c) $ is compatible with $ \mathcal {Z}$ then $Y=O$ , $Z=(h^{-1}(O))^c$ .⊣

Proof. First repeat the proof of Theorem 3.1 except for the last line. Then continue as follows.

We arrange all contours in a sequence $(\int {\mathcal {W}}_{\alpha })_{\alpha < \mathfrak {b}}$ and ${{}^{\omega }\omega }$ in a sequence $(f_{\beta })_{\beta < \mathfrak {b}}$ . We will build a family $\{({\mathcal {F}}_{\alpha }^{\beta })_{\alpha <\mathfrak {b}}\}_{\beta <\mathfrak {b}}$ of increasing $\mathfrak {b }$ -sequences $({\mathcal {F}}_{\alpha }^{\beta })_{\alpha <\mathfrak {b}}$ of filters such that:

1. 1) Each ${\mathcal {F}}_{\alpha }^{\beta }$ is generated by $\mathcal {A}$ together with some family of sets of cardinality  $<\mathfrak {b}$ ;

2. 2) ${\mathcal {F}}^\beta _0 = \mathcal {A}$ for each $\beta <\mathfrak {b}$ ;

3. 3) For each $\alpha ,\beta <\mathfrak {b}$ , there exists $F\in {\mathcal {F}} _{\alpha +1}^{\beta }$ such that $F^{c}\in \int {\mathcal {W}}_{\alpha }$ ;

4. 4) For every limit $\alpha $ and for each $\beta $ , let ${\mathcal {F}} _{\alpha }^{\beta }=\bigcup _{\gamma <\alpha }{\mathcal {F}}_{\gamma }^{\beta } $ ;

5. 5) For each $\alpha ,\gamma <\alpha ,\beta _{1}$ , $\beta _{2}<\alpha ,$ there exists a set $F\in {\mathcal {F}}_{\alpha +1}^{\beta _{1}}$ such that $ (f_{\gamma }(F))^{c}\in {\mathcal {F}}_{\alpha +1}^{\beta _{2}}$ .

The existence of such families follows by a standard induction with sub-inductions using Theorem 2.5 and Fact 3.3 for Condition 5. It follows from the proof of Fact 3.2 that ${\mathcal {F}} _{\alpha }^{\beta }$ is not an ultrafilter subbase for each $\alpha $ and $\beta $ . It suffices now to take for each $\beta <\mathfrak {c}$ , any ultrafilter extending $\bigcup _{\beta <\mathfrak {c}}{\mathcal {F}}_{\alpha }^{\beta }$ and note that, by Proposition 2.1, it is a P-point.⊣

Proof. For each $n<\omega $ we let $f_{n}$ to be a finite-to-one function that witnesses $ p_{n+1}>_{RK}p_{n}$ . Consider a sequence $(T_n)_{n<\omega }$ of disjoint trees such that for each $n<\omega $

1. 1) the root of $T_n$ is equal to $n\in \omega $ ;

2. 2) $\mathrm {Level\,}_m(T_n) = \{f_m^{-1}(k): k\in \mathrm {Level\,}_{m-1}(T_n)\}$ for $1\leq m\leq n$ ;

3. 3) $\mathrm {Level\,}_m(T_n)=\emptyset $ for $m>n$ .

Since $L_\infty =\bigcup _{n<\omega }\mathrm {max\,} T_n$ is countably infinite we treat it as $\omega $ as well as $L_m=\bigcup _{n<\omega }\mathrm {Level\,} _m(T_n)$ . Let $g_m: L_\infty \setminus \bigcup _{k<m}L_k \to L_m$ be a function defined by order of the trees $T_n$ .

On $L_\infty $ we define a family of sets: $\mathcal {B}=\bigcup _{n<\omega } g_m^{-1}(p_m)$ .

To conclude it suffices, by Corollary 2.6, to show that $\mathcal {B} $ is a $P^+$ -family, thus by Theorem 2.5 it suffices to prove that $g_{m}^{-1}(p_{m})$ is a $P^+$ -family, for any m. But this is an easier version of the fact which we established in the proof of Theorem 3.1.⊣

Fact 3.6. There exists a strictly $<^{\ast }$ -increasing sparse sequence $ \mathcal {F}=(f_{\alpha })_{\alpha <\mathfrak {b}}\subset {{}^{\omega }\omega }$ of nondecreasing functions such that $f_{\alpha }\left ( n\right ) \leq n$ for each $n<\omega $ and $\alpha <\mathfrak {b}$ .

Proof. First, we build, from the definition of $\mathfrak {b}$ , an $<^{\ast }$ -increasing sparse sequence $(g_{\alpha })_{\alpha <\mathfrak {b}}\subset { ^{\omega }\omega }$ of nondecreasing functions that fulfill the following condition: if $\alpha <\beta <\mathfrak {b}$ , then $g_{\alpha }(n)>g_{\beta }(2n)+n$ for almost all $n<\omega $ . Then a $\mathfrak {b}$ -sequence $(f_\alpha )_{\alpha <\mathfrak {b}}$ defined by $ f_\alpha (m)=m - \mathrm {max\,}\{n: g_\alpha (n)<m\}$ is as desired.⊣

Fact 3.7. If an ordinal number $\alpha $ can be sparsely embedded in ${{}^{\omega }\omega }$ , under identity, as nondecreasing functions, then $\alpha $ can be sparsely embedded in ${{}^{\omega }\omega }$ , under any function f that converges to $\infty $ .

Proof. Let $h<f$ be a nondecreasing function such that $h(n+1)-h(n)\leq 1$ . Let $ (g_\beta )_{\beta <\alpha }$ be an embedding of $\alpha $ . Define $ (f_\beta )_{\beta <\alpha }$ by: $f(\alpha )(k)=g_\alpha (n)$ if and only if $h(k)=n$ .⊣

Fact 3.8. If an ordinal number $\alpha $ can be sparsely embedded in ${ ^{\omega }\omega }$ as nondecreasing functions that are less than any function $f\in {{}^{\omega }\omega }$ , then $\alpha $ can be sparsely embedded as nondecreasing functions between any sparse pair of functions $g<^{\ast }h\in {{}^{\omega }\omega }$ .

Proof. Without loss of generality, we assume f to be nondecreasing. Let $(f_{\beta })_{\beta <\alpha }$ be an embedding of $\alpha $ under f. Clearly, it suffices to prove that there is an embedding under s defined by $s(n)=h(n)-g(n)$ if $ h(n)\geq g(n)$ and $s(n)=0$ otherwise.

Define a sequence $(k(n))_{n<\omega }$ by $k(0)=\mathrm {min\,}\{m:s(i)\geq f(0)$ for all $i\geq m\}$ , $k(n+1)=\mathrm {min\,}\{m:m>k(n)\text { }\&\text { } s(i)\geq f(n+1)$ for all $i\geq m\}$ . Finally define $g_\alpha $ as follows: $g_\alpha (n)=f_\alpha (m)$ if and only if $k(n) \leq m < k(n+1)$ .⊣

Fact 3.9. For each $\gamma <\mathfrak {b^{+}}$ , there exists a strictly $ <^{\ast }$ -increasing sparse sequence $\mathcal {F}=(f_{\alpha })_{\alpha <\gamma }\subset {{}^{\omega }\omega }$ of nondecreasing functions.

Proof. Facts 3.6 and 3.8 clearly imply that the first ordinal number which cannot be embedded as a sparse sequence in ${{}^{\omega }\omega } $ under $\mathrm {id\,}_{\omega }$ is equal to $\alpha $ or to $\alpha +1$ where $\alpha $ is a limit number. Facts 3.6 and 3.8 also imply that the set of ordinals less than $\alpha $ is closed under $ \mathfrak {b}$ sums.

Indeed, let $\beta $ be the minimal ordinal number $< \mathfrak {b}^+$ that may not be embedded under identity as an $<^*$ -increasing sparse sequence. Clearly $\mathrm {cof\, }(\beta ) \leq \mathfrak {b}$ . Take an increasing sequence $(\alpha _\delta )_{ \delta < \mathrm {cof\,} \beta }$ that converges to $\beta $ . Clearly for each $ \alpha < \beta $ there is $(g^\alpha _\eta )_{\eta <\alpha }$ —an embedding of $ \alpha $ into ${^\omega \omega }$ as a sparse sequence under identity. By Fact 3.8 for each $\alpha < \beta $ there is an $<^*$ -increasing sparse sequence of $(f^\alpha _\eta )_{\eta <\alpha }$ such that $f_\alpha <^* f^\alpha _\eta <^* f_{\alpha +1}$ (for $f_\alpha $ , $f_{\alpha +1}$ from the proof of Fact 3.6). Now $(f^\alpha _\eta )_{\alpha <\mathrm {cof\,} (\beta ), \eta <\alpha }$ with lexicographic order is a required embedding of $ \beta $ .

Thus, this number is not less than $\mathfrak {b}^{+}$ .⊣

Proof. Note that $\mathrm {cof\,}(\gamma )\leq \mathfrak {b}$ . Consider a set of pairwise disjoint trees $T_{n}$ such that each $T_{n}$ has a minimal element, each element of $T_{n}$ has exactly n immediate successors, and each branch has the highest $\omega $ .

Let $\{f_{\alpha }\}_{\alpha <\gamma }\subset {{}^{\omega }\omega }$ be a sparse, strictly $<^{\ast }$ -increasing sequence, the existence of which is demonstrated by Fact 3.9. For each $\alpha <\gamma $ , define

$$ \begin{align*} X_{\alpha }=\bigcup\nolimits_{n<\omega }\mathrm{Level\,}_{f_{\alpha }(n)}T_{n}. \end{align*} $$

For each $\alpha <\beta \leq \gamma $ , define

$$ \begin{align*} f_{\alpha }^{\beta }:\bigcup\nolimits_{\{n<\omega: f_{\alpha }(n)<f_{\beta }(n)\}}\mathrm{Level\,}_{f_{\beta }(n)}T_{n}\rightarrow \bigcup\nolimits_{\{n<\omega: f_{\alpha }(n)<f_{\beta }(n)\}}\mathrm{Level\,} _{f_{\alpha }(n)}T_{n} \end{align*} $$

that agrees with the order of trees $T_{n}$ for $n<\omega $ such that $ f_{\alpha }(n)<f_{\beta }(n)$ . Note that $\mathrm {dom\,}f_{\alpha }^{\beta }$ is cofinite on $X_{\beta }$ for each $\alpha <\beta $ .

Let $p=p_{0}$ be a P-point on $X_{0}=\bigcup _{n<\omega }\mathrm {Level\,} _{0}T_{n}$ . We proceed by recursively building a filter $p_{\beta }$ on $ X_{\beta }$ . Suppose that $p_{\alpha }$ are already defined for $\alpha <\beta $ . If $\beta $ is a successor, then it suffice to repeat a proof of Theorem 3.1 for $P_{\beta -1}$ and $f^{\beta }_{\beta -1}$ .

So suppose that $\beta $ is limit. Let $R\subset \beta $ be cofinite with $ \beta $ and of order type less than or equal to $\mathfrak {b}$ . Define a family

$$ \begin{align*} \mathcal{C}=\bigcup_{\alpha \in R }\{(f_{\alpha }^{\beta })^{-1}(p_{\alpha }) \}, \end{align*} $$

which is obviously strongly centered.

Clearly each filter that extends $\mathcal {C}$ is $RK$ -greater than each $ p_{\alpha }$ for $\alpha <\beta $ . But we need a P-point extension. Thus, by Corollary 2.6 it suffices to prove that $\mathcal {C}$ is a $P^+$ -family. Thus, by Theorem 2.5 it suffices to prove that $\bigcup _{\gamma \in R,\gamma \leq \alpha }\{(f_{\gamma }^{\beta })^{-1}(p_{\gamma })\}\subset (f_{\alpha }^{\beta })^{-1}(p_{\alpha })$ is a $P^+$ -family, for each $\alpha \in R$ . But it is (an easier version of) what we did in the proof of Theorem 3.1.⊣

Question 3.11. What is the least ordinal $\alpha $ such that there exists an unbounded embedding of $\alpha $ into the set of P-points?Footnote ⁵

Proof. ———————— (beginning of quotation) ————————

Let X be a set of all functions $x: \mathbb {Q} \to \omega $ such that $ x(r)=0$ for all but finitely many $r\in \mathbb {Q}$ ; here $\mathbb {Q}$ is the set of rational numbers. As X is denumerable, we may identify it with $ \omega $ via some bijection. For each $\xi \in \mathbb {R}$ , we define $h_\xi : X \to X$ by

$$ \begin{align*} h_{\xi }(x)(r)=\left\{ \begin{array}{c} x(r)\text{ if }r<\xi , \\ 0\text{ if }r\geq \xi. \end{array} \right. \end{align*} $$

Clearly, if $\xi < \eta $ , then $h_\xi \circ h_\eta = h_\eta \circ h_\xi = h_\xi $ . The embedding of $\mathbb {R}$ into P-points will be defined by $\xi \to D_\xi = h_\xi (D)$ for a certain ultrafilter D on X. If $\xi < \eta $ , then

$D_{\xi }=h_{\xi }(D)=h_{\xi }\circ h_{\eta }(D)=h_{\xi }(D_{\eta })\leq D_{\eta }$ .

We wish to choose D in such a way that

(a) $D_\xi \not \cong D_\eta $ (therefore, $D_\xi < D_\eta $ when $\xi < \eta $ ), and

(b) $D_\xi $ is a P-point.

Observe that it will be sufficient to choose D such that

(a’) $D_\xi \not \cong D_\eta $ when $\xi < \eta $ and both $\xi $ and $\eta $ are rational, and

(b’) D is a P-point.

Indeed, (a’) implies (a) because $\mathbb {Q}$ is dense in $\mathbb {R}$ . If (a) holds, then $D_{\xi -1} < D_\xi $ , so $D_\xi $ is a nonprincipal ultrafilter $\leq D$ ; hence (b’) implies condition (b).

Condition (a’) means that, for all $\xi <\eta \in \mathbb {Q}$ and all $ g:X\rightarrow X$ , $D_{\eta }\not =g(D_{\xi })=gh_{\xi }(D_{\eta })$ . By Theorem 2.7, this is equivalent to $\{x:gh_{\xi }(x)=x\}\not \in D_{\eta }$ , or by our definition of $D_{\eta }$ ,

(a”) $$ \begin{align} \{x:gh_{\xi }(x)=h_{\eta }(x)\}=h_{\eta }^{-1}\{x:gh_{\xi }(x)=x\}\not\in D. \end{align} $$

We now proceed to construct a P-point D satisfying (a’’) for all $\xi <\eta \in \mathbb {Q}$ and for all $g:X\rightarrow X$ ; this will suffice to establish the theorem.

————————– (end of quotation) —————————–

List all pairs $(\xi , \eta )$ , $\xi < \eta \in \mathbb {Q}$ in the sequence $ (\xi _i, \eta _i)_{i<\omega }$ . For each $g\in {{}^{X}X}$ , $\xi <\eta \in \mathbb { Q}$ , define ${A}_{g,\xi ,\eta }=\{x\in X:gh_{\xi }(x)\not =h_{\eta }(x)\}$ , $ \mathcal {A}_{i}=\mathcal {A}_{\xi _{i},\eta _{i}}=\{{A}_{g,\xi _{i},\eta _{i}}:g\in {{}^{X}X}\}$ , and $\mathcal {A}=\bigcup _{i<\omega }\mathcal {A}_{i}$ . Clearly, $\mathcal {A}$ is strongly centered.

We claim that $\mathcal {A}$ is a $P^+$ -family.

Indeed, by Theorem 2.5, it suffices to prove that for each $n<\omega $ , $\bigcup _{i<n}\mathcal {A}_{i}$ is a $P^+$ -family. Suppose not and take (by Remark 2.4) the witnesses $i_{0}$ and $ \int \mathcal {W}$ such that $\int {\mathcal {W}}\subset \langle \bigcup _{i<i_{0}}\mathcal {A}_{i}\cup \widetilde {\mathcal {W}}\rangle $ . For each $n<\omega $ , consider the condition $(S_{n})$ :

$$ \begin{align*} \exists x_{n}:\forall i<i_{0}:\left( x_{n}\in h_{\xi _{i}}({\widetilde{W}_{1} })\text{ }\&\text{ }\mathrm{card\,}(h_{\eta _{i}}(h_{\xi _{i}}^{-1}(x_{n}))\cap h_{\eta _{i}}({\widetilde{W}_{n}}))>n\right). \end{align*} $$

Case 1: $S_{n}$ is fulfilled for all $n<\omega $ . Then, for each $n<\omega $ , $j<n$ , choose $x_{n}^{j}\in \widetilde {W}_{n}$ such that $h_{\xi _i }(x_{n}^{j})=x_{n}$ and $h_{\eta _{i}}(x_{n}^{j_{0}})\not =h_{\eta _{i}}(x_{n}^{j_{1}})$ for $j_{0}\not =j_{1}$ . Define ${E}=\bigcup _{n<\omega }\bigcup _{j\leq n}\{x_{n}^{j}\}$ . Clearly ${E}^{c}\in \int {\mathcal {W}}$ , but ${E}\not \subset \bigcup _{i<i_{0}}\bigcup _{g\in \mathcal {G}}({A}_{g,\xi _{i},\eta _{i}})\cup \bigcup _{l<m}W_{l}$ for any choice of finite family $ \mathcal {G}\subset {{}^{X}X}$ and for any $m<\omega $ .

Case 2: $S_{n}$ is not fulfilled for some $n_{0}<\omega $ . Then, there exist functions $\{g_{n,i}\}_{n\leq n_{0},i<i_{0}}\subset {{}^{X}X}$ such that $ \widetilde {W}_{1}\subset \bigcup _{n\leq n_{0}}\bigcup _{i<i_{0}}({A} _{g_{n,i},\xi _{i},\eta _{i}})\cup \bigcup _{n\leq n_{0}}W_{n}$ , i.e., $\int \mathcal {W}\lnot\ \text{is not compatible with}\ \langle \bigcup _{i<i_{0}}\mathcal {A}_{i}\cup \widetilde { \mathcal {W}}\rangle $ .

Corollary 2.6 completes the proof.⊣

Proof. We will combine ideas form proofs of Theorems 3.1 and 3.12 with some new arguments.

Again, let X be a set of all functions $x: \mathbb {Q} \to \omega $ such that $x(r)=0$ for all but finitely many $r\in \mathbb {Q}$ . Since X is infinitely countable, we treat it as $\omega $ .

Let p be a P-point on X such that for each $q\in \mathbb {Q}$ and for each $ P\in p$ there exists $x \in P$ such that $\mathrm {max\,} \mathrm {supp\,}(x)< q$ . Let $f\in {{}^{X}X }$ be a finite-to-one function such that $ \limsup \nolimits _{x\in P}\mathrm {card\,}(f^{-1}(x))=\infty $ for all $P\in p$ and that $\mathrm {max\,} \mathrm {supp\,} x < \mathrm {max\,} \mathrm {supp\,} f(x)$ . Again, we define a family $\mathcal {A}$ as follows: $A\in \mathcal {A}$ if and only if there exist $i<\omega $ and $P\in p$ such that $\mathrm {card\,} (f^{-1}(x)\setminus A)\leq i$ for each $x\in P$ . For each $\xi \in \mathbb {R} $ , we again define functions $h_\xi : X \to X$ by

$$ \begin{align*} h_{\xi }(x)(r)=\left\{ \begin{array}{c} x(r)\text{ if }r<\xi , \\ 0\text{ if }r\geq \xi. \end{array} \right. \end{align*} $$

List all rational numbers in $\omega $ -sequence $\mathbb {Q}=(q_i)_{i<\omega }$ . Let $\mathcal {B}_i=h^{-1}_{q_i}(\mathcal {A} \cup f^{-1}(p))$ and let $ \mathcal {B} = \bigcup _{i<\omega } \mathcal {B}_i$ .

Our aim is to prove that $\mathcal {B}$ can be extended to such a P-point Q that $h_\xi (Q) \not = h_\eta (Q)$ for each $\xi \not = \eta \in \mathbb {Q}$ (and thus for each $\xi \not = \eta \in \mathbb {R}$ ). (4)

To this end, we add to $\mathcal {B}$ a family $\mathcal {C}$ defined as follows: list all pairs $(\xi , \eta )$ , $\xi < \eta \in \mathbb {Q}$ in the sequence $(\xi _i, \eta _i)_{i<\omega }$ . For each $g\in {{}^{X}X}$ , $\xi <\eta \in \mathbb {Q}$ , define ${C}_{g,\xi ,\eta }=\{x\in X:gh_{\xi }(x)\not =h_{\eta }(x)\}$ , $\mathcal {C}_{i}=\mathcal {C}_{\xi _{i},\eta _{i}}=\{{C}_{g,\xi _{i},\eta _{i}}:g\in {{}^{X}X}\}$ , and $\mathcal {C} =\bigcup _{i<\omega }\mathcal {C}_{i}$ .

Thus to prove (4), it suffices by Corollary 2.6 to prove that $ \mathcal {B}\cup \mathcal {C}$ is a $P^+$ -family. Thus, by Theorem 2.5, in order to prove (4), it suffices to prove that:

$\mathcal {D}_{i}$ is a $P^+$ -family for $\mathcal {D}_{i}$ defined for $i<\omega $ as follows:

$$\begin{align*}\mathcal{D}_{i}=\bigcup_{j\leq i}\mathcal{B}_{i}\cup \bigcup_{j\leq i} \mathcal{C}_{i}.\tag{5} \end{align*}$$

First, to prove it, we notice that $\mathcal {D}_i$ is strongly centered. Indeed, define $ q_m = \mathrm {min\,} \{q_j: j \leq i\}$ , $q_M = \mathrm {max\,} \{q_j: j \leq i\}$ , and $\xi _m=\mathrm {min\,}\{\xi _j: j\leq i\} $ and note that $ h^{-1}_{q_M}(x) \subset h^{-1}_{q_j}(x)$ for each $j \leq i$ and for each x such that $\mathrm {max\,}\mathrm {supp\,}(x) < \mathrm {min\,}\{ q_m, \xi _m \}$ . It is easy to see that $h^{-1}_{q_M}(x) \cup \bigcup _{j\leq i}\mathcal {C} _i $ is strongly centered, hence $\mathcal {D}_i$ is strongly centered.

Fix i, and suppose that (5) does not hold. So take a witness $\int \mathcal {W}$ . From Remark 2.4, without loss of generality, we may assume that $ \int {\mathcal {W}}\subset \langle \mathcal {D}_{i}\cup \widetilde {\mathcal {W}} \rangle $ .

Let $A_W \in \mathcal {A}$ , $P_W\in p$ , $n_W\in \omega $ , $C_n^W \in \mathcal {C }$ for $n\leq n_W$ , and $l_W \in \omega $ . Define

$$ \begin{align*} W^*(A_W, P_W, C_1^W, \ldots , C_{n_W}^W, \widetilde{W_{l_W}}) = \bigcap_{n\leq n_W} C_n^W \cap \bigcap_{j \leq i} h^{-1}_{\xi_j} (A_W \cap P_W) \cap \widetilde{W_{l_W}}. \end{align*} $$

Define $W \in \mathcal {W}^-$ if and only if $W \in \int \mathcal {W}$ and W is co-finite or empty on each $W_i$ . We will say that a set $W\in \mathcal {W}$ is $\mathit {attainable}$ (by $(n_A, P_W, n_W, l_W)$ ) if there exist $A_W \in \mathcal {A}_{n_A}$ , $\{C_k^W \in \bigcup _{j \leq i} \mathcal {C}_j: k \leq n_W\}$ such that the condition $W^*(A_W, P_W, C_1^W, \ldots , C_{n_W}^W, \widetilde {W_{l_W} }) \subset W$ is satisfied. The complement (to $\widetilde {W_1}$ ) of the attainable set is called removable and sometimes we indicate which variables, sets, functions.

Since $\int \mathcal {W} \subset \langle \mathcal {D}_i \rangle $ , thus each set $W \in \mathcal {W}^-$ is attainable.

Consider a sequence of possibilities:

1. 1) l cannot be fixed, i.e., for each $l\in \omega $ there exists $W \in { \mathcal {W}^-}$ such that W is not removable by any $(n_A, P_W, n_W, l)$ ;

2. 2) l can be fixed, but $n_A$ cannot, i.e., for each $W\in \mathcal {W}^-$ , W is attainable by some $(n^A_W, P_W, n^C_W, l)$ , but for each $n_A$ there exists $W^{\prime }\in {\mathcal {W}^-}$ such that $W^{\prime }$ is not attainable by any $(n^A, P_{W^{\prime }}, n^C_{W^{\prime }}, l)$ ;

3. 3) l and $n^A$ can be fixed, but $n_C$ cannot;

4. 4) l, $n^A$ and $n^C$ can be fixed, but P cannot;

5. 5) l, $n^A$ , $n^C$ , and P can be fixed.

Note that each set $W \in \mathcal {W}^-$ is attainable if and only if an alternative of cases 1) to 5) holds.

In case 1) for each l, let $W_l\subset \widetilde { {W}_l}$ and $W_l \in \mathcal {W}^-$ be a witness that l may not be fixed. Note that $\bigcup _{l<\omega }W_l \in \mathcal {W}^-$ and that $\bigcup _{l<\omega }W_l$ may not be removed by any $(n^A_W, P_W, n^C_W, l_W)$ .

In case 2) we proceed like in case 1). Note that if $l^{\prime }\geq l$ and ${ n^A}^{\prime }$ and a set $W \in \mathcal {W}^-$ is not removable by any $({ n^A}^{\prime }, P_W, n^C_W, l^{\prime })$ then the set W is also not removable by any $(n^A, P_W, n^C_W, l)$ . Thus it suffices to consider cases when $l=n^A$ . For each l, let $W_l\subset \widetilde { {W}_l}$ and $W_l \in \mathcal {W}^-$ be a witness that l and $n^A=l$ may not be fixed. Again note that $\bigcup _{l<\omega }W_l \in \mathcal {W}^-$ and that $\bigcup _{l<\omega }W_l$ may not be removed by any $(n^A_W, P_W, n^C_W, l_W)$ .

In case 3) we proceed just like in case 2), not using that $n^A$ is fixed.

In case 4) for $k<\omega $ , let $X_k$ be the set of those $x\in X$ that for all $U\subset f^{-1}(x)$ such that $\mathrm {card\,}(f^{-1}(x) \setminus U) \leq n^A$ , for all partitions of a set $\bigcup _{j=1}^i(\xi _j^{-1}(X) \cap \widetilde {(W_k)})$ on the sets $X_{m,n}$ , for $m<n\leq i$ , there exist $ m_0, n_0$ such that $m_0<n_0\leq i$ and there exist $x_1, \ldots , x_{n^C+1} \in X_{m_0,n_0}$ that for $\xi _{(min)}=\mathrm {min\,}\{\xi _{m_0}, \xi _{n_0}\} $ , $\xi _{(max)}=\mathrm {max\,}\{\xi _{m_0}, \xi _{n_0}\}$ there is $\xi _{(min)}(x_r)=\xi _{(min)}(x_j)$ for $r,j \leq n^C+1$ and $ \xi _{(max)}(x_r)\not =\xi _{(max)}(x_j)$ for $r,j \leq n^C+1$ , $r\not = j$ .

Clearly, $(X_{k})$ is a decreasing sequence. If there exists k such that $ X_{k}\not \in p$ then putting $l=k$ there exists a set $P=(X_{k})^{c}$ such that all $W\in \mathcal {W}^{-}$ may be attained by $(n^{A},P,n^{C},l)$ so we would be in case 5), so, without loss of generality, $X_{k}\in P$ for each $ k<\omega $ .

Thus take a partition of X by $(X_k\setminus X_{k+1})$ . Since p is a P-point, and since $X_k\in p$ thus there exists $P_0\in p$ such that $ P_k=P_0\cap (X_k \setminus X_{k+1})$ is finite for all $k<\omega $ . For each $ x\in P_k$ there exists a finite(!) set $K_{k,x}\subset \widetilde {(W_k)} \cap \bigcup _{j\leq i}h^{-1}_{\xi _i}$ that may not be removed by $ (n^A,X,n^C,k)$ . (The proof that $K_{k,x}$ may be chosen finite is analogical, but easier, to that of case 5)). Take $ K=\bigcup _{k<\omega , x\in P_k}K_{k,x}$ and notice that $\widetilde {(W_1)} \setminus K \in \mathcal {W}^-$ and that $\widetilde {(W_1)}\setminus K$ is not removable by $(n^A, P, n^C, l)$ for any $P\in p$ .

In case 5) arrange $\bigcup _{j \leq i} h^{-1}_{\xi _j} (P) \cap \widetilde { W_{l}}$ into a sequence $(x_k)_{k<\omega }$ . Let $R(x) = {\binom {{\mathrm { card\,}(f^{-1}(x))} }{{\mathrm {card\,}(f^{-1}(x)) - n^C}}}$ , where $\binom {{n } }{{k}}$ denotes a binomial coefficient, and let $(A_{x,r})_{r \leq R}$ be a sequence of all subsets of $f^{-1}(x)$ of cardinality equal to $\mathrm { card\,}(f^{-1}(x) - n^C)$ . Consider a tree T, where the root is $\emptyset $ and on a level k the nodes are pairs of natural numbers $j,r$ such that $ j\leq i$ and $r\leq R(f(x_k))$ and, for each branch $\tilde {T}$ of T, $ \pi _2(\tilde {T}(k_1))=\pi _2(\tilde {T}(k_2))$ if $f(x_{k_1})=f(x_{k_2})$ , where $\tilde {T}(k)$ is an element of level k of a branch $\tilde {T}$ and $ \pi _2$ is a projection on the second coordinate. We see j as a choice to which class $\mathcal {C}_j$ does a set $C_{(.)}^{(.)}$ belongs and we see r as a choice of one of sets $A_{f(x),r}$ that $C_{(.)}^{(.)}$ together with $ A_{f(x),r}$ removes $x_k$ .

Clearly, the complements of all finite sets belong to $\int \mathcal {W}$ , so each finite set is removable. (6)

The maximal element of the branch $\tilde {T}$ has no successors if and only if there is $j \leq i$ such that there is no n sets in $\mathcal {C}_j$ that remove all $ x_t$ such that $\tilde {T}(k)=j$ and $f(x_k)\in A_{f(x_k), \pi _2(\tilde {T} (k))}$ . It implies that the set $\{x_k: \tilde {T}(k)=j, f(x_k)\in A_{f(x_k), \pi _2(\tilde {T}(k))}\}$ contains more than n different elements, say $x^1, \ldots , x^{n+1}$ , such that $h_{\xi _j}(x_{s_1}) = h_{\xi _j}(x_{s_2})$ and $h_{\eta _j}(x_{s_1}) \not = h_{\eta _j}(x_{s_2})$ for $ s_1\not = s_2$ , $s_1, s_2 \in \{1, \ldots , n+1\}$ .

By K $\ddot {\mathrm {o}}$ nig Lemma if all branches are finite, then the height of the tree T is finite, and so there are irremovable finite sets in contrary to (6). Thus there is infinite branch and the whole set $\bigcap _{j \leq i} h^{-1}_{\xi _i} (P) \cap \widetilde {W_{l}}$ is removable.⊣

Question 4.7. Is $\mathfrak {q}$ equal to any already defined cardinal invariant? Is $ \mathfrak {b}<\mathfrak {q}$ consistent? Is $\mathfrak {q}<\mathfrak {d}$ consistent?

Proof. By taking $\mathcal {S}\in \mathbb {T}$ and $\mathcal {B}=\emptyset ,$ we infer that $\mathfrak {q}_{0}=0$ .

To see that $\mathfrak {q}_{1}=1,$ let A, B be disjoint countably infinite sets. Let $\mathcal {S}_{A}$ be a contour on A and let $\mathcal {S} _{B}$ be a cofinite filter on B. Define a filter $\mathcal {S}$ on $A\cup B$ by $S\in \mathcal {S}$ if and only if $S\cap A\in \mathcal {S}_{A}$ and $S\cap B\in \mathcal {S}_{B}$ . Clearly $\mathcal {S}$ is not finer than a contour (since is RK-smaller than a cofinite filter $\mathcal {S}_{B}$ ), and $\mathcal {S} \cup \{A\}$ is a subbase of a contour.

Clearly, $\mathfrak {q}_{2}$ cannot be finite. To see that $\mathfrak {q} _{2}=\aleph _{0},$ let $\mathcal {W}=(W_{n})_{n<\omega }$ be a partition of $ \omega $ into infinite sets. We define a family $\mathcal {S}$ so that $S\in \mathcal {S}$ if and only if S is cofinite on each $W_{n}$ . Suppose that there is a partition $\mathcal {V}=(V_{n})_{n<\omega }$ such that $\int {\mathcal {V}} \subset \langle \mathcal {S}\cup S_{0}\rangle $ for some set $S_{0}$ such that $\mathcal {S}\cup S_{0}$ is strongly centered. Let $N_{fin}=\{n<\omega : \mathrm {card\,}(W_{n}\cap S_{0})<\omega \}$ and $N_{\mathrm {\infty }}=\omega \setminus N_{fin}$ . Define $S_{fin}=S_{0}\cap \bigcup _{n\in N_{fin}}W_{n}$ and $S_{\infty }=S_{0}\cap \bigcup _{n\in N_{\infty }}W_{n}$ . Since $S_{fin}^{c}\in \mathcal {S }$ , without loss of generality, we can assume that $S_{0}=S_{\infty }$ and so without loss of generality we can assume that $S_{0}=\omega $ .

Note also that $\bigcup _{n<\omega }V_{n}\cap W_{i}$ is infinite for infinitely many i, and so $\int \mathcal{W}$ is compatible with $\int \mathcal{V}$ . Thus we meet the assumptions of Theorem 5.1, and in each of the three cases there exist $i,j<\omega $ such that $V_{i}\cap W_{j}$ is infinite. But $ \omega \setminus V_{i}\in \int \mathcal {V}$ and thus $\omega \setminus (V_{i}\cap W_{j})\in \mathcal {S}$ , contrary to the definition of $\mathcal {S} $ . On the other hand, by adding $\widetilde {\mathcal {W}}$ to $\mathcal {S}$ , we obtain a subbase of $\int \mathcal {W}.$

That $\mathfrak {q}_{3}=\mathfrak {q}$ follows directly from the definition of $\mathfrak {q}$ .

Finally, $\mathfrak {q}_{\alpha }\leq \mathfrak {d}$ since each contour has a base of cardinality $\mathfrak {d}$ , as we have shown in [Reference van Douwen, Kunen and Vaughan13, Theorem 5.2].

Fact 5.3. $\mathfrak {u}_0=0$ , $\mathfrak {u}_1=1$ , $\mathfrak {u}_2 \geq \aleph _1$ .