The higher Cichon diagram in the degenerate case

TL;DR

This paper investigates the relationships between various cardinal invariants related to the higher meager ideal at a regular uncountable cardinal, providing characterizations and models that clarify their possible values in the degenerate case.

Contribution

It offers a complete characterization of add(M_kappa) and cof(M_kappa) in terms of other invariants and constructs models showing the range of non(M_kappa) values, addressing open questions in the field.

Findings

Characterization of add(M_kappa) and cof(M_kappa) in terms of cov(M_kappa), non(M_kappa), and unbounding/dominating numbers.

Models demonstrating no restrictions on non(M_kappa) in the degenerate case 2^{<kappa} > kappa.

Open question remaining for cof(M_kappa) values.

Abstract

For a regular uncountable cardinal kappa, we discuss the order relationship between the unbounding and dominating numbers on kappa and cardinal invariants of the higher meager ideal M_kappa. In particular, we obtain a complete characterization of add(M_kappa) and cof(M_kappa) in terms of cov(M_kappa) and non(M_kappa) and unbounding and dominating numbers, and we provide models showing that there are no restrictions on the value of non(M_kappa) in the degenerate case 2^{<kappa} > kappa except 2^{<kappa} leq non(M_kappa) leq 2^kappa. The corresponding question for cof(M_kappa) remains open. Our results answer questions of joint work of the author with Brooke-Taylor, Friedman, and Montoya.

Equations

$$[\sigma ]=\{f\in 2^{\kappa }:\sigma \subseteq f\}$$

$$c_{\alpha }^{\gamma }=f(g_{\gamma }(c_{\alpha }(0)))\hat{}f(g_{\gamma }(c_{\alpha }(1)))\hat{}f(g_{\gamma }(c_{\alpha }(2)))\hat{}...\hat{}f(g_{\gamma }(c_{\alpha }(\zeta )))\hat{}...$$

$$\bar{f}(\tau )=f(\tau (0))\hat{}f(\tau (1))\hat{}f(\tau (2))\hat{}...\hat{}f(\tau (\zeta ))\hat{}...$$

$$h_{f,g,\alpha,\beta}(\gamma)=\left\{\begin{array}[]{ll}g(\alpha+\gamma)&\mbox{if }\gamma<\beta\\ f(\gamma)&\mbox{if }\gamma\geq\beta.\\ \end{array}\right.$$

$$h^{\prime }(i_{\zeta }+\xi )=h(\xi )$$

$$h_{f,g,i_{\eta },\beta }(\xi )=g(i_{\eta }+\xi )\ge h^{\prime }(i_{\eta }+\xi )=h(\xi ),$$

$${\Phi }_{-}(f)(\alpha )=\min \{\gamma <{\kappa }^{+}:\exists \zeta <\kappa {\exists }^{\kappa }\beta \in [\alpha ,\alpha +\gamma )\text{mbox}suchthatf(\beta )=\zeta \}$$

$$\alpha +{\Phi }_{-}(f)(\alpha )<\alpha +h(\alpha )\le {\alpha }_{\xi +1}+h({\alpha }_{\xi +1})={\alpha }_{\xi +2}.$$

Full text

1 Introduction

Cardinal invariants of the continuum, describing the combinatorial properties of the real numbers ($2^{\omega}$ or $\omega^{\omega}$) and taking values between $\omega_{1}$ and the continuum ${\mathfrak{c}}$, have been studied intensively for several decades, and a rich theory with ZFC-results and independence proofs about the order relationship between various cardinal invariants has evolved (see [BJ] and [Bl]). More recently, higher cardinal invariants, that is, cardinal invariants of the higher Cantor space $2^{\kappa}$ or the higher Baire space $\kappa^{\kappa}$, where $\kappa$ is an uncountable regular cardinal, have started to be investigated and our work is a contribution to this ongoing research.

Our focus lies on cardinal invariants in the higher Cichoń diagram (see [BBFM] and [BGS]). The original Cichoń diagram [BJ] describes the relationship between cardinal invariants related to measure and category as well as the unbounding and dominating numbers ${\mathfrak{b}}$ and ${\mathfrak{d}}$. The latter can be easily redefined in the context of regular uncountable $\kappa$, by

• •

${\mathfrak{b}}_{\kappa}=\min\{|F|:F\subseteq\kappa^{\kappa}$ and $\forall g\in\kappa^{\kappa}\;\exists f\in F\;(f\not\leq^{*}g)\}$,

the $\kappa$-unbounding number, and

• •

${\mathfrak{d}}_{\kappa}=\min\{|F|:F\subseteq\kappa^{\kappa}$ and $\forall g\in\kappa^{\kappa}\;\exists f\in F\;(g\leq^{*}f)\}$,

the $\kappa$-dominating number,

where $f\leq^{*}g$ if there is $\alpha<\kappa$ such that $f(\beta)\leq g(\beta)$ for all $\beta\geq\alpha$. Clearly ${\mathfrak{b}}_{\kappa}\leq{\mathfrak{d}}_{\kappa}$. Also, there is a natural analog of the meager ideal on the higher Cantor space $2^{\kappa}$: give $2$ the discrete topology and equip $2^{\kappa}$ with the $\kappa$-box topology. That is, basic clopen sets are of the form

[TABLE]

where $\sigma\in 2^{<\kappa}$, and open sets are arbitrary unions of such basic clopen sets. Thus a set $A\subseteq 2^{<\kappa}$ is nowhere dense in this topology if for all $\sigma\in 2^{<\kappa}$ there is $\tau\supseteq\sigma$ such that $[\tau]\cap A=\emptyset$. This implies that the nowhere dense ideal on $2^{\kappa}$, denoted by ${\cal{NWD}}_{\kappa}$, is $<\kappa$-closed (i.e. closed under unions of size $<\kappa$). Say that $A\subseteq 2^{\kappa}$ is $\kappa$-meager if it is a union of at most $\kappa$ many nowhere dense sets, and let ${\cal M}_{\kappa}$ denote the ($\kappa$-closed) ideal of $\kappa$-meager sets. It is much less clear how the null ideal should be generalized to regular uncountable $\kappa$; a very interesting candidate has been proposed (for weakly compact $\kappa$) by Baumhauer, Goldstern, and Shelah in [BGS]. We shall not pursue this here.

Let ${\cal I}$ be a non-trivial ideal on a set $X$, that is, all the singletons $\{x\}$, $x\in X$, belong to ${\cal I}$ and $X\notin{\cal I}$. Define

• •

${\mathsf{add}}({\cal I})=\min\{|{\cal F}|:{\cal F}\subseteq{\cal I}$ and $\bigcup{\cal F}\notin{\cal I}\}$, the additivity of ${\cal I}$,

• •

${\mathsf{cov}}({\cal I})=\min\{|{\cal F}|:{\cal F}\subseteq{\cal I}$ and $\bigcup{\cal F}=X\}$, the covering number of ${\cal I}$,

• •

${\mathsf{non}}({\cal I})=\min\{|Y|:Y\subseteq X$ and $Y\notin{\cal I}\}$, the uniformity of ${\cal I}$, and

• •

${\mathsf{cof}}({\cal I})=\min\{|{\cal F}|:{\cal F}\subseteq{\cal I}$ and $\forall A\in{\cal I}\;\exists B\in{\cal F}\;(A\subseteq B)\}$,

the cofinality of ${\cal I}$.

Easily ${\mathsf{add}}({\cal I})\leq{\mathsf{cov}}({\cal I}),{\mathsf{non}}({\cal I})\leq{\mathsf{cof}}({\cal I})$. In our earlier work [BBFM] we observed that ${\mathfrak{b}}_{\kappa}\leq{\mathsf{non}}({\cal M}_{\kappa})$ and ${\mathsf{cov}}({\cal M}_{\kappa})\leq{\mathfrak{d}}_{\kappa}$ [BBFM, Observation 17], and proved:

• (i)

${\mathsf{add}}({\cal M}_{\kappa})\leq{\mathfrak{b}}_{\kappa}$ and ${\mathfrak{d}}_{\kappa}\leq{\mathsf{cof}}({\cal M}_{\kappa})$ for strongly inaccessible $\kappa$ [BBFM, Corollary 28],

• (ii)

${\mathsf{add}}({\cal M}_{\kappa})\geq\min\{{\mathfrak{b}}_{\kappa},{\mathsf{cov}}({\cal M}_{\kappa})\}$ [BBFM, Corollary 31], and

• (iii)

${\mathsf{cof}}({\cal M}_{\kappa})\leq\max\{{\mathfrak{d}}_{\kappa},{\mathsf{non}}({\cal M}_{\kappa})\}$ in case $2^{<\kappa}=\kappa$ [BBFM, Corollary 31].

In particular, ${\mathsf{add}}({\cal M}_{\kappa})=\min\{{\mathfrak{b}}_{\kappa},{\mathsf{cov}}({\cal M}_{\kappa})\}$ and ${\mathsf{cof}}({\cal M}_{\kappa})=\max\{{\mathfrak{d}}_{\kappa},{\mathsf{non}}({\cal M}_{\kappa})\}$ for strongly inaccessible $\kappa$, and the cardinals can be displayed in the following diagram.

We asked whether the assumptions were necessary in (i) and (iii) above [BBFM, Questions 29 and 32], and these questions are the starting point of the present work.

Note that in the degenerate case $2^{<\kappa}>\kappa$, some of the cardinal invariants become trivial. Namely Landver [La, 1.3] (see also [BBFM, Observation 23(ii)]) observed that ${\mathsf{add}}({\cal M}_{\kappa})={\mathsf{cov}}({\cal M}_{\kappa})=\kappa^{+}$ and this accounts for (ii) above in case $2^{<\kappa}>\kappa$. Blass, Hyttinen, and Zhang [BHZ, 4.15] (see also [BBFM, Observation 23(iii)]) noticed that $2^{<\kappa}\leq{\mathsf{non}}({\cal M}_{\kappa})$. Finally, we proved [Br, Proposition 2 (c)] that $2^{<\kappa}<{\mathsf{cof}}({\cal M}_{\kappa})$ and this implies that ${\mathsf{cof}}({\cal M}_{\kappa})>\max\{{\mathfrak{d}}_{\kappa},{\mathsf{non}}({\cal M}_{\kappa})\}$ in any model of $2^{<\kappa}=2^{\kappa}$; in particular, (iii) may fail in the degenerate case. We will obtain better lower bounds for ${\mathsf{cof}}({\cal M}_{\kappa})$ (see in particular Corollary 2). The results about ${\mathsf{add}}({\cal M}_{\kappa})$ and ${\mathsf{cov}}({\cal M}_{\kappa})$ in the degenerate case suggest the problem of whether one can say more about ${\mathsf{non}}({\cal M}_{\kappa})$ and ${\mathsf{cof}}({\cal M}_{\kappa})$ in this situation. We shall see this is not the case for ${\mathsf{non}}({\cal M}_{\kappa})$ (see in particular Theorems 7 and 8), while for ${\mathsf{cof}}({\cal M}_{\kappa})$ some interesting questions remain open.

This paper is organized as follows. In Section 2, we obtain several ZFC-results about higher cardinal invariants which strengthen those of [BBFM]; in particular we will compute ${\mathsf{add}}({\cal M}_{\kappa})$ and ${\mathsf{cof}}({\cal M}_{\kappa})$ in terms of the other cardinals in most cases (Corollary 3). In Section 3, we present independence results for the values of ${\mathsf{non}}({\cal M}_{\kappa})$ and ${\mathsf{cof}}({\cal M}_{\kappa})$ in the degenerate context. Section 4 investigates dominating numbers naturally arising in the discussion of ${\mathsf{cof}}({\cal M}_{\kappa})$.

Preliminaries. Let $\kappa$ be regular and $\lambda\geq\kappa$. For $f,g\in\kappa^{\lambda}$ say that $g$ eventually dominates $f$ ($f\leq^{*}g$ in symbols) if the set $\{\gamma<\lambda:f(\gamma)>g(\gamma)\}$ has size less than $\kappa$. Let ${\mathfrak{b}}^{\lambda}_{\kappa}$ and ${\mathfrak{d}}^{\lambda}_{\kappa}$ be the unbounding and dominating numbers of $(\kappa^{\lambda},\leq^{*})$, respectively111As we will see in Proposition 12 in Section 4, it does not really matter whether we use domination everywhere, modulo $<\kappa$ or modulo $<\lambda$ for our results; however, for the proofs in Section 2 the present definition is most convenient.. That is,

• •

${\mathfrak{b}}^{\lambda}_{\kappa}=\min\{|F|:F\subseteq\kappa^{\lambda}$ and $\forall g\in\kappa^{\lambda}\;\exists f\in F\;(f\not\leq^{*}g)\}$

• •

${\mathfrak{d}}^{\lambda}_{\kappa}=\min\{|F|:F\subseteq\kappa^{\lambda}$ and $\forall g\in\kappa^{\lambda}\;\exists f\in F\;(g\leq^{*}f)\}$

For $\kappa=\lambda$ we have ${\mathfrak{b}}_{\kappa}^{\lambda}={\mathfrak{b}}_{\kappa}$ and ${\mathfrak{d}}_{\kappa}^{\lambda}={\mathfrak{d}}_{\kappa}$. In general ${\mathfrak{d}}_{\kappa}^{\lambda}\geq{\mathfrak{d}}_{\kappa}^{\mu}\geq{\mathfrak{d}}_{\kappa}$ where $\kappa\leq\mu\leq\lambda$. If $\lambda>\kappa$ then ${\mathfrak{b}}^{\lambda}_{\kappa}=\kappa$, as witnessed by the constant functions.

2 ZFC results

For this whole section, let $\kappa$ be regular uncountable and let $\lambda=|2^{<\kappa}|$.

Theorem 1**.**

There are functions $\Phi_{-}:\kappa^{\lambda}\to{\cal{NWD}}_{\kappa}$ and $\Phi_{+}:{\cal M}_{\kappa}\to[\kappa^{\lambda}]^{\lambda}$ such that if $A\in{\cal M}_{\kappa}$, $g\in\kappa^{\lambda}$, and $g$ is not eventually bounded by $\Phi_{+}(A)$, then $\Phi_{-}(g)\not\subseteq A$.

Proof.

Let $\Sigma=\{\tilde{\sigma}\}\cup\{\sigma_{\gamma}:\gamma<\lambda\}\subseteq 2^{<\kappa}$ be a maximal antichain in $2^{<\kappa}$.

Fix $g\in\kappa^{\lambda}$. We recursively define a nowhere dense tree $T_{g}\subseteq 2^{<\kappa}$. More explicitly, we define sets $C^{\alpha}_{g}\subseteq 2^{<\kappa}$ for $\alpha<\kappa$ such that

• •

$C^{\alpha}_{g}$ is an antichain in $2^{<\kappa}$,

• •

if $\alpha<\beta$ and $\tau\in C^{\beta}_{g}$ then there is (necessarily unique) $\sigma\in C^{\alpha}_{g}$ such that $\sigma\subsetneq\tau$,

• •

if $\alpha<\beta$ and $\sigma\in C^{\alpha}_{g}$ then there is $\tau\in C^{\beta}_{g}$ such that $\sigma\subsetneq\tau$,

• •

for each $\sigma\in C^{\alpha}_{g}$ there is $\tau\supsetneq\sigma$ incompatible with all members of $C^{\alpha+1}_{g}$.

Then let $T_{g}$ be the downward closure of $\bigcup_{\alpha<\kappa}C^{\alpha}_{g}$, i.e., $\sigma\in T_{g}$ if $\sigma\subseteq\tau$ for some $\tau\in C^{\alpha}_{g}$ and some $\alpha$. Clearly $T_{g}$ is a nowhere dense tree, and we let $\Phi_{-}(g)=[T_{g}]$.

Let $C^{0}_{g}=\{\langle\rangle\}$.

Assume $C^{\alpha}_{g}$ has been defined and let $\sigma\in C^{\alpha}_{g}$. Assume ${\mathrm{lh}}(\sigma)=\zeta$. Then $\tau\supsetneq\sigma$ belongs to $C^{\alpha+1}_{g}$ if for some $\gamma<\lambda$, $\sigma\,{}\hat{}\,\sigma_{\gamma}\subseteq\tau$ and ${\mathrm{lh}}(\tau)=\zeta+{\mathrm{lh}}(\sigma_{\gamma})+g(\gamma)$. Note that this implies that $\tau=\sigma\,{}\hat{}\,\tilde{\sigma}$ is incompatible with all of $C^{\alpha+1}_{g}$.

If $\alpha$ is a limit ordinal, put $\sigma$ into $C^{\alpha}_{g}$ if there is a strictly increasing sequence $(\tau_{\beta}:\beta<\alpha)$ such that $\sigma=\bigcup_{\beta<\alpha}\tau_{\beta}$ and each $\tau_{\beta}$ belongs to $C^{\beta}_{g}$. This completes the construction of the $C^{\alpha}_{g}$ and of $T_{g}$.

Next fix $A\in{\cal M}_{\kappa}$, $A=\bigcup_{\alpha<\kappa}A_{\alpha}$, where the $A_{\alpha}$ form an increasing sequence of nowhere dense sets. Define $h=h^{A}:2^{<\kappa}\to 2^{<\kappa}$ such that for all $\alpha<\kappa$ and all $\sigma\in 2^{\alpha}$, $\sigma\subseteq h(\sigma)$ and $[h(\sigma)]\cap A_{\alpha}=\emptyset$.

Fix $\sigma\in 2^{<\kappa}$. Say ${\mathrm{lh}}(\sigma)=\zeta$. Define $f^{A}_{\sigma}\in\kappa^{\lambda}$ such that ${\mathrm{lh}}(h(\sigma\,{}\hat{}\,\sigma_{\gamma}))=\zeta+{\mathrm{lh}}(\sigma_{\gamma})+f^{A}_{\sigma}(\gamma)$ for all $\gamma<\lambda$. Let $\Phi_{+}(A)=\{f^{A}_{\sigma}:\sigma\in 2^{<\kappa}\}$.

Now assume $g$ is not eventually bounded by $\Phi_{+}(A)$. We need to show $\Phi_{-}(g)\not\subseteq A$. To this end we recursively construct an increasing sequence $(\tau_{\alpha}:\alpha<\kappa)$ such that $\tau_{\alpha}\in C_{g}^{\alpha}$ and $[\tau_{\alpha+1}]\cap A_{\alpha}=\emptyset$ for all $\alpha<\kappa$. Letting $x=\bigcup\tau_{\alpha}$, we see $x\in\Phi_{-}(g)\setminus A$.

Let $\tau_{0}=\langle\rangle$.

Assume $\tau_{\alpha}$ has been defined as required. Since $g$ is not eventually bounded by $f_{\tau_{\alpha}}^{A}$, there is $\gamma<\lambda$ such that $f_{\tau_{\alpha}}^{A}(\gamma)<g(\gamma)$. Thus we can find $\tau_{\alpha+1}\in C_{g}^{\alpha+1}$ such that $h(\tau_{\alpha}\,{}\hat{}\,\sigma_{\gamma})\subsetneq\tau_{\alpha+1}$. $[\tau_{\alpha+1}]\cap A_{\alpha}=\emptyset$ follows because $[h(\tau_{\alpha}\,{}\hat{}\,\sigma_{\gamma})]\cap A_{{\mathrm{lh}}(\tau_{\alpha}\,{}\hat{}\,\sigma_{\gamma})}=\emptyset$, $[\tau_{\alpha+1}]\subseteq[h(\tau_{\alpha}\,{}\hat{}\,\sigma_{\gamma})]$, and ${\mathrm{lh}}(\tau_{\alpha}\,{}\hat{}\,\sigma_{\gamma})>{\mathrm{lh}}(\tau_{\alpha})\geq\alpha$ and thus $A_{\alpha}\subseteq A_{{\mathrm{lh}}(\tau_{\alpha}\,{}\hat{}\,\sigma_{\gamma})}$.

If $\alpha$ is a limit ordinal, simply let $\tau_{\alpha}=\bigcup_{\beta<\alpha}\tau_{\beta}$. ∎

Corollary 2**.**

${\mathsf{add}}({\cal M}_{\kappa})\leq{\mathfrak{b}}_{\kappa}$* and ${\mathsf{cof}}({\cal M}_{\kappa})\geq{\mathfrak{d}}^{\lambda}_{\kappa}$.* 2. 2.

In particular, ${\mathsf{cof}}({\cal M}_{\kappa})\geq{\mathfrak{d}}_{\kappa}$.

Proof.

(2) follows from (1) and ${\mathfrak{d}}^{\lambda}_{\kappa}\geq{\mathfrak{d}}_{\kappa}$. Also, if $2^{<\kappa}>\kappa$, then ${\mathsf{add}}({\cal M}_{\kappa})=\kappa^{+}\leq{\mathfrak{b}}_{\kappa}$, so the first inequality of (1) holds trivially.

To see the first inequality of (1) in case $2^{<\kappa}=\kappa$, let ${\cal F}\subseteq\kappa^{\kappa}$ be an unbounded family. If $A\in{\cal M}_{\kappa}$, there is $g\in{\cal F}$ not eventually bounded by $\Phi_{+}(A)$ because $|\Phi_{+}(A)|=\kappa$ and thus $\Phi_{+}(A)$ is bounded. So $\Phi_{-}(g)\not\subseteq A$. Thus we see that the union of the $\Phi_{-}(g)$, $g\in{\cal F}$, does not belong to ${\cal M}_{\kappa}$, and ${\mathsf{add}}({\cal M}_{\kappa})\leq{\mathfrak{b}}_{\kappa}$ follows.

For the second inequality of (1), let ${\cal A}\subseteq{\cal M}_{\kappa}$, and assume $|{\cal A}|<{\mathfrak{d}}_{\kappa}^{\lambda}$. Let ${\cal F}=\bigcup\{\Phi_{+}(A):A\in{\cal A}\}$. Since $|\Phi_{+}(A)|\leq\lambda$ for all $A$ and ${\mathfrak{d}}_{\kappa}^{\lambda}>\lambda$, we also have $|{\cal F}|<{\mathfrak{d}}_{\kappa}^{\lambda}$. Hence there is $g\in\kappa^{\lambda}$, which is not eventually bounded by ${\cal F}$. But then $\Phi_{-}(g)\not\subseteq A$ for all $A\in{\cal A}$, and ${\cal A}$ is not cofinal in ${\cal M}_{\kappa}$. Thus ${\mathsf{cof}}({\cal M}_{\kappa})\geq{\mathfrak{d}}^{\lambda}_{\kappa}$ follows. ∎

Corollary 3**.**

${\mathsf{add}}({\cal M}_{\kappa})=\min\{{\mathfrak{b}}_{\kappa},{\mathsf{cov}}({\cal M}_{\kappa})\}$* and ${\mathsf{cof}}({\cal M}_{\kappa})\geq\max\{{\mathfrak{d}}_{\kappa}^{\lambda},{\mathsf{non}}({\cal M}_{\kappa})\}$222An earlier version of this paper claimed equality here, but the proof was flawed..* 2. 2.

If $2^{<\kappa}=\kappa$, then ${\mathsf{cof}}({\cal M}_{\kappa})=\max\{{\mathfrak{d}}_{\kappa},{\mathsf{non}}({\cal M}_{\kappa})\}$.

Proof.

Since ${\mathsf{add}}({\cal M}_{\kappa})\leq{\mathsf{cov}}({\cal M}_{\kappa})$ is trivial and ${\mathsf{add}}({\cal M}_{\kappa})\leq{\mathfrak{b}}_{\kappa}$ holds by Corollary 2 (1), the first part of (1) follows from [BBFM, Corollary 31] (see also item (ii) in the Introduction). Similarly, the second part of (2) is immediate by Corollary 2 (1) and the obvious ${\mathsf{cof}}({\cal M}_{\kappa})\geq{\mathsf{non}}({\cal M}_{\kappa})$.

For (2), use (1) and  [BBFM, Corollary 31] (see also item (iii) in the Introduction). ∎

Assume additionally $\lambda$ is regular. With an argument similar to the proof of Theorem 1 we obtain:

Proposition 4**.**

There are functions $\Phi_{-}:\lambda^{\lambda}\to{\cal{NWD}}_{\kappa}$ and $\Phi_{+}:{\cal M}_{\kappa}\to\lambda^{\lambda}$ such that if $A\in{\cal M}_{\kappa}$, $g\in\lambda^{\lambda}$, and $g$ is not eventually bounded by $\Phi_{+}(A)$, then $\Phi_{-}(g)\not\subseteq A$.

Proof.

Let $\Sigma$ be as in the proof of Theorem 1. Also let $\mathrm{T}=\{\tau_{\delta}:\delta<\lambda\}=2^{<\kappa}$. Assume this enumeration satisfies additionally

• •

$\sigma\,{}\hat{}\,0=\tau_{\delta}$ implies $\sigma\,{}\hat{}\,1=\tau_{\delta+1}$,

• •

$\tau_{\delta}\subset\tau_{\epsilon}$ implies $\delta<\epsilon$.

Given $g\in\lambda^{\lambda}$, we define $C^{\alpha}_{g}\subseteq 2^{<\kappa}$, $T_{g}\subseteq 2^{<\kappa}$, and $\Phi_{-}(g)=[T_{g}]$ as in this proof, except for the successor step where for given $\sigma\in C^{\alpha}_{g}$, we first let $\tau\in D^{\alpha+1}_{g}$ if for some $\gamma<\lambda$, $\tau=\sigma\,{}\hat{}\,\sigma_{\gamma}\,{}\hat{}\,\tau_{\delta}$ for $\delta<g(\gamma)$ and then define $C^{\alpha+1}_{g}$ as the set of all $\tau$ such that

• •

either $\tau\in D^{\alpha+1}_{g}$ and no proper extension of $\tau$ belongs to $D^{\alpha+1}_{g}$

• •

or $\zeta:={\mathrm{lh}}(\tau)$ is a limit ordinal, for all $\xi<\zeta$ with $\xi\geq{\mathrm{lh}}(\sigma)$, $\tau{\upharpoonright}\xi\in D^{\alpha+1}_{g}$ and $\tau\notin D^{\alpha+1}_{g}$.

It is then easy to see that $C^{\alpha+1}_{g}$ is an antichain with the required properties. Note in particular that for all $\tau\in D^{\alpha+1}_{g}$ there is $\tau^{\prime}\supseteq\tau$ with $\tau^{\prime}\in C^{\alpha+1}_{g}$.

(For suppose this fails for some $\tau$. Assume $\tau^{\prime}\supseteq\tau$ with $\tau^{\prime}\in D^{\alpha+1}_{g}$. Then $\tau^{\prime}$ is not a maximal node in $D^{\alpha+1}_{g}$ and therefore $\tau^{\prime}\,{}\hat{}\,0\in D^{\alpha+1}_{g}$. Since the latter is not maximal either, also $\tau^{\prime}\,{}\hat{}\,1\in D^{\alpha+1}_{g}$. Similarly if $\tau^{\prime}\supseteq\tau$ and ${\mathrm{lh}}(\tau^{\prime})$ is a limit such that $\tau^{\prime}{\upharpoonright}\xi\in D^{\alpha+1}_{g}$ for all $\xi$ with ${\mathrm{lh}}(\tau)\leq\xi<{\mathrm{lh}}(\tau^{\prime})$ then $\tau^{\prime}\in D^{\alpha+1}_{g}$, for otherwise it would belong $C^{\alpha+1}_{g}$. This means that the full binary tree below $\tau$ belongs to $D^{\alpha+1}_{g}$, a contradiction.)

Given $A=\bigcup_{\alpha<\kappa}A_{\alpha}\in{\cal M}_{\kappa}$ as in the proof of Theorem 1, $\alpha<\kappa$, and $\sigma\in 2^{\alpha}$, define $f_{\sigma}^{A}\in\lambda^{\lambda}$ such that $[\sigma\,{}\hat{}\,\sigma_{\gamma}\,{}\hat{}\,\tau_{f_{\sigma}^{A}(\gamma)}]\cap A_{\alpha}=\emptyset$. Let $\Phi_{+}(A)$ eventually dominate all $f_{\sigma}^{A}$.

If $g$ is not eventually bounded by $\Phi_{+}(A)$, we construct an increasing sequence $(\tau_{\alpha}:\alpha<\kappa)$ in $2^{<\kappa}$ with $\tau_{\alpha}\in C^{\alpha}_{g}$ and $[\tau_{\alpha+1}]\cap A_{\alpha}=\emptyset$ as in the proof of Theorem 1. ∎

As a consequence we immediately get (though this will follow from Corollary 2 if Question 14 has a positive answer):

Corollary 5**.**

${\mathsf{cof}}({\cal M}_{\kappa})\geq{\mathfrak{d}}_{\lambda}$.

A family ${\cal F}\subseteq\lambda^{\lambda}$ of functions is called almost disjoint if given any distinct $f,g\in{\cal F}$, the set $\{\gamma<\lambda:f(\gamma)=g(\gamma)\}$ has size less than $\lambda$. Yet another modification of the proof of Theorem 1 gives us an absolute result under certain assumptions.

Theorem 6**.**

Assume $\lambda=|2^{<\kappa}|\geq\kappa^{++}$. Let ${\cal F}\subseteq\lambda^{\lambda}$ be an almost disjoint family of functions. Then there is ${\cal A}\subseteq{\cal{NWD}}_{\kappa}$ of size $|{\cal F}|$ such that for all $B\in{\cal M}_{\kappa}$, the set $\{A\in{\cal A}:A\subseteq B\}$ has size at most $\kappa$. In particular, if there is an almost disjoint family of functions of size $2^{\lambda}$, then ${\mathsf{cof}}({\cal M}_{\kappa})=2^{\lambda}$.

Proof.

Let $\Sigma=\{\sigma_{\gamma}:\gamma<\lambda\}\subseteq 2^{<\kappa}$ be a maximal antichain in $2^{<\kappa}$. By $\kappa<2^{<\kappa}$, it is easy to see that we may additionally assume that $\Sigma$ forms a front, that is, for each $x\in 2^{\kappa}$ there is a (necessarily unique) $\sigma\in\Sigma$ with $\sigma\subseteq x$333This may be false if $\kappa=2^{<\kappa}$. For example, if $\kappa$ is weakly compact, it is easy to see there is no front of size $\kappa$. This was pointed out to us by Tristan van der Vlugt..

Recursively define fronts $C^{\alpha}$, $\alpha<\kappa$, such that

• (i)

$C^{0}=\{\langle\rangle\}$,

• (ii)

$C^{\alpha+1}=\{\sigma\,{}\hat{}\,\sigma_{\gamma}:\sigma\in C^{\alpha}$ and $\gamma<\lambda\}$, and

• (iii)

$C^{\alpha}=\{\sigma\in 2^{<\kappa}:\exists(\tau_{\beta}:\beta<\alpha)$ strictly increasing such that $\sigma=\bigcup_{\beta<\alpha}\tau_{\beta}$ and $\tau_{\beta}\in C^{\beta}\}$ for limit ordinals $\alpha$.

Construe $f\in{\cal F}$ as a function from $2^{<\kappa}$ to $\lambda$. Define a nowhere dense tree $T_{f}\subseteq 2^{<\kappa}$ by recursion on $\alpha<\kappa$ by producing sets $C^{\alpha}_{f}\subseteq C^{\alpha}$ such that $C^{0}_{f}=C^{0}=\{\langle\rangle\}$, $C^{\alpha}_{f}$ is obtained from the previous $C^{\beta}_{f}$ as in (iii) for limit ordinals $\alpha$, and

• (ii’)

$C^{\alpha+1}_{f}=\{\sigma\,{}\hat{}\,\sigma_{\gamma}:\sigma\in C^{\alpha}_{f}$ and $f(\sigma)\neq\gamma\}$.

Let $T_{f}$ be the downward closure of $\bigcup_{\alpha<\kappa}C^{\alpha}_{f}$ and put $A_{f}=[T_{f}]$. Since for $\sigma\in C^{\alpha}_{f}$, $\sigma\,{}\hat{}\,\sigma_{f(\sigma)}$ is incompatible with all elements of $C^{\alpha+1}_{f}$, we see that $[\sigma\,{}\hat{}\,\sigma_{f(\sigma)}]\cap[T_{f}]=\emptyset$, and $A_{f}$ is indeed nowhere dense. Put ${\cal A}=\{A_{f}:f\in{\cal F}\}$.

Assume $B=\bigcup_{\alpha<\kappa}B_{\alpha}$ is a $\kappa$-meager set, where the $B_{\alpha}$ form an increasing sequence of nowhere dense sets. We claim that $\{f\in{\cal F}:A_{f}\subseteq B\}$ has size at most $\kappa$.

Indeed, assume that $\{f_{\zeta}:\zeta<\kappa^{+}\}\subseteq{\cal F}$. We need to find $\zeta<\kappa^{+}$ and $x\in A_{f_{\zeta}}\setminus B$. By almost disjointness and $\lambda>\kappa^{+}$, we can find $\sigma\in\bigcap_{\zeta<\kappa^{+}}C^{1}_{f_{\zeta}}$ such that for all $\alpha\geq 1$, all $\sigma^{\prime}\in C^{\alpha}$ with $\sigma\subseteq\sigma^{\prime}$ and all distinct $\zeta,\zeta^{\prime}<\kappa^{+}$, we have $f_{\zeta}(\sigma^{\prime})\neq f_{\zeta^{\prime}}(\sigma^{\prime})$. Recursively build $\tau_{\alpha}\in C^{\alpha}$, $\alpha<\kappa$, such that

$\tau_{0}=\langle\rangle$, $\tau_{1}=\sigma$, 2. 2.

$\tau_{\alpha}=\bigcup_{\beta<\alpha}\tau_{\beta}$ for limit ordinals $\alpha$, and 3. 3.

if $\tau_{\alpha}$ has been constructed for $\alpha\geq 1$, find $\tau^{*}\supseteq\tau_{\alpha}$ such that $[\tau^{*}]\cap B_{\alpha}=\emptyset$, let $\beta>\alpha$ and $\tau_{\beta}\supseteq\tau^{*}$ be such that $\tau_{\beta}\in C^{\beta}$, and define $\tau_{\gamma}\in C^{\gamma}$ for $\alpha<\gamma<\beta$ such that $\tau_{\alpha}\subseteq\tau_{\gamma}\subseteq\tau_{\beta}$.

Put $x=\bigcup_{\alpha<\kappa}\tau_{\alpha}$. Clearly $x\notin B$.

Define a function $f$ on all $\tau_{\alpha}$, $\alpha<\kappa$, as follows.

$f(\tau_{0})=f(\langle\rangle)$ is the unique $\gamma$ such that $\tau_{1}=\sigma=\sigma_{\gamma}$ and 2. 2.

$f(\tau_{\alpha})$ is the unique $\gamma$ such that $\tau_{\alpha}\,{}\hat{}\,\sigma_{\gamma}=\tau_{\alpha+1}$ for $\alpha\geq 1$.

Now notice that by disjointness of the functions $f_{\zeta}$, there is $\zeta<\kappa^{+}$ such that for all $\alpha\geq 1$, we have $f_{\zeta}(\tau_{\alpha})\neq f(\tau_{\alpha})$. This means, however, that for all $\alpha$, $\tau_{\alpha}$ belongs to $C^{\alpha}_{f_{\zeta}}$: indeed $\tau_{1}=\sigma\in C^{1}_{f_{\zeta}}$, $\tau_{\alpha}\in C^{\alpha}_{f_{\zeta}}$ implies $\tau_{\alpha+1}=\tau_{\alpha}\,{}\hat{}\,\sigma_{f(\tau_{\alpha})}\in C^{\alpha+1}_{f_{\zeta}}$ because $f_{\zeta}(\tau_{\alpha})\neq f(\tau_{\alpha})$, and $\tau_{\alpha}\in C^{\alpha}_{f_{\zeta}}$ for limit $\alpha$ follows trivially. Therefore $x\in[T_{f_{\zeta}}]=A_{f_{\zeta}}$, as required, and the proof is complete. ∎

By an old result of Baumgartner, almost disjoint families of functions of size $2^{\lambda}$ may not exist. Indeed, if $\kappa<\lambda$ are regular and GCH holds, and we first add at least $\lambda^{++}$ Cohen subsets of $\kappa$ and then add $\lambda$ Cohen reals, then $\lambda=|2^{<\kappa}|$, $2^{\lambda}\geq\lambda^{++}$ and there is no almost disjoint family of functions in $\lambda^{\lambda}$ of size $\lambda^{++}$, by [Ba, Theorem 5.4 (a)].

3 Models

Models for ${\mathsf{non}}({\cal M}_{\kappa})$. We know [BBFM, Observation 23 (iii)] that $2^{<\kappa}\leq{\mathsf{non}}({\cal M}_{\kappa})\leq 2^{\kappa}$. We shall now see that this is all we can say, even if $2^{<\kappa}>\kappa$.

In the model obtained by adding $\kappa^{+}$ Cohen reals over a model of GCH, we have $\kappa<2^{<\kappa}={\mathsf{non}}({\cal M}_{\kappa})=2^{\kappa}$.

For a model with $\kappa<2^{<\kappa}<{\mathsf{non}}({\cal M}_{\kappa})=2^{\kappa}$, simply add $\kappa^{++}$ many $\kappa$-Hechler functions (see [BBFM, Subsection 4.2]) followed by $\kappa^{+}$ many Cohen reals to a model of GCH. In the intermediate model, ${\mathfrak{b}}_{\kappa}=\kappa^{++}$. Since Cohen forcing is $\kappa^{\kappa}$-bounding, it does not change the value of ${\mathfrak{b}}_{\kappa}$. Also ${\mathfrak{b}}_{\kappa}\leq{\mathsf{non}}({\cal M}_{\kappa})$ in ZFC. Therefore the final model satisfies ${\mathsf{non}}({\cal M}_{\kappa})=2^{\kappa}=\kappa^{++}$.

Theorem 7**.**

It is consistent that $\kappa<2^{<\kappa}={\mathsf{non}}({\cal M}_{\kappa})<2^{\kappa}$.

Proof.

Assume GCH in the ground model $V$. Add first $\kappa^{++}$ many Cohen subsets of $\kappa$ to obtain the model $V[G]$. Denote the forcing by ${\mathbb{C}}_{\kappa}^{\kappa^{++}}$ and, more generally, for $A\subseteq\kappa^{++}$, use ${\mathbb{C}}_{\kappa}^{A}$ for the forcing adding the Cohen sets with index in $A$. Next add $\kappa^{+}$ many Cohen reals to obtain the model $V[G][H]$. Work in the model $V[H]$. The forcing $({\mathbb{C}}_{\kappa}^{\kappa^{++}})^{V}$ is still $<\kappa$-distributive in this model (though not $<\kappa$-closed anymore) and $\kappa^{+}$-cc; in particular, it does not add new sequences of length $<\kappa$. In $V[H]$, $2^{<\kappa}=2^{\kappa}=\kappa^{+}$, and we claim that $2^{\kappa}\cap V[H]$ is a witness for ${\mathsf{non}}({\cal M}_{\kappa})$ in $V[G][H]$.

Let $(\dot{A}_{\alpha}:\alpha<\kappa)$ be a $({\mathbb{C}}_{\kappa}^{\kappa^{++}})^{V}$-name for an increasing sequence of nowhere dense sets. Thus there is a name $\dot{f}$ for a function from $2^{<\kappa}$ to $2^{<\kappa}$ such that for $\alpha<\kappa$ and $\sigma\in 2^{\alpha}$, the trivial condition forces $\sigma\subseteq\dot{f}(\sigma)$ and $[\dot{f}(\sigma)]\cap\dot{A}_{\alpha}=\emptyset$ Without loss, ${\mathrm{lh}}(\dot{f}(\sigma))$ is forced to be at least $\alpha+1$. Let $p\in({\mathbb{C}}_{\kappa}^{\kappa^{++}})^{V}$. Recursively produce sets $C_{\alpha}\subseteq\kappa^{++}$ with $C_{\alpha}\in V$, conditions $p_{\alpha},q_{\alpha}\in({\mathbb{C}}_{\kappa}^{\kappa^{++}})^{V}$, and sequences $\tau_{\alpha}\in 2^{<\kappa}$, $\alpha<\kappa$, such that

• •

$|C_{\alpha}|=\kappa$, and the $C_{\alpha}$ are an increasing chain,

• •

$p_{\alpha}\leq p,q_{\alpha}\leq p_{\alpha}$,

• •

$p_{\alpha}\in({\mathbb{C}}_{\kappa}^{C_{\alpha}})^{V},q_{\alpha}\in({\mathbb{C}}_{\kappa}^{C_{\alpha+1}})^{V}$, and for every $q\in({\mathbb{C}}_{\kappa}^{C_{\alpha}})^{V}$ there is a $\beta\geq\alpha$ such that $q=p_{\beta}$,

• •

$\tau_{\alpha}\in 2^{\geq\alpha}$ and $\tau_{\alpha}\subseteq\tau_{\beta}$ for $\alpha\leq\beta$,

• •

$q_{\alpha}\Vdash[\tau_{\alpha+1}]\cap\dot{A}_{\alpha}=\emptyset$.

Note that for $C\subseteq\kappa^{++}$ of size $\kappa$, $|({\mathbb{C}}_{\kappa}^{C})^{V}|=(2^{<\kappa})^{V}=\kappa$, and thus the second clause of the third item can easily be achieved by a book-keeping argument.

Let $C_{0}\subseteq\kappa^{++}$ be such that $C_{0}\in V$ and $p\in({\mathbb{C}}_{\kappa}^{C_{0}})^{V}$. Put $\tau_{0}=\langle\rangle$. Let $p_{0}\leq p$ be the condition in $({\mathbb{C}}_{\kappa}^{C_{0}})^{V}$ handed down by the book-keeping, and find $q_{0}\leq p_{0}$ and $\tau_{1}$ such that $q_{0}$ forces $\tau_{1}=\dot{f}(\tau_{0})$ (note that this is possible because no new $<\kappa$-sequences are added). In particular, $q_{0}\Vdash[\tau_{1}]\cap\dot{A}_{0}=\emptyset$. Let $C_{1}\supseteq C_{0}$ be such that $C_{1}\in V$ and $q_{0}\in({\mathbb{C}}_{\kappa}^{C_{1}})^{V}$.

Assume we are at stage $\alpha$, and everything has been constructed for $\beta<\alpha$. In case $\alpha$ is successor, we also assume $C_{\alpha}$ and $\tau_{\alpha}$ have been produced, and if $\alpha$ is limit we let $C_{\alpha}\supseteq\bigcup_{\beta<\alpha}C_{\beta}$ with $C_{\alpha}\in V$ and $\tau_{\alpha}=\bigcup_{\beta<\alpha}\tau_{\beta}$. Let $p_{\alpha}\leq p$ be the condition in $({\mathbb{C}}_{\kappa}^{C_{\alpha}})^{V}$ given by the book-keeping, and proceed as in the basic step to get $q_{\alpha}\leq p_{\alpha}$, $\tau_{\alpha+1}\supseteq\tau_{\alpha}$, and $C_{\alpha+1}\supseteq C_{\alpha}$ such that $C_{\alpha+1}\in V$ and $q_{\alpha}\in({\mathbb{C}}_{\kappa}^{C_{\alpha+1}})^{V}$ forces $\tau_{\alpha+1}=\dot{f}(\tau_{\alpha})$ and thus $[\tau_{\alpha+1}]\cap\dot{A}_{\alpha}=\emptyset$. This completes the recursive construction.

Now let $C=\bigcup_{\alpha<\kappa}C_{\alpha}$ and $x=\bigcup_{\alpha<\kappa}\tau_{\alpha}\in 2^{\kappa}$. We claim that $p$ forces that $x\notin\dot{A}$ where $\dot{A}=\bigcup_{\alpha<\kappa}A_{\alpha}$. Let $\beta<\kappa$. Take any $q^{\prime}\leq p$ in $({\mathbb{C}}_{\kappa}^{\kappa^{++}})^{V}$ and let $q=q^{\prime}{\upharpoonright}C$. Thus $q=p_{\alpha}$ for some $\alpha\geq\beta$ (by the book-keeping). By construction, $q_{\alpha}\leq p_{\alpha}$ forces that $[\tau_{\alpha}]\cap\dot{A}_{\alpha}=\emptyset$. Clearly $q_{\alpha}$ and $q^{\prime}$ are compatible. Let $q^{\prime\prime}=q_{\alpha}\cup q^{\prime}$ be the smallest common extension. Then $q^{\prime\prime}$ forces $x\notin\dot{A}_{\beta}$. Since this holds for any $q^{\prime}\leq p$ and any $\beta$, $p$ forces $x\notin\dot{A}$.

Hence $2^{\kappa}\cap V[H]$ is indeed non-meager in $V[G][H]$. ∎

The proof of the following result is somewhat more complicated.

Theorem 8**.**

It is consistent that $\kappa<2^{<\kappa}<{\mathsf{non}}({\cal M}_{\kappa})<2^{\kappa}$.

Proof.

Again assume GCH. First add $\kappa^{+++}$ Cohen subsets of $\kappa$. Then perform a $\kappa^{++}$-stage iteration of $\kappa$-Hechler forcing (see [BBFM, Subsection 4.2]). In the resulting model $V[G_{0}][G_{1}]$, $2^{<\kappa}=\kappa<{\mathfrak{b}}_{\kappa}={\mathsf{non}}({\cal M}_{\kappa})=\kappa^{++}<2^{\kappa}=\kappa^{+++}$. Next add $\kappa^{+}$ Cohen reals to make $2^{<\kappa}=\kappa^{+}$ in the final model $V[G_{0}][G_{1}][H]$. Again it is clear this model will satisfy ${\mathfrak{b}}_{\kappa}=\kappa^{++}$, so that ${\mathsf{non}}({\cal M}_{\kappa})\geq\kappa^{++}$, and it suffices to show ${\mathsf{non}}({\cal M}_{\kappa})\leq\kappa^{++}$. In $V[H]$ the remainder forcing is $<\kappa$-distributive and $\kappa^{+}$-cc.

It is well-known (see e.g. [BBFM, Subsection 4.2]) that $\kappa$-Hechler forcing also adjoins a $\kappa$-Cohen function. In $V[G_{0}][G_{1}]$ let $(c_{\alpha}\in\kappa^{\kappa}:\alpha<\kappa^{++})$ be the Cohen functions decoded from the $\kappa^{++}$ many Hechler functions added in the iteration. We will use the $c_{\alpha}$ to code $c_{\alpha}^{\gamma}\in 2^{\kappa}$, $\kappa\leq\gamma<\kappa^{+}$, in such a way that the set $C=\{c_{\alpha}^{\gamma}:\alpha<\kappa^{++},\kappa\leq\gamma<\kappa^{+}\}$ is non-meager in $V[G_{0}][G_{1}][H]$.

More explicitly, let $f:\kappa^{+}\to 2^{<\kappa}$ be a bijection with $f\in V[H]$. By distributivity of the remainder forcing, $f$ is still a bijection in $V[G_{0}][G_{1}][H]$. Assume $f(0)=\langle\rangle$. Next, for each $\gamma$ with $\kappa\leq\gamma<\kappa^{+}$, let $g_{\gamma}:\kappa\to\gamma$ be a bijection with $g_{\gamma}\in V$ and $g_{\gamma}(0)=0$. Define $c_{\alpha}^{\gamma}\in V[G_{0}][G_{1}][H]$ to be the concatenation of the $f(g_{\gamma}(c_{\alpha}(\zeta)))$ where $\zeta<\kappa$, i.e.

[TABLE]

Note that if $c_{\alpha}(\zeta)\neq 0$ then $f(g_{\gamma}(c_{\alpha}(\zeta)))\neq\langle\rangle$ so that by genericity $\kappa$ many of the $f(g_{\gamma}(c_{\alpha}(\zeta)))$ are non-empty sequences and $c_{\alpha}^{\gamma}$ is indeed an element of $2^{\kappa}$.

To see that $C$ is non-meager, let $A$ be a meager set in $V[G_{0}][G_{1}][H]$, say $A=\bigcup_{\zeta<\kappa}A_{\zeta}$ where the $A_{\zeta}$ form an increasing sequence of nowhere dense sets. Thus there is $h:2^{<\kappa}\to 2^{<\kappa}$ such that for all $\zeta<\kappa$ and all $\sigma\in 2^{\zeta}$, $[\sigma\,{}\hat{}\,h(\sigma)]\cap A_{\zeta}=\emptyset$. By the $\kappa^{+}$-cc of the remainder forcing, there is $\alpha<\kappa^{++}$ such that $h\in V[H][G_{0}][G_{1}^{\alpha}]$, where $G_{1}^{\alpha}$ is the generic for the $\alpha$ first stages of the iteration of $\kappa$-Hechler forcing. That is, $c_{\alpha}$ is still ${\mathbb{C}}_{\kappa}^{V}$-generic over $V_{\alpha}:=V[H][G_{0}][G_{1}^{\alpha}]$. Work in $V_{\alpha}$. We shall show that for some $\gamma<\kappa^{+}$, $\dot{c}_{\alpha}^{\gamma}$ is forced to be outside $A$.

Let $\gamma<\kappa^{+}$ with $\gamma\geq\kappa$. For $\tau\in\gamma^{<\kappa}$ let $\bar{f}(\tau)$ be the concatenation of the $f(\tau(\zeta))$, $\zeta<{\mathrm{lh}}(\tau)$, i.e.

[TABLE]

Say that $\gamma$ is $h$-good if

• •

$\delta<\gamma$ implies that $f^{-1}(h(f(\delta)))<\gamma$,

• •

$\tau\in(\gamma^{<\kappa}\cap V)$ implies that $f^{-1}(\bar{f}(\tau))<\gamma$.

Claim 9**.**

There are $h$-good $\gamma$.

Proof.

This is a standard closure argument. Let $\gamma_{0}$ be arbitrary with $\kappa\leq\gamma_{0}<\kappa^{+}$. Recursively construct an increasing continuous sequence $(\gamma_{\zeta}:\zeta\leq\kappa)$ of ordinals below $\kappa^{+}$ such that

• •

if $\delta<\gamma_{\zeta}$ then $f^{-1}(h(f(\delta)))<\gamma_{\zeta+1}$,

• •

if $\tau\in(\gamma_{\zeta}^{<\kappa}\cap V)$ then $f^{-1}(\bar{f}(\tau))<\gamma_{\zeta+1}$.

Since $|\gamma_{\zeta}^{<\kappa}\cap V|=\kappa$, this is possible. Clearly $\gamma_{\kappa}$ is $h$-good. ∎

Claim 10**.**

If $\gamma$ is $h$-good, then $\dot{c}_{\alpha}^{\gamma}$ is forced to be outside $A$.

Proof.

Let $\zeta<\kappa$ and let $\upsilon\in{\mathbb{C}}_{\kappa}^{V}=\kappa^{<\kappa}\cap V$ be a $\kappa$-Cohen condition. We need to find $\upsilon^{\prime}\leq\upsilon$ such that $\upsilon^{\prime}\Vdash\dot{c}_{\alpha}^{\gamma}\notin A_{\zeta}$. Let $\tau\in\gamma^{<\kappa}\cap V$ be the image of $\upsilon$ under $g_{\gamma}$, that is, ${\mathrm{lh}}(\tau)={\mathrm{lh}}(\upsilon)$ and $\tau(\zeta)=g_{\gamma}(\upsilon(\zeta))$ for all $\zeta<{\mathrm{lh}}(\upsilon)$. Let $\sigma:=\bar{f}(\tau)$. Without loss of generality we may assume that $\eta:={\mathrm{lh}}(\sigma)\geq\zeta$; otherwise extend the condition $\upsilon$. Clearly $\upsilon\Vdash\dot{c}_{\alpha}^{\gamma}\in[\sigma]$. Since $\gamma$ is $h$-good, we know that $\delta:=f^{-1}(\sigma)<\gamma$. Note $f(\delta)=\sigma$. We also have that $\epsilon:=f^{-1}(h(\sigma))<\gamma$. Note $f(\epsilon)=h(\sigma)$. By definition of $h$, we obtain $[\sigma\,{}\hat{}\,h(\sigma)]\cap A_{\eta}=\emptyset$ and therefore also $[\sigma\,{}\hat{}\,h(\sigma)]\cap A_{\zeta}=\emptyset$. Let $\tau^{\prime}=\tau\,{}\hat{}\,\epsilon$ and $\upsilon^{\prime}=\upsilon\,{}\hat{}\,g_{\gamma}^{-1}(\epsilon)$. Thus $\bar{f}(\tau^{\prime})=\bar{f}(\tau)\,{}\hat{}\,f(\epsilon)=\sigma\,{}\hat{}\,h(\sigma)$ and $\upsilon^{\prime}\Vdash\dot{c}_{\alpha}^{\gamma}\in[\sigma\,{}\hat{}\,h(\sigma)]$. In particular $\upsilon^{\prime}\Vdash\dot{c}_{\alpha}^{\gamma}\notin A_{\zeta}$. ∎

This completes the proof of the theorem. ∎

Models for ${\mathsf{cof}}({\cal M}_{\kappa})$. As before let $\lambda=|2^{<\kappa}|$. By the results of Section 2 we know in particular that $\lambda<{\mathsf{cof}}({\cal M}_{\kappa})\leq 2^{\lambda}$. We are interested in models with $\kappa<\lambda$.

If we add $\kappa^{+}$ Cohen reals over a model of GCH, we obtain a model of $\kappa<\lambda=2^{\kappa}=\kappa^{+}<{\mathsf{cof}}({\cal M}_{\kappa})=2^{\lambda}=\kappa^{++}$.

For the consistency of $\kappa<\lambda<{\mathsf{cof}}({\cal M}_{\kappa})=2^{\kappa}$, use the model of Theorem 7.

To obtain a model of $\kappa<\lambda<2^{\kappa}<{\mathsf{cof}}({\cal M}_{\kappa})$, first add $\kappa^{+++}$ Cohen subsets of $\kappa^{+}$, then $\kappa^{++}$ Cohen subsets of $\kappa$, and finally $\kappa^{+}$ Cohen reals over a model for GCH. For $\lambda=\kappa^{+}$, ${\mathfrak{d}}_{\lambda}=\kappa^{+++}$ in the first extension, and this is preserved. Therefore the final model satisfies ${\mathfrak{d}}_{\lambda}={\mathsf{cof}}({\cal M}_{\kappa})=2^{\lambda}=\kappa^{+++}$, by Corollary 5.

Question 11**.**

Is $\kappa<2^{<\kappa}$ together with ${\mathsf{cof}}({\cal M}_{\kappa})<2^{2^{<\kappa}}$ consistent?

In view of the results Section 2, this is related to questions about the dominating numbers in the next section.

4 Dominating numbers

For this section, let $\kappa$ be regular (not necessarily uncountable) and $\lambda\geq\kappa$ arbitrary. Let us first see that in the definition of ${\mathfrak{d}}^{\lambda}_{\kappa}$ it does not matter whether we use everywhere domination or domination modulo $<\kappa$ or $<\lambda$ (in case ${\mathrm{cf}}(\lambda)\geq\kappa$). For $f,g\in\kappa^{\lambda}$, say $f\leq_{\lambda}g$ if there is $\delta<\lambda$ such that for all $\gamma\geq\delta$, $f(\gamma)\leq g(\gamma)$. Let ${\mathfrak{d}}^{\lambda}_{\kappa}(\leq)$ be the dominating number of $\kappa^{\lambda}$ with the everywhere domination ordering, and let ${\mathfrak{d}}_{\kappa}^{\lambda}(\leq_{\lambda})$ be the dominating number of $\kappa^{\lambda}$ with the ordering $\leq_{\lambda}$.

Proposition 12**.**

${\mathfrak{d}}_{\kappa}^{\lambda}(\leq)={\mathfrak{d}}_{\kappa}^{\lambda}(\leq_{\lambda})$. In particular, if ${\mathrm{cf}}(\lambda)\geq\kappa$, ${\mathfrak{d}}_{\kappa}^{\lambda}={\mathfrak{d}}_{\kappa}^{\lambda}(\leq)={\mathfrak{d}}_{\kappa}^{\lambda}(\leq_{\lambda})$.

Proof.

The second statement follows from the first because ${\mathrm{cf}}(\lambda)\geq\kappa$ obviously implies ${\mathfrak{d}}_{\kappa}^{\lambda}(\leq_{\lambda})\leq{\mathfrak{d}}_{\kappa}^{\lambda}\leq{\mathfrak{d}}_{\kappa}^{\lambda}(\leq)$. So it suffices to see that ${\mathfrak{d}}_{\kappa}^{\lambda}(\leq)\leq{\mathfrak{d}}_{\kappa}^{\lambda}(\leq_{\lambda})$.

Take ${\cal F}\subseteq\kappa^{\lambda}$ dominating in $(\kappa^{\lambda},\leq_{\lambda})$. For $f,g\in{\cal F}$ and $\alpha,\beta<\lambda$ define the function $h_{f,g,\alpha,\beta}\in\kappa^{\lambda}$ by

[TABLE]

Since $|\{h_{f,g,\alpha,\beta}:f,g\in{\cal F},\alpha,\beta<\lambda\}|=|{\cal F}|$, it suffices to show that this family is dominating everywhere. To this end let $h\in\kappa^{\lambda}$ be arbitrary. There are $f\in{\cal F}$ and $\beta<\lambda$ such that $f(\gamma)\geq h(\gamma)$ for all $\gamma\geq\beta$. Now partition $\lambda$ into intervals $I_{\zeta}$, $\zeta<\lambda$, so that each $I_{\zeta}$ has length exactly $\beta$. Let $i_{\zeta}=\min I_{\zeta}$ for all $\zeta$. Define $h^{\prime}\in\kappa^{\lambda}$ by

[TABLE]

for all $\zeta<\lambda$ and all $\xi<\beta$. There are $g\in{\cal F}$ and $\eta<\lambda$ such that $g(\delta)\geq h^{\prime}(\delta)$ for all $\delta\geq i_{\eta}$. Then, for $\xi<\beta$,

[TABLE]

and we see that $h_{f,g,i_{\eta},\beta}$ dominates $h$ everywhere. ∎

Since ${\mathrm{cf}}(2^{<\kappa})\geq\kappa$ the second statement is true when $\lambda=|2^{<\kappa}|$ as in Section 2.

Theorem 13**.**

Let $\kappa$ be regular and $\lambda>\kappa$. Then ${\mathfrak{d}}^{\lambda}_{\kappa}\geq{\mathfrak{d}}^{\lambda}_{\kappa^{+}}$.

Proof.

We do the proof using Tukey connections. First define $\Phi_{-}(f):\kappa^{\lambda}\to(\kappa^{+})^{\lambda}$ by

[TABLE]

for $f\in\kappa^{\lambda}$ and $\alpha<\lambda$. Here $\exists^{\kappa}\beta$ denotes the quantifier “there are $\kappa$ many $\beta$”. This is clearly well-defined by the pigeonhole principle.

For defining $\Phi_{+}:(\kappa^{+})^{\lambda}\to\kappa^{\lambda}$, fix $h\in(\kappa^{+})^{\lambda}$. Assume without loss that $h(\alpha)>0$ for all $\alpha$. Let $(\alpha_{\xi}:\xi<\lambda)$ be the strictly increasing club sequence in $\lambda$ recursively defined by $\alpha_{0}=0$, $\alpha_{\xi+1}=\alpha_{\xi}+h(\alpha_{\xi})$, and $\alpha_{\eta}=\bigcup_{\xi<\eta}\alpha_{\xi}$ for limit ordinals $\eta$. Notice that for $\eta<\lambda$ we always have $\alpha_{\eta}<\lambda$ (even for singular $\lambda$) because the sequence increases only by ordinals of size at most $\kappa$. Replacing $h$ recursively by a larger function, if necessary, we may also assume that for every $\xi<\lambda$ and every $\alpha\in[\alpha_{\xi},\alpha_{\xi+1})$, $h(\alpha_{\xi+1})\geq h(\alpha)$. Now define $\Phi_{+}(h)$ such that for every $\xi<\lambda$, $\Phi_{+}(h){\upharpoonright}[\alpha_{\xi},\alpha_{\xi+1})$ is a one-to-one function into $\kappa$.

We first claim that if $h\not\leq^{*}\Phi_{-}(f)$ then $\Phi_{+}(h)\not\leq^{*}f$.

To see this let $\alpha<\lambda$ be such that $\Phi_{-}(f)(\alpha)<h(\alpha)$. There are at least $\kappa^{+}$ many such $\alpha$. Let $\xi<\lambda$ be such that $\alpha\in[\alpha_{\xi},\alpha_{\xi+1})$. By choice of $h$ we have $\Phi_{-}(f)(\alpha)<h(\alpha)\leq h(\alpha_{\xi+1})$ so that also

[TABLE]

In particular there is $\zeta<\kappa$ such that $f$ assumes value $\zeta$ exactly $\kappa$ many times in the interval $[\alpha_{\xi},\alpha_{\xi+2})$. Since $\Phi_{+}(h)$ is one-to-one on both intervals $[\alpha_{\xi},\alpha_{\xi+1})$ and $[\alpha_{\xi+1},\alpha_{\xi+2})$, it follows that there are $\kappa$ many places where $\Phi_{+}(h)$ is above $f$. Thus $\Phi_{+}(h)\not\leq^{*}f$.

As a consequence, it now readily follows that if ${\cal F}\subseteq\kappa^{\lambda}$ is dominating then so is $\{\Phi_{-}(f):f\in{\cal F}\}$ in $(\kappa^{+})^{\lambda}$. ∎

Question 14**.**

Let $\kappa$ and $\mu$ be regular and $\kappa<\mu\leq\lambda$. Does ${\mathfrak{d}}^{\lambda}_{\kappa}\geq{\mathfrak{d}}^{\lambda}_{\mu}$ hold? In particular, is ${\mathfrak{d}}^{\lambda}_{\kappa}\geq{\mathfrak{d}}_{\lambda}$ for regular $\lambda$?

The inequality in Theorem 13 is consistently strict.

Observation 15**.**

Assume GCH. Let $\kappa_{0}<\kappa_{1}<...<\kappa_{n}$ be regular cardinals, let $\lambda\geq\kappa_{n}$, and let $\mu_{n}<...<\mu_{1}<\mu_{0}$ be cardinals with $\mu_{n}>\lambda$ and ${\mathrm{cf}}(\mu_{i})>\lambda$. Then there is a forcing extension with ${\mathfrak{d}}_{\kappa_{i}}^{\lambda}=\mu_{i}$.

Proof.

Start by adding $\mu_{n}$ many Cohen subsets of $\kappa_{n}$. By backwards recursion add $\mu_{i}$ many Cohen subsets of $\kappa_{i}$. Finish by adding $\mu_{0}$ many Cohen subsets of $\kappa_{0}$. The first forcing forces $2^{\kappa_{n}}=2^{\lambda}={\mathfrak{d}}_{\kappa_{n}}^{\lambda}=\mu_{n}$. Since the remainder forcing is $\kappa_{n-1}^{+}$-cc and thus $\kappa_{n}^{\kappa_{n}}$-bounding, it preserves the value of ${\mathfrak{d}}^{\lambda}_{\kappa_{n}}$. Iterating this argument we see that $2^{\kappa_{i}}=2^{\lambda}=\mu_{0}$ and ${\mathfrak{d}}_{\kappa_{i}}^{\lambda}=\mu_{i}$ in the final model. ∎

In particular ${\mathfrak{d}}^{\lambda}_{\kappa}<2^{\lambda}$ is consistent for $\kappa<\lambda$, but in the model provided by the observation $2^{<\kappa}>{\mathfrak{d}}^{\lambda}_{\kappa}>\lambda$ holds. On the other hand, for an affirmative answer to Question 11 we would need a model with $\lambda:=2^{<\kappa}>\kappa$ and ${\mathfrak{d}}^{\lambda}_{\kappa}<2^{\lambda}$. Let us formulate this question somewhat more generally:

Question 16**.**

Assume $\kappa$ is regular and $\lambda>\kappa$ with $\lambda\geq 2^{<\kappa}$. Is ${\mathfrak{d}}^{\lambda}_{\kappa}<2^{\lambda}$ consistent?

For $\kappa=\omega$ and $\lambda=\omega_{1}$ this is a famous old question of Jech and Prikry [JP] (see also [Mi, Problem 8.1]):

Question 17** (Jech, Prikry).**

Is ${\mathfrak{d}}^{\omega_{1}}_{\omega}<2^{\omega_{1}}$ consistent?

The dominating numbers ${\mathfrak{d}}^{\lambda}_{\kappa}$ have since been used in [BS].