Base matrices of various heights

Abstract

A classical theorem of Balcar, Pelant, and Simon says that there is a base matrix of height ${\mathfrak h}$, where ${\mathfrak h}$ is the distributivity number of ${\cal P} (\omega ) / {\mathrm {fin}}$. We show that if the continuum ${\mathfrak c}$ is regular, then there is a base matrix of height ${\mathfrak c}$, and that there are base matrices of any regular uncountable height $\leq {\mathfrak c}$ in the Cohen and random models. This answers questions of Fischer, Koelbing, and Wohofsky.

1 Introduction

A collection ${\mathfrak A} = \{ {\cal A}_\gamma : \gamma < \vartheta \}$ of mad (maximal almost disjoint) families of subsets of the natural numbers $\omega $ is called a refining matrix of height $\vartheta $ if:

• • ${\cal A}_\delta $ refines ${\cal A}_\gamma $ for $\delta \geq \gamma $ , i.e., for all $A \in {\cal A}_\delta $ , there is $B \in {\cal A}_\gamma $ with $A \subseteq ^* B$ , and

• • there is no common refinement of the ${\cal A}_\gamma $ , i.e., no mad family ${\cal A}$ refining all the ${\cal A}_\gamma $ .

${\mathfrak A}$ is a base matrix if it is a refining matrix and $\bigcup _{\gamma < \vartheta } {\cal A}_\gamma $ is dense in ${\cal P} (\omega ) / {\mathrm {fin}}$ , i.e., for all $B \in [\omega ]^\omega $ , there are $\gamma < \vartheta $ and $A \in {\cal A}_\gamma $ with $A \subseteq ^* B$ . The distributivity number ${\mathfrak h}$ of ${\cal P} (\omega ) / {\mathrm {fin}}$ is the least cardinal $\kappa $ such that ${\cal P} (\omega ) / {\mathrm {fin}}$ as a forcing notion is not $\kappa $ -distributive; equivalently, it is the least $\kappa $ such that there is a collection ${\mathfrak A}$ of size $\kappa $ of mad families without common refinement. Clearly, a refining matrix must have height at least ${\mathfrak h}$ , and it is easy to see that there is one of height ${\mathfrak h}$ and none of regular height $> {\mathfrak c}$ . Furthermore, if there is a refining matrix of height $\vartheta $ , then there is one of height $cf (\vartheta )$ so that it suffices to consider regular heights. A famous theorem of Balcar, Pelant, and Simon [BPS] (see also [Bl, Theorem 6.20]) says that there is even a base matrix of height ${\mathfrak h}$ . It is natural to ask whether there can consistently be refining (base) matrices of other heights, and in interesting recent work, Fischer, Koelbing, and Wohofsky [FKW1] proved that it is consistent that ${\mathfrak h} = \omega _1$ and there is a refining matrix of height $\vartheta \leq {\mathfrak c}$ , where $\vartheta> \omega _1$ is regular, all of whose maximal branches are cofinal (see Section 2 for a formal definition). We show the following theorems.

Theorem A If ${\mathfrak c}$ is regular, then there is a base matrix of height ${\mathfrak c}$ .

Theorem B In the Cohen and random models, there are base matrices of any regular uncountable height $\leq {\mathfrak c}$ .

This answers Questions 7.5 and 7.7 of [FKW1]. Note that our results are incomparable with the one of the latter work. Their construction does not give a base matrix (in fact, by another result of Fischer, Koelbing, and Wohofsky [FKW2], a base matrix of height $> {\mathfrak h}$ always has some non-cofinal maximal branches, though one may still ask whether one can get such a base matrix in which some maximal branches are cofinal), whereas ours necessarily gives non-cofinal maximal branches. In fact, in the Cohen and random models, ${\mathfrak h} = \omega _1$ is the only cardinal $\vartheta $ for which there is a refining (base) matrix of height $\vartheta $ all of whose maximal branches are cofinal, and higher refining matrices have no cofinal branches at all (this follows from Fact 1).

2 Preliminaries

The Cohen model (resp. random model) is the model obtained by adding at least $\omega _2$ many Cohen (resp. random) reals to a model of the continuum hypothesis CH [BJ].

For $A,B \subseteq \omega $ , we say A is almost contained in B, and write $A \subseteq ^* B$ , if $A \setminus B$ is finite. $A \subsetneq ^* B$ if $A \subseteq ^* B$ and $B \setminus A$ is infinite. For an ordinal $\vartheta _0$ , $\{ A_\gamma : \gamma < \vartheta _0 \}$ is a $\subseteq ^*$ -decreasing chain of length $\vartheta _0$ if $A_\delta \subseteq ^* A_\gamma $ for all $\gamma < \delta < \vartheta _0$ . $\subsetneq ^*$ -decreasing chains are defined analogously. For a refining matrix ${\mathfrak A} = \{ {\cal A}_\gamma : \gamma < \vartheta \}$ and an ordinal $\vartheta _0 \leq \vartheta $ , $\{ A_\gamma : \gamma < \vartheta _0 \}$ is a branch in ${\mathfrak A}$ if it is a $\subseteq ^*$ -decreasing chain and $A_\gamma \in {\cal A}_\gamma $ for $\gamma < \vartheta _0$ . A branch is maximal if it cannot be properly extended to a longer branch. A branch is cofinal if $\vartheta _0 = \vartheta $ . Every cofinal branch is maximal, but there may be maximal branches that are not cofinal.

Fact 1 (Folklore)

There are no $\subsetneq ^*$ -decreasing chains of length $\omega _2$ in ${\cal P} (\omega )$ in the Cohen and random models.

This is proved by an isomorphism-of-names argument using the homogeneity of the Cohen or random algebra.

For $A,B \in [\omega ]^\omega $ , A splits B if both $A \cap B$ and $B \setminus A$ are infinite. ${\cal X} \subseteq [\omega ]^\omega $ is a splitting family if every $B \in [\omega ]^\omega $ is split by a member of ${\cal X}$ . The splitting number ${\mathfrak s}$ is the least size of a splitting family. It is well known that ${\mathfrak h} \leq {\mathfrak s}$ ([Bl] or [Ha]).

Fact 2 (Folklore [see [Bl]; see also [Ha, Proposition 22.13] for Cohen forcing])

After adding at least $\omega _1$ Cohen or random reals to a model of ZFC, ${\mathfrak s} = \omega _1$ . (In fact, the first $\omega _1$ generics are a witness for ${\mathfrak s}$ .)

We will prove the following.

Main Theorem 3 Assume $\vartheta \leq {\mathfrak c}$ is a regular cardinal and

1. (A) either there is no $\subsetneq ^*$ -decreasing chain of length $\vartheta $ in ${\cal P} (\omega )$ ,

2. (B) or ${\mathfrak s} \leq \vartheta $ .

Then there is a base matrix of height $\vartheta $ .

Clearly, Theorem A follows from part (B) of the main theorem. (We note, however, that splitting families and ${\mathfrak s} \leq {\mathfrak c}$ are not needed in this case [see the comment at the beginning of the proof of Main Claim 5].) Theorem B follows from either (A) or (B) in view of Facts 1 and 2. Note that part (B) implies that in many other models of set theory there are base matrices of height $\vartheta $ for any regular $\vartheta $ between ${\mathfrak h}$ and ${\mathfrak c}$ , e.g., in the Hechler model (this satisfies ${\mathfrak s} = \omega _1$ by [BD]; see also [Bl]), or in any extension by at least $\omega _1$ Cohen or random reals (Fact 2). The former is, and the latter may be (depending on the ground model), a model for the failure of (A). We do not know whether (A) $+ \neg $ (B) is consistent but conjecture that it is. This clearly implies ${\mathfrak s} \geq {\mathfrak b}^{++}$ , where ${\mathfrak b}$ is the unbounding number (which is known to be consistent; see [BF]).

3 Proof of main theorem

By recursion on $\alpha < {\mathfrak c}$ , we shall construct sets $\Omega _\gamma \subseteq {\mathfrak c}$ and families ${\cal A}_\gamma = \{ A_{\gamma , \alpha } : \alpha \in \Omega _\gamma \}$ , $\gamma < \vartheta $ , such that:

1. (I) All ${\cal A}_\gamma $ are mad.

2. (II) If $\gamma < \delta < \vartheta $ and $\beta \in \Omega _\delta $ , then there is $\alpha \leq \beta $ in $\Omega _\gamma $ such that $A_{\delta ,\beta } \subseteq ^* A_{\gamma ,\alpha }$ .

3. (III) For all $B \in [\omega ]^\omega $ , there are $\gamma < \vartheta $ and $\alpha \in \Omega _\gamma $ such that $A_{\gamma ,\alpha } \subseteq ^* B$ .

This is clearly sufficient. In case (B), let $\{ S_\zeta : \zeta < \nu \}$ be a splitting family with $\nu \leq \vartheta $ . Let $\{ (X_\alpha , \xi _\alpha ) : \alpha < {\mathfrak c} \}$ list all pairs $(X,\xi ) \in [\omega ]^\omega \times \vartheta $ . At stage $\alpha $ of the construction, we will have sets $\{ \Omega _\gamma \cap \alpha : \gamma < \vartheta \}$ , ordinals $\{ \eta _\beta : \beta < \alpha \}$ below $\vartheta $ , and families $\{ \{ A_{\gamma ,\beta } : \beta \in \Omega _\gamma \cap \alpha \} : \gamma < \vartheta \}$ such that:

• (i_α) ${\cal A}^\alpha _\gamma : = \{ A_{\gamma ,\beta } : \beta \in \Omega _\gamma \cap \alpha \}$ is almost disjoint for $\gamma < \vartheta $ .

• (ii_α) For all $\beta < \alpha $ , the set $\{ \gamma : \beta \in \Omega _\gamma \}$ is the interval of ordinals $[\eta _\beta , \max ( \eta _\beta , \xi _\beta ) ]$ and

  • – for $\gamma \in [ \eta _\beta , \max ( \eta _\beta , \xi _\beta ) ]$ , $A_{\gamma ,\beta } = A_{\eta _\beta ,\beta }$ , and

  • – for $\gamma < \eta _\beta $ , there is $\beta ' < \beta $ in $\Omega _\gamma $ such that $A_{\eta _\beta , \beta } \subsetneq ^* A_{\gamma , \beta '}$ .

• (iii_α) For all $\beta < \alpha $ , $A_{\eta _\beta , \beta } \subsetneq ^* X_\beta $ and, in case (B), $A_{\eta _\beta ,\beta } \subseteq ^* S_\zeta $ or $A_{\eta _\beta ,\beta } \subseteq ^* \omega \setminus S_\zeta $ , where $\zeta $ is minimal such that $S_\zeta $ splits $A_{\gamma , \beta '}$ whenever $\gamma < \eta _\beta $ and $\beta ' \in \Omega _\gamma \cap \beta $ are such that $A_{\eta _\beta , \beta } \subsetneq ^* A_{\gamma , \beta '}$ .

Let us first see that this suffices for completing the proof: indeed, (II) and (III) follow from (ii $_\alpha $ ) and (iii $_\alpha $ ), respectively. To see (I), fix $\gamma < \vartheta $ and $Y \in [\omega ]^\omega $ . Then there is $\alpha < {\mathfrak c}$ such that $(Y,\gamma ) = (X_\alpha , \xi _\alpha )$ . So $A_{\max (\eta _\alpha , \xi _\alpha ), \alpha } = A_{\eta _\alpha , \alpha } \subseteq ^* Y$ by (ii $_{\alpha + 1}$ ) and (iii $_{\alpha + 1}$ ) and $A_{\max (\eta _\alpha , \xi _\alpha ), \alpha } \subseteq ^* A_{\gamma , \beta } $ for some $\beta \leq \alpha $ by (ii $_{\alpha + 1}$ ). Thus $Y \cap A_{\gamma , \beta }$ is infinite, as required.

Next, we notice that, for $\alpha = 0$ and for limit $\alpha $ , there is nothing to show. Hence it suffices to describe the successor step, that is, the construction at stage $\alpha + 1$ , and to prove that (i $_{\alpha + 1}$ ) through (iii $_{\alpha + 1}$ ) still hold. Assume $Y \subseteq ^* X_\alpha \cap A_{\gamma ,\beta }$ for some $\gamma < \vartheta $ and $\beta \in \Omega _\gamma \cap \alpha $ , and let $\delta $ be such that $\gamma < \delta < \vartheta $ . We say that Y splits at $\delta $ if:

• • for all $\gamma ' $ with $\gamma \leq \gamma ' < \delta $ , there is $\beta \in \Omega _{\gamma '} \cap \alpha $ such that $Y \subseteq ^* A_{\gamma ' , \beta }$ , and

• • there is no $\beta \in \Omega _\delta \cap \alpha $ such that $Y \subseteq ^* A_{\delta ,\beta }$ .

We say Y splits below $\gamma _0> \gamma $ if there is $\delta $ with $\gamma < \delta < \gamma _0$ such that Y splits at $\delta $ . For infinite $Y \subseteq X_\alpha $ , call ${\cal A}^\alpha _\gamma {\upharpoonright } Y$ mad if $\{ Y \cap A_{\gamma ,\beta } : \beta \in \Omega _\gamma \cap \alpha $ and $| Y \cap A_{\gamma , \beta } | = \aleph _0 \}$ is a mad family below Y. The following is crucial for our construction.

Crucial Lemma 4 Let $\gamma _0 \leq \vartheta $ be an ordinal, and let $Y_0 \subseteq X_\alpha $ be infinite. Assume

• (mad) ${\cal A}^\alpha _\gamma {\upharpoonright } Y_0$ is mad for all $\gamma < \gamma _0$ .

Then there are $\gamma < \gamma _0$ , $\beta \in \Omega _\gamma \cap \alpha $ , and an infinite $Y \subseteq ^* Y_0 \cap A_{\gamma ,\beta }$ that does not split below $\gamma _0$ .

Proof We make a proof by contradiction. Assume

• (split) if $Z \subseteq ^* Y_0 \cap A_{\gamma ,\beta } $ , for some $\gamma < \gamma _0$ and $\beta \in \Omega _\gamma \cap \alpha $ , then Z splits below $\gamma _0$ .

By recursion on $n\in \omega $ , we construct infinite sets $( Y^0_s : s \in 2^{<\omega } )$ and $(Y_s : s \in 2^{<\omega } )$ , as well as ordinals $(\delta ^0_s : s \in 2^{<\omega } )$ and $(\delta _n : n \in \omega )$ such that:

1. (a) $Y_s \subseteq Y^0_s$ and $Y^0_{s \,{}\hat {}\, i} \subseteq Y_s$ for $i \in \{ 0,1 \}$ .

2. (b) $\delta _n = \max \{ \delta ^0_s : |s| = n \} < \gamma _0$ and $\delta ^0_{s \,{}\hat {}\, i}> \delta _{|s|}$ for $i \in \{ 0, 1 \}$ .

3. (c) $Y_s$ splits at $\delta ^0_s$ and there are distinct $\beta , \beta ' \in \Omega _{\delta ^0_s} \cap \alpha $ such that $Y^0_{s \,{}\hat {}\, 0} = Y_s \cap A_{\delta ^0_s, \beta }$ and $Y^0_{s \,{}\hat {}\, 1} = Y_s \cap A_{\delta ^0_s, \beta '}$ (in particular, $Y^0_{s \,{}\hat {}\, 0} \cap Y^0_{s \,{}\hat {}\, 1}$ is finite).

4. (d) $Y_{s \,{}\hat {}\, i} = Y^0_{s \,{}\hat {}\, i} \cap A_{\delta _{|s|} , \beta }$ for some $\beta \in \Omega _{\delta _{|s|}} \cap \alpha $ , for $i \in \{ 0,1 \}$ .

We verify that we can carry out the construction. In the basic step $n = 0$ and $s = \langle \rangle $ , by (mad), let $Y_{\langle \rangle } = Y^0_{\langle \rangle } : = Y_0 \cap A_{0,\beta }$ for some $\beta \in \Omega _0 \cap \alpha $ such that this intersection is infinite. By clause (split), we know that there is $\delta _0 = \delta ^0_{\langle \rangle }$ with $0 < \delta _0 < \gamma _0$ such that $Y_{\langle \rangle }$ splits at $\delta _0$ .

Suppose $Y_s^0, Y_s$ , and $\delta ^0_s$ have been constructed for $|s| = n$ and $\delta _n = \max \{ \delta ^0_s: |s| = n \} < \gamma _0$ are such that (a) through (d) hold. We thus know that $Y_s$ splits at $\delta ^0_s$ and, by the definition of splitting and clause (mad), we can find distinct $\beta , \beta ' \in \Omega _{\delta ^0_s} \cap \alpha $ such that $Y^0_{s \,{}\hat {}\, 0}: = Y_s \cap A_{\delta ^0_s, \beta }$ and $Y^0_{s \,{}\hat {}\, 1} : = Y_s \cap A_{\delta ^0_s, \beta '}$ are infinite. Using again (mad), we see that for $i \in \{ 0,1 \}$ there is $\beta \in \Omega _{\delta _n} \cap \alpha $ such that $Y_{s \,{}\hat {}\, i} : = Y^0_{s \,{}\hat {}\, i} \cap A_{\delta _{n} , \beta }$ is infinite. Again by (split), there is $\delta ^0_{s\,{}\hat {}\, i}$ , $i \in \{ 0,1\}$ , with $\delta _n < \delta ^0_{s\,{}\hat {}\, i} < \gamma _0$ such that $Y_{s \,{}\hat {}\, i} $ splits at $\delta ^0_{s \,{}\hat {}\, i}$ . Finally, let $\delta _{n+1} : = \max \{ \delta ^0_{s \,{}\hat {}\, i} : |s| = n$ and $i \in \{ 0,1 \} \} < \gamma _0$ . This completes the construction.

Let $\delta _\omega = \bigcup _n \delta _n$ . Clearly $\delta _\omega \leq \gamma _0$ is a limit ordinal of countable cofinality. Next, for $f \in 2^\omega $ , let $Y_f$ be a pseudointersection of the $Y_{f {\upharpoonright } n}$ , $n \in \omega $ . If possible, choose $\beta _f \in \Omega _{\delta _\omega } \cap \alpha $ such that $Y_f \cap A_{\delta _\omega , \beta _f}$ is infinite. By (a) and (c) in this construction and by (ii $_\alpha $ ), we see that if $f \neq f'$ then $\beta _f \neq \beta _{f'}$ . However, $\Omega _{\delta _\omega } \cap \alpha $ has size strictly less than ${\mathfrak c}$ , and therefore there is $f \in 2^\omega $ for which there is no such $\beta _f$ . Since $Y_f \subseteq ^* Y_0$ by construction, this implies that ${\cal A}^\alpha _{\delta _\omega } {\upharpoonright } Y_0$ is not mad and, by (mad), $\gamma _0 = \delta _\omega $ . This means, however, that any $Y_f$ contradicts (split). This completes the proof of the crucial lemma.

We next show:

Main Claim 5 There is $\gamma < \vartheta $ such that ${\cal A}^\alpha _\gamma {\upharpoonright } X_\alpha $ is not mad.

Proof Note that, in case $\vartheta = {\mathfrak c}$ , there is nothing to show because by (ii $_\alpha $ ) we see that a tail of the sequence $(\Omega _\gamma \cap \alpha : \gamma < \vartheta )$ is empty, and therefore so is ${\cal A}^\alpha _\gamma $ (in fact, the proof of Theorem A is quite a bit simpler than the general argument: there is no need to list the $\xi _\alpha $ , we may simply let $\xi _\alpha = \alpha $ , $\eta _\alpha $ will always be $\leq \alpha $ , and the splitting family is unnecessary).

Hence assume $\vartheta < {\mathfrak c}$ . By way of contradiction, suppose all ${\cal A}^\alpha _\gamma {\upharpoonright } X_\alpha $ are mad. By the crucial lemma with $\gamma _0 = \vartheta $ and $Y_0 = X_\alpha $ , we know that there are $\gamma < \vartheta $ , $\beta \in \Omega _\gamma \cap \alpha $ and an infinite $Y \subseteq ^* X_\alpha \cap A_{\gamma ,\beta }$ that does not split below $\vartheta $ . This means for all $\delta $ with $\gamma \leq \delta < \vartheta $ there is $\beta \in \Omega _\delta \cap \alpha $ such that $Y \subseteq ^* A_{\delta ,\beta }$ . By (i $_\alpha $ ) and (ii $_\alpha $ ), we see that there must be a strictly increasing sequence $(\beta _\varepsilon : \varepsilon < \vartheta )$ of ordinals below $\alpha $ such that for $\varepsilon '> \varepsilon $ ,

• • $\eta _{\beta _{\varepsilon '}}> \max ( \eta _{\beta _\varepsilon } , \xi _{\beta _\varepsilon } )$ and $Y \subsetneq ^* A_{\eta _{\beta _{\varepsilon '}} , \beta _{\varepsilon '}} \subsetneq ^* A_{\eta _{\beta _{\varepsilon }} , \beta _{\varepsilon }}$ .

In case (A), this contradicts the initial assumption that there are no $\subsetneq ^*$ -decreasing chains of length $\vartheta $ in ${\cal P} (\omega )$ . So assume we are in case (B). Define a sequence $(\zeta _\varepsilon : \varepsilon < \vartheta )$ of ordinals below $\nu $ such that

• • $\zeta _\varepsilon $ is minimal such that $S_{\zeta _\varepsilon }$ splits all $A_{\eta _{\beta _{\varepsilon '}} , \beta _{\varepsilon '}}$ for $\varepsilon ' < \varepsilon $ .

Using (iii $_\alpha $ ), we see that $S_{\zeta _\varepsilon }$ does not split $A_{\eta _{\beta _{\varepsilon }} , \beta _{\varepsilon }}$ . Therefore, the sequence must be strictly increasing, which is impossible (and thus contradictory) in case $\nu < \vartheta $ . If $\nu = \vartheta $ note that there cannot be any $\zeta $ such that $S_\zeta $ splits Y, contradicting the initial assumption that the $S_\zeta $ form a splitting family. This final contradiction establishes the main claim.

We now let $\eta _\alpha : = \min \{ \gamma : {\cal A}^\alpha _\gamma {\upharpoonright } X_\alpha $ is not mad $\} < \vartheta $ . Choose $Y_0 \subseteq X_\alpha $ infinite and almost disjoint from all members of ${\cal A}^\alpha _{\eta _\alpha }$ . Note that ${\cal A}^\alpha _\gamma {\upharpoonright } Y_0$ is mad for all $\gamma < \eta _\alpha $ . Thus, by the crucial lemma with $\gamma _0 = \eta _\alpha $ , we know there are $\gamma < \eta _\alpha $ , $\beta \in \Omega _\gamma \cap \alpha $ , and an infinite $Y \subseteq ^* Y_0 \cap A_{\gamma ,\beta }$ that does not split below $\eta _\alpha $ . Then,

• (⋆) for all $\delta $ with $\gamma \leq \delta < \eta _\alpha $ , there is $\beta = \beta _\delta \in \Omega _\delta \cap \alpha $ such that $Y \subseteq ^* A_{\delta ,\beta }$ .

Choose infinite $A_{\eta _\alpha ,\alpha } \subsetneq ^* Y$ . In case (B), choose $\zeta < \nu $ minimal such that $S_\zeta $ splits all $A_{\delta , \beta _\delta }$ with $\gamma \leq \delta < \eta _\alpha $ . If $Y \cap S_\zeta $ is infinite, additionally require $A_{\eta _\alpha ,\alpha } \subsetneq ^* Y \cap S_\zeta $ . (If not, we will automatically have $A_{\eta _\alpha ,\alpha } \subsetneq ^* \omega \setminus S_\zeta $ .)

Next, for all $\gamma $ with $\eta _\alpha \leq \gamma \leq \max (\eta _\alpha ,\xi _\alpha )$ , we let $A_{\gamma ,\alpha } = A_{\eta _\alpha , \alpha }$ . Also put

$$\begin{align*}\Omega_\gamma \cap ( \alpha + 1) = \left\{ \begin{array}{ll} \Omega_\gamma \cap \alpha, & \mbox{ if } \gamma < \eta_\alpha \mbox{ or } \gamma> \max (\eta_\alpha,\xi_\alpha), \\ (\Omega_\gamma \cap \alpha) \cup \{ \alpha \}, & \mbox{ if } \eta_\alpha \leq \gamma \leq \max (\eta_\alpha,\xi_\alpha). \end{array} \right. \end{align*}$$

Then clauses (i $_{\alpha +1}$ ) and (iii $_{\alpha + 1}$ ) are immediate, and (ii $_{\alpha +1}$ ) follows from $(\star )$ . This completes the proof of the main theorem.

4 Further remarks and questions

Obviously, the main remaining problem is whether the spectrum of heights of base matrices can be non-convex on regular cardinals.

Question 6 Is it consistent that for some regular $\vartheta $ with ${\mathfrak h} < \vartheta < {\mathfrak c}$ there is no base (refining) matrix of height $\vartheta $ ?

The simplest instance would be ${\mathfrak h} = \omega _1$ and ${\mathfrak c} = \omega _3$ with no base (refining) matrix of height $\omega _2$ . By (B) in Main Theorem 3, this would imply ${\mathfrak s} = \omega _3$ .

As the referee remarked, another constellation for a nontrivial spectrum, which would be convex, might be a model where ${\mathfrak s} = {\mathfrak c}$ is singular and there is a regular cardinal $\kappa \geq {\mathfrak h}$ with $\kappa < {\mathfrak c}$ such that the spectrum consists exactly of the regular cardinals in the interval $[{\mathfrak h} , \kappa ]$ . It is unknown, however, whether ${\mathfrak s} = {\mathfrak c} $ singular is consistent at all. The consistency of singular ${\mathfrak s}$ was shown by Dow and Shelah [DS], but in their model, ${\mathfrak c}$ is at least ${\mathfrak s}^+$ .

The proof of Main Theorem 3 may look a little like cheating because we do not refine our mad families everywhere when going to the next level. Thus, let us say ${\mathfrak A} = \{ {\cal A}_\gamma : \gamma < \vartheta \}$ is a strict base (refining) matrix if it is a base (refining) matrix and for any $\gamma < \delta < \vartheta $ and any $A \in {\cal A}_\delta $ there is $B \in {\cal A}_\gamma $ with $A \subsetneq ^* B$ . We then obtain the following.

Proposition 7 Assume $\vartheta \leq {\mathfrak c}$ is a regular cardinal such that there are $\subsetneq ^*$ -decreasing chains of length $\alpha $ in ${\cal P} (\omega )$ for any $\alpha < \vartheta $ and

1. (A) either there is no $\subsetneq ^*$ -decreasing chain of length $\vartheta $ in ${\cal P} (\omega )$ ,

2. (B) or ${\mathfrak s} \leq \vartheta $ .

Then there is a strict base matrix of height $\vartheta $ .

Proof sketch

Modify the proof of Main Theorem 3 by attaching a $\subsetneq ^*$ -decreasing chain of length $\zeta _\beta + 1$ to the set $\{ \gamma : \beta \in \Omega _\gamma \} = [ \eta _\beta , \max (\eta _\beta , \xi _\beta ) ]$ , where $\eta _\beta + \zeta _\beta = \max (\eta _\beta , \xi _\beta )$ . This is clearly possible by assumption.

To analyze this a bit further, let ${\mathfrak {ds}}$ denote the least ordinal $\alpha $ such that there is no $\subsetneq ^*$ -decreasing chain of length $\alpha $ in ${\cal P} (\omega )$ . It is easy to see that ${\mathfrak {ds}}$ is a regular cardinal with ${\mathfrak b}^+ \leq {\mathfrak {ds}} \leq {\mathfrak c}^+$ . Put $\vartheta _0 = \min \{ {\mathfrak {ds}} , {\mathfrak c} \}$ , and assume $\vartheta _0$ is regular. Then:

1. (1) there are strict base matrices of heights ${\mathfrak h}$ and $\vartheta _0$ , and

2. (2) all strict refining matrices have height between ${\mathfrak h}$ and $\vartheta _0$ .

To see (1), use the previous proposition for height $\vartheta _0$ , and note that the original construction of [BPS] gives a strict base matrix of height ${\mathfrak h}$ . (2) is obvious. We leave it to the reader to verify that Proposition 7 implies the corresponding versions of Theorems A and B.

Corollary 8 If ${\mathfrak c} \leq \omega _2$ , then there is a strict base matrix of height ${\mathfrak c}$ .

Corollary 9 Let $\vartheta $ be a regular uncountable cardinal. In the Cohen and random models, the following are equivalent:

1. (i) $\vartheta \in \{ \omega _1 , \omega _2 \}$ .

2. (ii) There is a strict base matrix of height $\vartheta $ .

3. (iii) There is a strict refining matrix of height $\vartheta $ .

To see, e.g., Corollary 9, note that by Fact 1, ${\mathfrak {ds}} = \omega _2$ in either model, and use (1) and (2) above.