Free sequences in ${\mathscr {P}}\left( \omega \right) /\text {fin}$

Abstract

We investigate maximal free sequences in the Boolean algebra ${\mathscr {P}}\left( \omega \right) {/}\text {fin}$, as defined by Monk (Comment Math Univ Carol 52(4):593–610, 2011). We provide some information on the general structure of these objects and we are particularly interested in the minimal cardinality of a free sequence, a cardinal characteristic of the continuum denoted ${\mathfrak {f}}$. Answering a question of Monk, we demonstrate the consistency of $\omega _1 = {\mathfrak {i}} = {\mathfrak {f}} < {\mathfrak {u}} = \omega _2$. In fact, this consistency is demonstrated in the model of Shelah for ${\mathfrak {i}}< {\mathfrak {u}}$ (Shelah in Arch Math Log 31(6):433–443, 1992). Our paper provides a streamlined and mostly self contained presentation of this construction.

Appendix: Preservation theorem for the iteration

The forcing iteration argument in Sect. 5 uses a typical preservation theorem for countable support forcing iteration, in this instance the preservation of a filter–co-filter pair. This theorem follows the usual pattern described in [8, 17]. However, as specific instances of preservation theorems are sometimes difficult to derive from the general statements given in these sources, we decided to provide the proof of the relevant preservation theorem in this appendix, making the paper more self-contained.

Let ${\mathscr {F}}$ be a filter on $\omega$. We will use the following game $\mathrm G({{\mathscr {F}}})$. Players I and II alternate for $\omega$ many rounds. In the n-th round player I plays a set $F_n \in {\mathscr {F}}$, and player II responds with $a_n \in F_n$. Player II wins if $\left\{ \, a_n \shortmid n \in \omega \, \right\} \in {\mathscr {F}}$. The following is well known.

Fact 32

Player I does not have a winning strategy in the game $\mathrm G({{\mathscr {F}}})$ iff ${\mathscr {F}}$ is a rare P-filter.

Theorem 33

Let ${\mathscr {F}}$ be a P-filter on $\omega$, denote ${\mathscr {K}} = {\mathscr {P}}\left( \omega \right) \smallsetminus {\mathscr {F}}$. For $\delta$ limit let $P_\delta = \left\langle P_\alpha , Q_\alpha \shortmid \alpha < \delta \right\rangle$ be a countable support iteration of proper forcing notions such that for each $\alpha < \delta$

$$\begin{aligned} P_\alpha \Vdash {\mathscr {F}} \text{ is } \text{ a } \text{ rare } \text{ filter } \text{ and } \left\langle {\mathscr {F}} \right\rangle \cup \left\langle {\mathscr {K}} \right\rangle = {\mathscr {P}}\left( \omega \right) . \end{aligned}$$

Then also $P_\delta \Vdash \left\langle {\mathscr {F}} \right\rangle \cup \left\langle {\mathscr {K}} \right\rangle = {\mathscr {P}}\left( \omega \right)$.

By $\left\langle {\mathscr {F}} \right\rangle$ and $\left\langle {\mathscr {K}} \right\rangle$ we denote the upwards, respectively downwards closure of ${\mathscr {F}}$ and ${\mathscr {K}}$ in the appropriate models. The assumption for $\alpha = 0$ states that ${\mathscr {F}}$ is a rare P-filter in the ground model V. Standard arguments shows that $\left\langle {\mathscr {F}} \right\rangle$ is a P-filter in any generic extension via a proper forcing, and $\left\langle {\mathscr {F}} \right\rangle$ is rare in any generic extension via a bounding forcing.

Proof

If the cofinality of $\delta$ is uncountable, no new reals are added at stage $\delta$ of the iteration, and the conclusion of the theorem holds true. Therefore we will assume that the cofinality of $\delta$ is countable, and by passing to a cofinal sequence of $\delta$, it is sufficient to prove the theorem in case $\delta = \omega$. In the following $G_\alpha$ denotes exclusively generic filters on $P_\alpha$. We use P to denote posets $P_\delta / G_\alpha$ in the intermediate generic extensions $V[G_\alpha ]$. Let X be a P-name for a subset of $\omega$. For $r \in P$ let $X_r = \left\{ \, n \in \omega \shortmid r \not \Vdash n \notin X \, \right\}$. $\square$

Lemma 34

Let ${\mathscr {H}}$ be a rare P-filter and $p \in P$ a condition. If $X_r \in {\mathscr {H}}$ for each $r < p$, then there exists $H \in {\mathscr {H}}$ and a sequence $\left\langle r_i \in P \shortmid i \in \omega \right\rangle$, $r_0 = p$, $r_{i+1} < r_i$ such that $r_i \Vdash H \cap i \subset X$ for each $i\in \omega$.

Proof

Put $p_0 = p$ and play the game $\mathrm G({{\mathscr {H}}})$ as follows. In the n-th round player I plays the set $X_{p_n} \in {\mathscr {H}}$, player II responds with some $a_n \in X_{p_n}$. Player I then chooses $p_{n+1} \in P$, $p_{n+1} < r_n$ such that $p_{n+1} \Vdash a_n \in X$ and proceeds to the next round. Since ${\mathscr {H}}$ is a rare P-filter, this strategy is not winning for player I. Thus there is a sequence of moves of player II and conditions $\left\langle p_n \shortmid n \in \omega \right\rangle$ such that player II wins the game, i.e. $H = \left\{ \, a_n \shortmid n \in \omega \, \right\} \in {\mathscr {H}}$. A sequence of conditions $\left\langle r_i \shortmid i \in \omega \right\rangle$ such that $r_i = p_{a_n}$ if $a_n < i \le a_{n+1}$ is as required in the lemma. $\square$

Let p be a condition in $P_\omega$. The goal is to find a stronger condition which forces either $X \in \left\langle {\mathscr {F}} \right\rangle$ or $X \in \left\langle {\mathscr {K}} \right\rangle$. In case there exists an intermediate extension $V[G_\alpha ]$, $p \in G_\alpha$ and $r \in P/G_\alpha$, $r < p/G_\alpha$ such that $X_r \notin \left\langle {\mathscr {F}} \right\rangle$ (in $V[G_\alpha ]$), then $r \Vdash X \in \left\langle {\mathscr {K}} \right\rangle$ due to the assumption of the theorem, and there exists a condition in $P_\omega$ stronger than p forcing the same statement. Therefore we will assume in the rest of the proof that this is not the case.

For a sufficiently large $\theta$ fix a countable elementary submodel $N \prec H(\theta )$ such that $X, p, {\mathscr {F}}, P_\omega \in N$. Use Lemma 34 in N for ${\mathscr {H}} = {\mathscr {F}}$ and $P = P_\omega$ to get $H \in {\mathscr {F}} \cap N$ and a sequence $\left\langle r^0_n \in P_\omega \shortmid n \in \omega \right\rangle \in N$. Since ${\mathscr {F}}$ is a P-filter, there exists $A^* \in {\mathscr {F}}$ such that $A^* \subset H$, and $A^* \subset ^* F$ for each $F \in {\mathscr {F}} \cap N$.

Lemma 35

Let q be a $(P_i,N)$-master condition, and let $\left\langle F_n \shortmid n \in \omega \right\rangle \in N[G_i]$ be a sequence of elements of ${\mathscr {F}}$. Then

$$\begin{aligned} q \Vdash \text { There are infinitely many } n \in \omega \text { such that } A^* \smallsetminus n \subset F_n. \end{aligned}$$

Proof

Since $N[G_i] \prec H(\theta )[G_i]$ and ${\mathscr {F}}$ generates a non-meager filter in $H(\theta )[G_i]$, there is $F \in {\mathscr {F}} \cap N[G_i]$ such that $F \smallsetminus n \subset F^n$ for infinitely many n (Fact 12). Now $q \Vdash F \in N$ and we can use that $A^* \subset ^* F$.

We will inductively construct a condition $q < p$ such that $q \Vdash A^* \subset X$. Specifically, we construct two sequences of conditions $p_i, q_i$ for $i \in \omega$ with the following properties;

• (1)

  • $p_i \in P_\omega$,

  • $p_{i+1} < p_i$,

  • $p_{i+1} \mathbin {\!\upharpoonright }i = p_{i} \mathbin {\!\upharpoonright }i$,

  • $q_i \in P_i$,

  • $q_{i+1} \mathbin {\!\upharpoonright }i = q_{i}$,

  • $q_i < p_{i} \mathbin {\!\upharpoonright }i$,

  • $q_i$ is a $(N, P_i)$-master condition;

• (2) $q_i \Vdash (p_i/G_i \Vdash A^* \cap i \subset X)$,

• (3) $q_i \Vdash \big (\text {There is a sequence } \left\langle r^i_n \in P_\omega /G_i \shortmid n \in \omega \right\rangle \in N[G_i], r^i_n {<} p_i/G_i \text { such that } r^i_n \Vdash A^* \cap n \subset X\big )$.

The construction starts with putting $p_0 = p$ and let $q_0$ be a trivial condition (in the trivial forcing $P_0$). Existence of the sequence $\left\langle r^0_n \in P_\omega \shortmid n \in \omega \right\rangle$ follows from the choice of $A^*$.

Suppose that $p_i, q_i$ are defined, work in $N[G_i]$ assuming $q_i \in G_i$. For each $n \in \omega$ consider a model $N[G_{i+1}]$ such that $r^i_n \mathbin {\!\upharpoonright }(i+1) \in G_{i+1}/G_i$. Use Lemma 34 in $N[G_{i+1}]$ for $\left\langle {\mathscr {F}} \right\rangle$ and $r^i_n/G_{i+1}$ to get $H_n \in \left\langle {\mathscr {F}} \right\rangle \cap N[G_{i+1}]$ and a sequence $\left\langle s^n_k \in P_\omega /G_{i+1} \shortmid k \in \omega \right\rangle \in N[G_{i+1}]$ as in the lemma. We can assume that $H_n \in {\mathscr {F}} \cap N[G_{i+1}]$, and by strengthening $r^i_n \mathbin {\!\upharpoonright }\left\{ \, i \, \right\}$ to $t^i_n \mathbin {\!\upharpoonright }\left\{ \, i \, \right\} \in N[G_i]$ we can decide $H_n$ to be some $F_n \in {\mathscr {F}} \cap N[G_i]$. Since $q_i$ is $(N, P_i)$-master, Lemma 35 implies that there is $m > i$ such that $A^* \smallsetminus m \subset F_m$.

Define $p_{i+1} = p_i \mathbin {\!\upharpoonright }i \mathbin {\!{}^\smallfrown \!}t^i_m$, and let $q_{i+1} < p_{i+1}\mathbin {\!\upharpoonright }i+1$ be any $(N, P_{i+1})$-master condition such that $q_{i+1} \mathbin {\!\upharpoonright }i = q_i$. Property (1) is obviously satisfied. Property (2) follows from $m > i$, the inductive hypothesis for $r_m^i$, and from ${{q_{i+1}} \mathbin {\!{}^\smallfrown \!}(p_{i+1}/G_{i+1})} < q_i \mathbin {\!{}^\smallfrown \!}r_m^i$. To justify (3) notice that $q_{i+1}$ forces that the sequence $\left\langle s^m_k \shortmid k \in \omega \right\rangle$ satisfies the condition required for $\left\langle r^i_n \shortmid n \in \omega \right\rangle$; for $y \in A^* \cap m$ this follows from the inductive hypothesis on $r^i_m$, and for $y \in A^*, x \ge m$ from the choice of $\left\langle s^m_k \shortmid k \in \omega \right\rangle$ and $A^* \smallsetminus m \subset F_m$.

Once the inductive construction is done, the condition $q = \bigcup \left\{ \, q_i \shortmid i \in \omega \, \right\}$ forces that $A^* \subset X$. The inclusion $A^* \cap i \subset X$ is guaranteed by property (2) and $q < q_i \mathbin {\!{}^\smallfrown \!}{(p_i/G_i)}$. $\square$