Recall that Martin’s Axiom asserts that for every partial order $\mathbb P$ satisfying c.c.c., and for any family $\mathcal D$ of $<2^{\aleph_0}$ many dense subsets of $\mathbb P$, there exists a directed subset $G$ of $\mathbb P$ such that $G\cap D\neq\emptyset$ for all $D\in\mathcal D$.
Solovay and Tennenbaum proved that Martin’s Axiom is consistent with $2^{\aleph_0}>\aleph_1$.
In [Sh:80], Shelah proved that the following weak generalization of Martin’s Axiom is consistent with $2^{\aleph_1}>\aleph_2$ and CH. If $\mathbb P$ is a partial order satisfying:
- $|\mathbb P|<2^{\aleph_1}$;
- $\mathbb P$ is $\sigma$-closed;
- $\mathbb P$ is $\aleph_2$-stationary-c.c.: if $\{ p_\alpha\mid\alpha<\aleph_2\}$ is an injective enumeration of conditions, then there exists a regressive map $f:\omega_2\rightarrow\omega_2$ and a club $D\subseteq\omega_2$, so that $\{ p_\alpha\mid \alpha\in D\cap E^{\omega_2}_{\omega_1}, f(\alpha)=i\}$ is directed for all $i<\omega_2$;
- well-met condition: for every $p,q\in\mathbb P$, if $\{ r\in\mathbb P\mid p\le r\ \&\ q\le r\}$ is nonempty (i.e., $p$ and $q$ are compatible), then the set has a least element,
then for any family $\mathcal D$ of $<2^{\aleph_1}$ many dense subsets of $\mathbb P$, there exists a directed subset $G$ of $\mathbb P$ such that $G\cap D\neq\emptyset$ for all $D\in\mathcal D$.
Let us look at requirements (1)–(4).
Requirement (1) is a rather natural restriction (though, this restriction may sometime be waived via tail-catching, or by reflection arguments in the presence of large cardinals).
Requirement (2) has to do with the fact that any forcing axiom for meeting more than $\aleph_1$ many dense sets which will be compatible with CH, will have to be applicable only to forcing notions that do not add reals.
Requirement (3): Recall that Martin’s Axiom implies that the product of any two c.c.c. posets is again c.c.c., whereas there are ZFC examples of two $\aleph_2$-c.c. posets whose product is not anymore $\aleph_2$-c.c. (see here). So, plain $\aleph_2$-c.c. must be replaced by a stronger variation.
Today, I learned from Ashutosh Kumar that [Sh:1036] offers a nice and relatively simple justification of Requirement (4). Luckily, the example builds upon a theorem of Shelah which I already presented in a previous blog post (see also Section 2 of this survey paper, for motivation):
Theorem (Shelah). If $\langle C_\delta\mid \delta\in E^{\omega_2}_{\omega_1}\rangle$ is a sequence such that $C_\delta$ is a club in $\delta$ of order-type $\omega_1$ for each $\delta$, then there exists a sequence of local colorings $\langle f_\delta:C_\delta\rightarrow2\mid \delta\in E^{\omega_2}_{\omega_1}\rangle$, such that for every global coloring $f:\omega_2\rightarrow\omega_2$, there exists $\delta\in E^{\omega_2}_{\omega_1}$ for which $\{ \alpha\in C_\delta\mid f(\alpha)\neq f_\delta(\alpha)\}$ is not bounded in $\delta$.
The very same proof shows that “not bounded” may be strengthened to “is stationary”. Thus, fix a sequence $\langle f_\delta\mid \delta\in E^{\omega_2}_{\omega_1}\rangle$ witnessing the latter. For simplicity, we may assume that $\min(C_\delta)=1$ for all $\delta\in E^{\omega_2}_{\omega_1}$. Define a notion of forcing $\mathbb P$, as follows. $p\in\mathbb P$ iff it is a partial function from $E^{\omega_2}_{\omega_1}$ to $\mathcal P_{\omega_1}(\omega_2)$, such that:
- $\text{dom}(p)$ is a countable subset of $E^{\omega_2}_{\omega_1}$;
- $p(\delta)$ is a closed bounded subset of $C_\delta$;
- if $\delta_1<\delta_2$ are in $\text{dom}(p)$ and $\alpha\in p(\delta_1)\cap p(\delta_2)$, then $f_{\delta_1}(\alpha)=f_{\delta_2}(\alpha)$;
- if $\delta_1\neq\delta_2$ are in $\text{dom}(p)$, then $\beta<\max(p(\delta_1))$ for all $\beta\in C_{\delta_1}\cap C_{\delta_2}$.
A condition $q$ extends $p$, iff
- $\text{dom}(q)\supseteq\text{dom}(p)$;
- $q(\delta)$ is an end-extension of $p(\delta)$ for all $\delta\in\text{dom}(p)$.
Let $\mathcal D:=\{ D_{\alpha,\delta}\mid \alpha<\delta, \delta\in E^{\omega_2}_{\omega_1}\}$, where $$D_{\alpha,\delta}:=\{ p\in\mathbb P\mid \delta\in\text{dom}(p)\ \&\ \max(p(\delta))>\alpha\}.$$
Then $|\mathcal D|=\aleph_2$ and if $G$ happens to be a directed subset of $\mathbb P$ such that $G\cap D$ for all $D\in \mathcal D$, then for all $\delta\in E^{\omega_2}_{\omega_1}$, $c_\delta:=\bigcup\{ p(\delta)\mid p\in G\}$ is a club subset of $C_\delta$, and $f:=\bigcup\{f_\delta\restriction c_\delta\mid \delta\in E^{\omega_2}_{\omega_1}\}$ is a function! In particular, $\{\alpha\in C_\delta\mid f(\alpha)\neq f_\delta(\alpha)\text{ or }\alpha\not\in\text{dom}(f)\}$ is nonstationary for all $\delta\in E^{\omega_2}_{\omega_1}$.
Claim 1. $D_{\alpha,\delta}$ is dense for every $\delta\in E^{\omega_2}_{\omega_1}$ and $\alpha<\delta$.
Proof. Given $q\in\mathbb P\setminus D_{\alpha,\delta}$, we distinguish two cases.
If $\delta\in\text{dom}(q)$, then let $\alpha’:=\min(C_\delta\setminus(\max(q(\delta)\cup\{\alpha\})+1))$, and set $$p:=(q\restriction(\text{dom}(p)\setminus\{\delta\}))\cup\{(\delta,q(\delta)\cup\{\alpha’\})\}.$$If $\delta\not\in\text{dom}(q)$, let $t:=\bigcup\{ C_\beta\cap C_\gamma\mid \beta\neq\gamma\text{ in dom}(q)\cup\{\delta\}\}$. As $t$ and $\text{dom}(q)$ are countable, it is easy to construct an injective sequence $$\langle \alpha_\beta\mid \beta\in \text{dom}(q)\cup\{\delta\}\rangle\in\prod_{\beta\in \text{dom}(q)\cup\{\delta\}}C_\beta$$ such that $\alpha_\beta>\sup(q(\beta)\cup(t\cap\beta))$ for all $\beta\in\text{dom}(q)$, and $\alpha_\delta>\sup(\alpha\cup(t\cap\delta))$.
Finally, let $p\in\mathbb P$ be such that $\text{dom}(p)=\text{dom}(q)\cup\{\delta\}$ and, $$p(\beta)=\begin{cases}q(\beta)\cup\{\alpha_\beta\},&\beta\in\text{dom}(q)\\\{\alpha_\delta\},&\text{otherwise}\end{cases}. \square$$
Claim 2. $\mathbb P$ is $\sigma$-closed.
Proof. Given an increasing sequence $\langle p_n\mid n<\omega\rangle$ in $\mathbb P$, define $p\in\mathbb P$ such that $\text{dom}(p):=\bigcup_{n<\omega}\text{dom}(p_n)$ and $p(\delta)$ is the closure of $\bigcup\{p_n(\delta)\mid n<\omega, \delta\in\text{dom}(p_n)\}$ for all $\delta\in\text{dom}(p)$. To see that $p\in\mathbb P$, we only need to verify that if $\delta_1<\delta_2$ are in $\text{dom}(p)$ and $\alpha\in p(\delta_1)\cap p(\delta_2)$, then $f_{\delta_1}(\alpha)=f_{\delta_2}(\alpha)$. Fix a large enough $N<\omega$ such that $\delta_1$ and $\delta_2$ are in $\text{dom}(p_N)$. Put $\beta:=\sup(C_{\delta_1}\cap C_{\delta_2})$, then $\beta<\delta_1<\delta_2$, and by clause (4) above, $\beta<\max(p_N(\delta_i))$ for $i<2$. By $\alpha\in C_{\delta_1}\cap C_{\delta_2}$, we have $\alpha\le\beta<\max(p_N(\delta_i))$ for $i<2$. As $ p_n(\delta_i)$ end-extends $p_N(\delta_i)$ whenever $N\le n<\omega$ and $i<2$, we infer that $\alpha\in p_N(\delta_1)\cap p_N(\delta_2)$, and so by Clause (3), $f_{\delta_1}(\alpha)=f_{\delta_2}(\alpha)$. $\square$
Notation. For every $q\in\mathbb P$, denote $$g_q:=\bigcup\{ f_\delta\restriction q(\delta)\mid \delta\in\text{dom}(q)\}.$$
Claim 3. CH entails that $\mathbb P$ has size $\aleph_2$ and satisfies the $\aleph_2$-stationary-c.c.
Proof. It is a basic counting argument that shows that CH entails a bijection $\psi:\omega_2\leftrightarrow\mathbb P$. Next, given a sequence of conditions $\langle p_\alpha\mid\alpha\in E^{\omega_2}_{\omega_1}\rangle$, we first extend each $p_\alpha$ to a condition $q_\alpha$ satisfying $\alpha\in\text{dom}(q_\alpha)$, and $\max(q_\alpha(\delta))>\alpha$ for all $\delta\in\text{dom}(q_\alpha)$ above $\alpha$. As $g_{q_\alpha}$ is a countable partial function from $\omega_2$ to $2$, we infer from CH and the Engelking-Karlowicz theorem the exisetnce of a sequence $\langle \varphi_i:\omega_2\rightarrow2\mid i<\omega_1\rangle$ such that for each $\alpha\in E^{\omega_2}_{\omega_1}$, there is $i_\alpha<\omega_1$ with $g_{q_\alpha}\subseteq \varphi_{i_\alpha}$. Let $D$ be the set of all $\delta<\omega_2$ such that for all $\alpha<\delta$, $\{ p\in\mathbb P\mid \text{dom}(p)\subseteq\alpha\}\subseteq \psi[\delta]$. Then $D$ is a club, and for all $\delta\in D\cap E^{\omega_2}_{\omega_1}$, $\{ p\in\mathbb P\mid \text{dom}(p)\subseteq\delta\}\subseteq \psi[\delta]$. Define a function $h:E^{\omega_2}_{\omega_1}\rightarrow\omega_2$ by stipulating $h(\delta):=\sup(\bigcup\{\text{dom}(q_\alpha)\mid\alpha\in E^\delta_{\omega_1}\})$, and let $E:=\{\delta\in D\mid h[\delta]\subseteq\delta\}$. Finally, define a regressive function $f:E\cap E^{\omega_2}_{\omega_1}\rightarrow\omega_1\times\omega_2$ by letting $f(\delta):=(i_\delta,\psi^{-1}(q_\delta\restriction\delta))$.
To see that $\{ p_\alpha\mid \alpha\in E\cap E^{\omega_2}_{\omega_1}, f(\alpha)=i\}$ is directed for all $i\in\omega_1\times\omega_2$, fix arbitrary $\delta_1<\delta_2$ in $E\cap E^{\omega_2}_{\omega_1}$ with $f(\delta_1)=f(\delta_2)$. As $q_{\delta_1}\restriction\delta_1=q_{\delta_2}\restriction\delta_2$ and $\delta_1\in\delta_2\in E$, we have $$q_{\delta_1}\restriction(\text{dom}(q_{\delta_1})\cap\text{dom}(q_{\delta_2}))=q_{\delta_2}\restriction(\text{dom}(q_{\delta_1})\cap\text{dom}(q_{\delta_2})).$$
Set $p:=q_{\delta_1}\cup q_{\delta_2}$. Then $p(\alpha)$ is a closed bounded subset of $C_\alpha$ for all $\alpha\in\text{dom}(p)$. Also, $g_p$ is a function, since $g_p\subseteq\varphi_{i_{\delta_1}}=\varphi_{i_{\delta_2}}$. So, $p$ satisfies Clauses (1)–(3) above. Now, as in the end of the proof of Claim 1, it is easy to find $p’\in\mathbb P$ with $\text{dom}(p’)=\text{dom}(p)$ such that $p'(\alpha)$ is an end-extension of $p(\alpha)$ by a singleton for all $\alpha\in\text{dom}(p’)$. Clearly, $p’$ is stronger than $q_{\delta_1}$ and $q_{\delta_2}$, let alone $p_{\delta_1}$ and $p_{\delta_2}$. $\square$
Note that any condition $p\in\mathbb P$ admits $\mathfrak c$ many pairwise incompatible extensions, hence, CH is also necessary for $\mathbb P$ to satisfy the $\aleph_2$-c.c.
Claim 4. $\mathbb P$ does not satisfy the well-met condition.
Proof. By iterating the pressing down lemma, we can find a stationary subset $S$ of $E^{\omega_2}_{\omega_1}$ along with ordinals $\alpha_0<\alpha_1<\alpha_2$ such that for all $\delta\in S$, the first three elements of $C_\delta$ are $\{\alpha_0,\alpha_1,\alpha_2\}$. Pick $\delta_0<\delta_1$ from $S$. Let $p_i:=\{(\delta_i,\{\alpha_i\})\}$ for $i<2$. Then $p_0$ and $p_1$ are compatible elements of $\mathbb P$, but due to Clause (4) above, $p_0\cup p_1$ is not in $\mathbb P$, and so the set $\{ r\in\mathbb P\mid p_0\le r\ \&\ p_1\le r\}$ does not admit a least element. $\square$
