Proper Forcing, Countable-Support Iterations, and PFA: Examples and Counterexamples

Fix a dense set $D\subseteq P$ with $D\in M$. By elementarity, inside $M$ choose a maximal antichain $A\subseteq D$. It is also maximal in $P$: maximality is the first-order assertion that every $r\in P$ is compatible with some $a\in A$, and all witnesses to compatibility are conditions in the ambient $H_{\theta }$. Since $P$ is ccc, $A$ is countable in the universe. Elementarity then puts in $M$ a surjection $e:\omega \to A$ (or a finite enumeration), and every $n<\omega$ belongs to $M$; hence $A\subseteq M$.

Let $r\le p$ be arbitrary. Maximality of $A$ gives $a\in A$ compatible with $r$. By step 1.1, $a\in A\subseteq D\cap M$. Thus $D\cap M$ is predense below $p$. Since this holds for every dense $D\in M$, $p$ is $(M,P)$-generic; the reflexive inequality $p\le p$ makes it a master below the original $p$.

The calculation works unchanged when $P$, $A$, or $D$ is finite. A dense subset of the stipulated nonempty $P$ cannot be empty, and for a one-condition order its unique condition is the required antichain member and master. No stronger condition than $p$ was constructed: ccc makes the starting condition itself sufficient.

Let $M$ be a relevant countable elementary submodel, let $p\in \mathbb{B}\cap M$, and put $\delta =M\cap {\omega }_1$. Then $M\cap {\omega }_1$ is an initial segment with no largest member, so $\delta$ is a countable limit ordinal. By elementarity choose in $M$ a normal $h:{\omega }_1\to {\omega }_1$ extending $p$. For every $\alpha <\delta$, both $\alpha$ and $h(\alpha )$ belong to $M\cap {\omega }_1$, while $h(\alpha )\ge \alpha$; hence $\sup h\mathrm{‘}\mathrm{‘}\delta =\delta$ and continuity gives $h(\delta )=\delta$. Thus $q=p\cup \{(\delta ,\delta )\}$ belongs to $\mathbb{B}$ and satisfies $q\le p$.

For each $\alpha <{\omega }_1$, the set $E_{\alpha }=\{p:\alpha \in \mathrm{dom}(p)\}$ is dense: extend a witness normal function for $p$ and add its value at $\alpha$. Directedness of $G$ makes $F=\bigcup G$ a function, and meeting all $E_{\alpha }$ makes it total. Given $\alpha <\beta$, take two filter conditions specifying the two values and a common stronger condition; its normal extension shows $F(\alpha )<F(\beta )$.

Fix $r\le q$. Any normal extension of $r$ contains $(\delta ,\delta )$, so strict increase gives $\alpha <\delta ⟺r(\alpha )<\delta$ for every $\alpha \in \mathrm{dom}(r)$. Consequently $s=r\cap M$ is exactly the part of $r$ whose coordinates and values lie below $\delta$. It is a finite condition, belongs to $M$, and is extended by $r$. If $D\in M$ is dense, elementarity supplies $r^{\mathrm{\prime }}\in D\cap M$ with $r^{\mathrm{\prime }}\le s$. Every coordinate and value of the finite $r^{\mathrm{\prime }}$ lies below $\delta$.

Let $\delta <{\omega }_1$ be a nonzero limit and put $\gamma =\sup F\mathrm{‘}\mathrm{‘}\delta \le F(\delta )=\beta$. Suppose $\gamma <\beta$, and choose $p\in G$ containing $(\delta ,\beta )$. Below $p$, conditions which specify some $(\alpha ,\rho )$ with $\alpha <\delta$ and $\gamma <\rho <\beta$ are dense. Indeed, from any $s\le p$, take a normal extension $u$; continuity at $\delta$ gives an $\alpha <\delta$, beyond the finite lower domain of $s$, with $\gamma <u(\alpha )<\beta$, and add $(\alpha ,u(\alpha ))$. A generic containing $p$ meets this dense-below-$p$ set (equivalently, adjoin the conditions incompatible with $p$ to make it globally dense), contradicting the definition of $\gamma$. Hence $F(\delta )=\sup F\mathrm{‘}\mathrm{‘}\delta$, and $F$ is normal.

The conditions $r$ and $r^{\mathrm{\prime }}$ are compatible. To verify the point suppressed by the usual proof, choose a normal $u$ extending $r$. By elementarity choose a normal $v\in M$ extending $r^{\mathrm{\prime }}$. Both satisfy $u(\delta )=v(\delta )=\delta$: for $u$ this follows from $(\delta ,\delta )\in r$, and for $v$ by the calculation in step 1.1. Splice $v$ below and at $\delta$ with $u$ above $\delta$. The result is normal: both pieces agree at $\delta$, their values on the lower piece are below $\delta$, and replacing the lower piece by another sequence cofinal in $\delta$ does not change continuity at any later limit. It extends $r\cup r^{\mathrm{\prime }}$, so that finite union is a common condition. Therefore $D\cap M$ is predense below $q$. By F1 and F2, $q$ is an $(M,\mathbb{B})$-master below $p$, and $\mathbb{B}$ is proper.

The range $C=F\mathrm{‘}\mathrm{‘}{\omega }_1$ is unbounded because strict increase implies $F(\alpha )\ge \alpha$. It is closed: if $\eta <{\omega }_1$ is a limit point of $C$, then $\xi =\sup \{\alpha :F(\alpha )<\eta \}$ is a nonzero limit below ${\omega }_1$, and continuity and cofinality of the selected values give $F(\xi )=\eta$. Thus $C$ is club by F3.

Finally fix any ground-model normal function $a:{\omega }_1\to {\omega }_1$. The set $E_a=\{p\in \mathbb{B}:(\mathrm{\exists }\alpha \in \mathrm{dom}(p))p(\alpha )\mathrm{\ne }a(\alpha )\}$ is dense. Given $p$, choose a normal extension $u$, a successor $\alpha$ above its finite domain, and two successive values above $u(\alpha )$; at least one differs from $a(\alpha )$, and replacing the value at that new successor by the chosen larger value and continuing normally witnesses an extension in $E_a$. Genericity makes $F\mathrm{\ne }a$ for every ground normal $a$. If $C$ were in the ground model, its increasing enumeration would be a ground normal function and, as the unique increasing bijection from ${\omega }_1$ onto $C$, would equal $F$. This contradiction proves that the club $C$ is new. The forcing is nonempty (the empty map is greatest), and all finite, singleton, zero-coordinate, and limit-coordinate cases used above are included.

Because $\mathrm{cf}(\eta )=\omega$ and $\eta \in M$, choose in $M$ an increasing cofinal sequence $0={\eta }_0<{\eta }_1<\cdots <\eta ,$ and enumerate the dense subsets of $P_{\eta }$ which belong to $M$ as $⟨D_n:n<\omega ⟩$, repeating a dense set if necessary. Put $q_0=1_{P_0}$ and let ${\dot{p}}_0=\check{p}$. Then $q_0$ is the trivial $(M,P_0)$-master and forces $p_0\upharpoonright {\eta }_0\in G_0$.

Recursively suppose that $q_n$ is an $(M,P_{{\eta }_n})$-master and forces that $p_n\in P_{\eta }\cap M$ with $p_n\upharpoonright {\eta }_n\in G_{{\eta }_n}$. Work in a $P_{{\eta }_n}$-generic extension containing $q_n$ and resolve $p_n$. This value is a ground-model condition in $M$, so the ground set $E_n=\{u\in P_{{\eta }_n}:u\perp p_n\upharpoonright {\eta }_n\text{ or }(\mathrm{\exists }r\le p_n)[r\in D_n\mathrm{\&}u\le r\upharpoonright {\eta }_n]\}\mathrm{.}$ This set belongs to $M$ and is dense: below a condition compatible with $p_n\upharpoonright {\eta }_n$, take a common extension, paste it to the tail of $p_n$, and strengthen the resulting $P_{\eta }$-condition into $D_n$. By F2 the generic below $q_n$ meets $E_n\cap M$. Its member cannot take the incompatible alternative because $p_n\upharpoonright {\eta }_n$ is in the same generic. Elementarity therefore gives a name ${\dot{p}}_{n+1}$ forced to satisfy $p_{n+1}\in D_n\cap M,p_{n+1}\le p_n,p_{n+1}\upharpoonright {\eta }_n\in G_{{\eta }_n}\mathrm{.}$ Apply F1 from ${\eta }_n$ to ${\eta }_{n+1}$ to obtain an $(M,P_{{\eta }_{n+1}})$-master $q_{n+1}$ such that $q_{n+1}\upharpoonright {\eta }_n=q_n$ and $q_{n+1}$ forces $p_{n+1}\upharpoonright {\eta }_{n+1}\in G_{{\eta }_{n+1}}$.

Define $q={\bigcup }_{n<\omega }{q}_n\mathrm{.}$ The displayed coherence makes this a function whose restriction to every ${\eta }_n$ is exactly $q_n$. Its support is contained in the countable union of the countable supports of the $q_n$, hence is countable; cofinality of the ${\eta }_n$ leaves no unfilled coordinate below $\eta$. Thus F3 gives $q\in P_{\eta }$. This is the fusion step. It takes no lower bound of the sequence $⟨q_n(\xi ):n<\omega ⟩$ inside a single iterand: after coordinate $\xi$ first appears, later conditions preserve the already constructed initial segment containing it.

The conclusion that $q$ forces each $p_{n+1}$ into the full generic is the limit conclusion of F1 applied to exactly the recursion in steps 1.1--3.1. It does not follow merely from compatibility of all bounded restrictions, and no such inverse-limit compactness is asserted here. F1 therefore gives $q⊩p_{n+1}\in D_n\cap M\cap {\dot{G}}_{\eta }$ for every $n$. It follows that every $D_n\cap M$ is predense below $q$, so $q$ is an $(M,P_{\eta })$-master. Also F1 gives $q⊩\check{p}\in {\dot{G}}_{\eta }$, so $q$ is compatible with $p$. Choose a common extension $q^{\mathrm{\prime }}\le q,p$; mastery and all displayed forced conclusions persist below $q^{\mathrm{\prime }}$. Hence $q^{\mathrm{\prime }}$ is the promised master literally below $p$. The index $n=0$, the empty initial stage, one-coordinate supports, and a finite list of dense sets are all covered by the same recursion.

Write $T_{\alpha }$ for the $\alpha$th level. Under A1 choose for every $\alpha <{\omega }_1$ an injection $e_{\alpha }:T_{\alpha }\to \omega$. Then $t⟼(\mathrm{ht}(t),e_{\mathrm{ht}(t)}(t))$ injects $T$ into ${\omega }_1\times \omega$. Since $\omega \subseteq {\omega }_1$, F5 bounds this product by $\mathrm{\mid }{\omega }_1\times {\omega }_1\mathrm{\mid }={\omega }_1$. Hence $\mathrm{\mid }T\mathrm{\mid }\le {\omega }_1$, so the family $\mathcal{D}=\{D_t:t\in T\}$ has cardinality at most ${\omega }_1$.

By F3, $P(T)$ is ccc, and F2 makes it proper. It is nonempty because the empty finite function is its greatest condition. By F4 every member of $\mathcal{D}$ is dense. Reindex the distinct members of $\mathcal{D}$ along an ordinal $\lambda \le {\omega }_1$ using step 1.1, and apply F1 to obtain a filter $G\subseteq P(T)$ meeting every $D_t$. Since the family is nonempty, so is $G$; by the filter convention it is downward directed.

Put $f=\bigcup G$. If two conditions in $G$ assign a value to the same node, a common stronger member of $G$ extends both, so the values agree and $f$ is a function. Meeting $D_t$ puts every $t\in T$ in its domain. If $s{<}_{T}t$, choose members of $G$ mentioning $s$ and $t$ and then a common stronger member; its specializing-condition inequality gives $f(s)\mathrm{\ne }f(t)$. Thus $f:T\to \omega$ is total and specializes $T$, exactly as F4 asserts.

The dense family may have repetitions, but step 2.1 reindexes its distinct members and loses no requirement. A one-node level, the label $0$, and the empty initial condition are all allowed by F4. PFA itself chooses the filter; no generic filter over the universe is postulated. AC is used exactly in step 1.1 and in the reindexing in step 2.1, and is retained through A1.

Let $P=\mathrm{Fn}({\omega }_1,2,<{\omega }_1),$ ordered by reverse inclusion. By F2 its conditions are the countable partial functions from ${\omega }_1$ to $2$, and the empty function is its greatest condition. In particular, $P$ is nonempty. (It is not being identified with $\mathrm{Col}({\omega }_1,2)$, whose displayed parameters would violate that definition's requirement $\kappa \le \lambda$.)

For every $\alpha <{\omega }_1$, define $p_{\alpha }\in P$ on $\alpha +1$ by $p_{\alpha }(\xi )=\{\begin{array}{ll}0,& \xi <\alpha ,\\ 1,& \xi =\alpha \mathrm{.}\end{array}$ The domain is countable because $\alpha$ is a countable ordinal. If $\alpha <\beta <{\omega }_1$, then $p_{\alpha }(\alpha )=1$ whereas $p_{\beta }(\alpha )=0$. No function can extend both, so $p_{\alpha }$ and $p_{\beta }$ are incompatible. Consequently $\{p_{\alpha }:\alpha <{\omega }_1\}$ is an uncountable antichain, and F3 shows that $P$ is not ccc. Notice that $p_0=\{(0,1)\}$, so the zero endpoint also obeys the displayed definition.

Suppose $⟨q_n:n<\omega ⟩$ is descending in $P$. Reverse inclusion means $q_n\subseteq q_{n+1}$, so $q={\bigcup }_{n<\omega }{q}_n$ is a function extending every $q_n$. Each $\mathrm{dom}(q_n)$ is countable, and F4 with A1 makes their union countable. Hence $q\in P$ and $q\le q_n$ for every $n$. Thus $P$ is countably closed. Constant sequences, including the constant empty-condition sequence, are covered by the same union calculation.

By F1, the countably closed forcing $P$ is proper.

The forward implication, ccc implies proper, is true by F1. Steps 3.1 and 1.2 give one proper forcing that is not ccc, so the reverse implication and therefore the advertised equivalence are false. AC is spent only through the countable-union assertion in step 2.1; no generic filter or further choice is used in the antichain witness.