Proper Forcing, Countable-Support Iterations, and PFA

Fix $D\in M$ dense. If $D\cap M$ is predense below $q$, then conditions below $q$ that extend a condition of $D\cap M$ are dense below $q$; F2 gives $q⊩\dot{G}\cap D\cap M\mathrm{\ne }\mathrm{\varnothing }$. Conversely, if some $r\le q$ were incompatible with every member of $D\cap M$, then $r$ would force that intersection empty. This proves (1) if and only if (2).

Assume (1). The inclusion $M\subseteq M[G]\cap V$ follows from check names. For the reverse inclusion, let $r\le q$ force that a name $\dot{x}\in M$ equals a ground object $x$. Define $D_{\dot{x}}$ to contain (a) every $s$ for which some ground object $y$ satisfies $s⊩\dot{x}=\check{y}$, and (b) every $s$ below which no condition has property (a). This set is dense: from any condition, either an extension has property (a), or the original condition has property (b). By definability of forcing it belongs to $M$. Since $D_{\dot{x}}\cap M$ is predense below $q$, some $s\in D_{\dot{x}}\cap M$ is compatible with $r$. It cannot have property (b), because a common extension with $r$ would force $\dot{x}=\check{x}$ while admitting no ground-value extension. Hence $s$ has property (a); by elementarity its witness may be taken as some $y\in M$. A common extension of $r$ and $s$ forces both $\dot{x}=\check{x}$ and $\dot{x}=\check{y}$, so $x=y\in M$. Thus $q$ forces every ground member of $M[\dot{G}]$ to lie in $M$, proving (3). Statement (3) immediately implies (4), since ordinals are ground objects and check names give the opposite inclusion.

Assume (4), and let $A\in M$ be a maximal antichain. In $M$, use A1 to fix a bijection $e:\xi \to A$ from an ordinal $\xi$, and form by mixing the name $\dot{\beta }$ for the unique index of the member of $A\cap \dot{G}$. Then $\dot{\beta }\in M$ and $q⊩\dot{\beta }\in M\cap \mathrm{O}\mathrm{r}\mathrm{d}$ by (4). Consequently $q$ forces $e(\dot{\beta })\in A\cap M\cap \dot{G}$, so $A\cap M$ is predense below $q$. Every dense $D\in M$ contains, by elementarity and A1, such a maximal antichain $A\in M$; hence $D\cap M$ is predense below $q$ and (1) follows.

The all-model definition immediately gives the club formulation, since the countable elementary submodels of a fixed well-ordered $H_{\mu }$ structure form a club by F4 and A1. Conversely, suppose the good models contain a club in $[H_{\mu }{]}^{\omega }$, where $\mu >2^{\mathrm{\mid }P\mathrm{\mid }}$. Represent a subclub as the models closed under a function $F:H_{\mu }^{<\omega }\to H_{\mu }$. Choose $\lambda >\mu$ and a well-order ${<}_{\lambda }$ so that $F$ may be taken as the ${<}_{\lambda }$-least such witness. Every countable $M\prec (H_{\lambda },\in ,{<}_{\lambda },P,\dots )$ is then closed under $F$, so $N=M\cap H_{\mu }$ is a good club model. Every subset of $P$, and hence every dense set or maximal antichain in $M$, belongs to $H_{\mu }$; therefore $D\cap N=D\cap M$. An $(N,P)$-master below $p\in P\cap M$ is thus also an $(M,P)$-master. This proves the all-model formulation and completes both claimed equivalences. AC is used exactly for A1; no countable transitive model or generic filter is selected.

Suppose first that $P$ is ccc, let $M$ be a relevant countable elementary model, and fix $p\in P\cap M$. For each dense $D\in M$, elementarity and A1 give a maximal antichain $A\in M$ with $A\subseteq D$. By F2, $A$ is externally countable. Any externally countable set $A\in M$ is a subset of $M$: elementarity supplies in $M$ a surjection from $\omega$ onto $A$, and every natural number belongs to $M$. Hence $A\subseteq D\cap M$ is predense below every condition, so $p$ itself is $(M,P)$-generic and is a master below $p$. F1 proves that $P$ is proper.

Suppose instead that $P$ is countably closed. Enumerate all dense subsets of $P$ belonging to $M$ as $⟨D_n:n<\omega ⟩$, repeating one if the family is finite. Starting with $p_0=p$, use elementarity and A1 to choose $p_{n+1}\in D_n\cap M$ with $p_{n+1}\le p_n$; every $p_n$ stays in $M$. By F3 there is $q\le p_n$ for all $n$. For every $D_n$, the condition $p_{n+1}\in D_n\cap M$ lies above $q$, so $D_n\cap M$ is predense below $q$. Thus $q$ is an $(M,P)$-master below $p$, and F1 again makes $P$ proper.

The two arguments cover the ccc and countably closed hypotheses independently and use no converse. AC is spent exactly in the maximal-antichain and enumeration/recursive-choice operations identified in steps 1.1 and 1.2. Therefore every forcing in either class is proper.

Choose a sufficiently large well-ordered $H_{\theta }$ structure containing $P,p,S,\dot{C}$, and fix Skolem functions for it. We first derive the elementary-model trace fact needed here. For $\beta <{\omega }_1$, let $M_{\beta }$ be the Skolem hull of $\beta \cup \{P,p,S,\dot{C}\}$ and put $f(\beta )=\sup (M_{\beta }\cap {\omega }_1)+1<{\omega }_1\mathrm{.}$ The set $E$ of nonzero limit ordinals $\delta <{\omega }_1$ closed under $f$ is club. If $\delta \in E$, finite character of Skolem terms gives $M_{\delta }={\bigcup }_{\beta <\delta }{M}_{\beta }$, so $M_{\delta }\cap {\omega }_1=\delta$. By stationarity choose $\delta \in S\cap E$ and set $M=M_{\delta }$. Then $M$ is countable elementary, contains all the required parameters, and has trace $M\cap {\omega }_1=\delta$. Properness supplies an $(M,P)$-master $q\le p$.

In $M$ choose a name $\dot{f}$ which $p$ forces to be the increasing continuous enumeration of $\dot{C}$. For every $\alpha <\delta$, one has $\alpha \in M$ and hence the ordinal name $\dot{f}(\check{\alpha })$ belongs to $M$. By F1, $q$ forces its value into $M\cap {\omega }_1=\delta$. Thus $q$ forces $\dot{f}\mathrm{‘}\mathrm{‘}\delta \subseteq \delta$. Since an increasing enumeration satisfies $\dot{f}(\alpha )\ge \alpha$, its first $\delta$ values are cofinal in $\delta$; closure of $\dot{C}$ then gives $q⊩\delta \in \dot{C}$. As $\delta \in S$ is a ground ordinal, $q⊩\dot{C}\cap \check{S}\mathrm{\ne }\mathrm{\varnothing }$.

The choices of $p$ and the club name were arbitrary, so no condition can force a ground stationary $S$ to become nonstationary. To see preservation of ${\omega }_1$ without a hidden cofinality inference, let $p$ force that $\dot{g}:\omega \to {\omega }_1^V$ is any function, choose a relevant countable model $M$ containing $p,\dot{g}$, and use properness to choose an $(M,P)$-master $q\le p$. F1 then forces each $\dot{g}(n)$ into the fixed countable ordinal $M\cap {\omega }_1$, so the range is bounded and $\dot{g}$ is not cofinal, hence not surjective. Therefore ${\omega }_1^V$ remains uncountable and equals the extension's ${\omega }_1$. AC is used exactly in A1.

We prove the displayed extension property by transfinite induction on $\alpha$. At $\alpha =\gamma$ take $q=q_0$: the hypothesis already says $q_0$ forces $\dot{p}\in {\dot{G}}_{\gamma }$. Assume as induction hypothesis that the property holds at every smaller iteration length.

We record the name-selection argument used below. Suppose $u$ forces that there is a set $x$ satisfying a fixed formula $\phi (x)$. By the existential forcing clause, the conditions below $u$ that force $\phi (\dot{x})$ for some name $\dot{x}$ are dense below $u$. Use A1 to choose a maximal antichain $A$ of such conditions and, for each $a\in A$, one witness name ${\dot{x}}_a$. The usual mixed name $\dot{x}={\bigcup }_{a\in A}({\dot{x}}_a\upharpoonright a)$ agrees with ${\dot{x}}_a$ below $a$. Thus the conditions forcing $\phi (\dot{x})$ are dense below $u$, and the forcing definition gives $u⊩\phi (\dot{x})$. This derives the needed maximum principle from the forcing clauses and AC rather than attributing it to F4.

Let $\alpha =\beta +1$. Apply the induction hypothesis at $\beta$ to obtain an $(M,P_{\beta })$-generic $q_{\beta }$ extending $q_0$ and forcing $\dot{p}\upharpoonright \beta \in {\dot{G}}_{\beta }$. In a $P_{\beta }$-extension containing $q_{\beta }$, the last coordinate $p(\beta )$ belongs to $M[G_{\beta }]\cap Q_{\beta }$. Since $Q_{\beta }$ is proper there, choose an $(M[G_{\beta }],Q_{\beta })$-master $q_{\beta }^{\ast }$ below it, and apply step 1.2 to choose a name for this condition. By F3, $(q_{\beta },{\dot{q}}_{\beta }^{\ast })$ forces $\dot{p}$ into the two-step generic. It is $(M,P_{\beta +1})$-generic: for any ordinal-valued $P_{\beta +1}$-name in $M$, the quotient master forces its value into $M[G_{\beta }]$, and the first-stage master then forces that ground ordinal into $M$; F2 applies. This gives the successor case.

Now let $\alpha$ be limit. The case $\gamma =\alpha$ was settled at step 1.1, so assume $\gamma <\alpha$ and put $\rho =\sup (M\cap \alpha )$. Choose an increasing sequence $⟨{\gamma }_n:n<\omega ⟩$ from $M\cap (\alpha +1)$ with ${\gamma }_0=\gamma$ and supremum $\rho$, and enumerate the dense subsets of $P_{\alpha }$ in $M$ as $⟨D_n:n<\omega ⟩$. Recursively construct $(M,P_{{\gamma }_n})$-generic $q_n$ and $P_{{\gamma }_n}$-names ${\dot{p}}_n$, beginning with the given pair, so that $q_{n+1}\upharpoonright {\gamma }_n=q_n$ and $q_n$ forces: $p_n\in P_{\alpha }\cap M$; $p_n\le p_{n-1}$ and $p_n\in D_{n-1}$ for $n>0$; and $p_n\upharpoonright {\gamma }_n\in G_{{\gamma }_n}$. For the recursive step, work in a $P_{{\gamma }_n}$-generic extension containing $q_n$ and resolve $p_n\in P_{\alpha }\cap M$. In the ground model define $E=\{u\in P_{{\gamma }_n}:u\perp p_n\upharpoonright {\gamma }_n\text{ or }(\mathrm{\exists }r\le p_n)[r\in D_n\mathrm{\&}u\le r\upharpoonright {\gamma }_n]\}\mathrm{.}$ The set $E$ belongs to $M$ and is dense: below a condition compatible with $p_n\upharpoonright {\gamma }_n$, first take a common extension, paste it to the tail of $p_n$, and then strengthen the resulting $P_{\alpha }$-condition into $D_n$. Since $q_n$ is an $(M,P_{{\gamma }_n})$-master, the generic meets $E\cap M$. Its member cannot take the incompatible alternative because $p_n\upharpoonright {\gamma }_n$ is in the same generic. Elementarity therefore supplies $p_{n+1}\in D_n\cap M$ below $p_n$ whose restriction lies in the generic. Apply step 1.2 to name that choice, then apply the induction hypothesis at ${\gamma }_{n+1}<\alpha$ to obtain $q_{n+1}$.

Define $q$ on $\rho$ by $q={\bigcup }_n{q}_n$ and fill every coordinate in $[\rho ,\alpha )$ with its supplied top name. This is a condition: the equalities $q_{n+1}\upharpoonright {\gamma }_n=q_n$ make the union a coherent function, and F6 makes its support, a subset of ${\bigcup }_n\mathrm{supp}(q_n)$, countable. This is the only fusion operation; no coordinatewise lower bound in an arbitrary proper iterand is used. To check what $q$ forces, take any $P_{\alpha }$-generic $G$ containing it and resolve the names $p_n$. For $k\ge n$, the construction and truth lemma give $p_n\upharpoonright {\gamma }_k\in G_{{\gamma }_k}$. Also $p_n\in M$, so the countable set $\mathrm{supp}(p_n)$ belongs to $M$ and is a subset of $M\cap \alpha \subseteq \rho$; hence $p_n\upharpoonright \rho =p_n$. The inverse-limit generic is determined on a condition by these cofinal projections, so $p_n\in G_{\rho }\subseteq G_{\alpha }$. Thus $q$ forces $p_n\in G_{\alpha }$ for every $n$, in particular $p_0=p$.

Step 3.1 shows that $q$ forces $p_{n+1}\in D_n\cap M\cap G_{\alpha }$ for every $n$, so every $D_n\cap M$ is predense below $q$; F2 makes $q$ $(M,P_{\alpha })$-generic. Its restriction to ${\gamma }_0$ is $q_0$, and step 3.1 gives $q⊩\dot{p}\in {\dot{G}}_{\alpha }$. The base, successor, and limit cases exhaust the induction, so the lemma holds for every $\alpha$. AC is used exactly in A1, including the countable-support union through F6.

Fix $\alpha \le \delta$ and a sufficiently large well-ordered $H_{\theta }$ containing the full iteration and $\alpha$. The countable elementary submodels containing these fixed parameters form a club. Fix one such $M$ and $p\in P_{\alpha }\cap M$; then $P_{\alpha }$ and the restricted iteration belong to $M$. At the trivial stage $P_0$, its unique condition $q_0$ is $(M,P_0)$-generic, and the canonical $P_0$-name $\check{p}$ is forced to belong to $P_{\alpha }\cap M$ with trivial restriction in $G_0$. Apply F1 with $\gamma =0$ to obtain an $(M,P_{\alpha })$-generic $q$ such that $q⊩\check{p}\in {\dot{G}}_{\alpha }$. Hence $q$ and $p$ are compatible: otherwise directedness of a generic filter would make $q$ force $\check{p}\mathrm{\notin }{\dot{G}}_{\alpha }$. Choose a common extension $q^{\mathrm{\prime }}\le q,p$. Predensity below $q$ persists below the stronger condition $q^{\mathrm{\prime }}$, so $q^{\mathrm{\prime }}$ is still $(M,P_{\alpha })$-generic and is now literally below $p$.

Step 1.1 proves the master condition on the club of models containing the full iteration and $\alpha$; F3 converts this to the all-model formulation in F2. Thus $P_{\alpha }$ is proper. Since $\alpha \le \delta$ was arbitrary and the hypotheses restrict to every initial segment, every $P_{\alpha }$, including $P_{\delta }$, is proper. Successor lengths, limits of countable cofinality, and limits where $M\cap \alpha$ is bounded are already the exhaustive cases in F1; no closure of the individual iterands is assumed. AC is used only as recorded in A1 and in the supplier F1.

Let $P$ be ccc and let $\mathcal{D}$ be a family of at most ${\omega }_1$ dense subsets of $P$. By F2, $P$ is proper, so F1 gives a filter meeting all members of $\mathcal{D}$. This is precisely $\mathrm{M}\mathrm{A}({\mathrm{\aleph }}_1)$.

Applying F3 to step 1.1 shows that no Suslin tree exists.

By F4, nonexistence of Suslin trees is equivalent to nonexistence of Suslin lines, which is the Suslin Hypothesis. Thus PFA implies both asserted conclusions. No value of the continuum was used.

Fix a large regular $\theta$. For every countable $\mathcal{X}\subseteq \mathcal{I}$, use F2 and A1 to fix $I_{\mathcal{X}}\in \mathcal{I}$ with $a{\subseteq }^{\ast }{I}_{\mathcal{X}}$ for all $a\in \mathcal{X}$, and put $I_N=I_{N\cap \mathcal{I}}$ for a countable $N\prec H_{\theta }$. Define $Q_{\mathcal{I}}$ as follows. A condition $p=(Z_p,{\mathcal{N}}_p)$ has finite $Z_p\subseteq S$ and a finite membership-chain ${\mathcal{N}}_p$ of countable elementary submodels containing $\mathcal{I}$; distinct points of $Z_p$ are separated by some $N\in {\mathcal{N}}_p$; and if $N\in {\mathcal{N}}_p$ and $X\in N\cap {\mathcal{I}}^{\perp }$, then $X\cap Z_p\subseteq N$. Put $p\le q$ when $Z_q\subseteq Z_p$, ${\mathcal{N}}_q\subseteq {\mathcal{N}}_p$, and $(Z_p\setminus Z_q)\cap N\subseteq I_N$ for every $N\in {\mathcal{N}}_q$. These clauses are preserved by extension and make the empty pair a greatest condition.

We verify properness, including the combinatorial compatibility step. Let $M$ be suitable, $p_0\in Q_{\mathcal{I}}\cap M$, and add $N=M\cap H_{\theta }$ to its side chain; the result $q$ is a condition below $p_0$. Fix once and for all the well-order of $H_{\theta }$ carried by the elementary structure. For a condition $u$ and a model $K\in {\mathcal{N}}_u$, define the condition trace $u\upharpoonright K=(Z_u\cap K,{\mathcal{N}}_u\cap K),$ and enumerate the finite set $Z_u\setminus K$ in the fixed well-order, writing $t_{u,K}\in S^{\mathrm{\mid }{Z}_u\setminus K\mathrm{\mid }}$ for the resulting tuple. It suffices by F3 to take $r\le q$ and dense $D\in M$, first strengthen $r$ into $D$, and then find a member of $D\cap M$ compatible with this strengthening; rename the strengthened condition $r$. Put $r_0=r\upharpoonright N$ and $n=\mathrm{\mid }{Z}_r\setminus N\mathrm{\mid }$. By elementarity, restrict $D$ to the conditions $s$ carrying a distinguished $N_s\in {\mathcal{N}}_s$ such that $s\upharpoonright N_s=r_0$ and $\mathrm{\mid }{Z}_s\setminus N_s\mathrm{\mid }=n$; the witnesses for $r$ are $N_r=N$ and $t_{r,N}$. Let $T=\{t_{s,N_s}:s\in D\}\subseteq S^n$ and let $\mathcal{J}$ be the sigma-ideal generated by ${\mathcal{I}}^{\perp }$. For $U\subseteq S^n$, let $\mathrm{\partial }U$ retain the tuples $u$ for which, at every coordinate $k<n$, the fibre of possible $k$th entries above $u\upharpoonright k$ is $\mathcal{J}$-positive. The derivative claim in Moore's cited tutorial says that $T^{\mathrm{\prime }}={\mathrm{\partial }}^{n}T$ is a nonempty ${\mathcal{J}}^{+}$-splitting member of $M$ and contains the external tuple $t_{r,N}$. Its finite induction uses the membership chain and clause 4 of the forcing: if $t_{r,N}$ first disappeared, the least bad fibre's countable ${\mathcal{I}}^{\perp }$ decomposition, coded in the relevant side model, would put one of the corresponding outside points of $Z_r$ in a member of ${\mathcal{I}}^{\perp }$ from that model, contrary to clause 4. In particular, this claim does not assume that $t_{r,N}$ itself belongs to $M$. Starting with the empty tuple, choose successively in $M$ an initial segment extendible in $T^{\mathrm{\prime }}$. Its next-coordinate set $C\in M$ is not in $\mathcal{J}$, hence is not orthogonal to $\mathcal{I}$; elementarity gives an infinite $H\in \mathcal{I}\cap M$ with $H\subseteq C$. For every one of the finitely many outer models $P\in {\mathcal{N}}_r\setminus M$, membership-chain coherence gives $H\in P\cap \mathcal{I}$, so $H{\subseteq }^{\ast }{I}_P$. Choose the next coordinate in $H\cap {\bigcap }_{P\in {\mathcal{N}}_r\setminus M}{I}_P$. After $n$ choices, elementarity supplies $s\in D\cap M$ whose tuple $t_{s,N_s}$ is the chosen one, and hence $Z_s\setminus N_s$ is contained in every $I_P$. Then $(Z_s\cup Z_r,{\mathcal{N}}_s\cup {\mathcal{N}}_r)$ is a common extension. Thus $q$ is an $(M,Q_{\mathcal{I}})$-master and $Q_{\mathcal{I}}$ is proper.

If $S$ is a countable union of members of ${\mathcal{I}}^{\perp }$, PID's second alternative holds. Otherwise choose a suitable countable $M$ and $x\in S\setminus \bigcup (M\cap {\mathcal{I}}^{\perp })$. Then $q=(\{x\},\{M\cap H_{\theta }\})$ is a condition and, by step 2.1, an $M$-master. For a $Q_{\mathcal{I}}$-generic $G$ containing $q$, set $\dot{Z}={\bigcup }_{p\in \dot{G}}{Z}_p$ and $\dot{\mathcal{N}}={\bigcup }_{p\in \dot{G}}{\mathcal{N}}_p$. The master condition forces $\dot{Z}$ uncountable: if an $M$-name enumerated it countably, F3 would put all of its ground points in $M$, contrary to $x\in \dot{Z}\setminus M$. Properness gives the ground-model countable-covering property by the same master-name argument, so every countable subset of $\dot{Z}$ is contained in some side model $N\in \dot{\mathcal{N}}$. The order clause gives $N\cap \dot{Z}{\subseteq }^{\ast }{I}_N$, and therefore $N\cap \dot{Z}\in \mathcal{I}$. Hence $q$ forces every countable subset of $\dot{Z}$ to lie in $\mathcal{I}$.

Work in the proper cone $Q_{\mathcal{I}}\upharpoonright q$. Choose names $\dot{f}:{\omega }_1\to \dot{Z}$ and $\dot{g}:{\omega }_1\to \mathcal{I}$ such that $q$ forces $\dot{f}$ injective and $\dot{f}\mathrm{‘}\mathrm{‘}\xi \subseteq \dot{g}(\xi )$ for every $\xi <{\omega }_1$; step 3.1 supplies them. For each $\xi$, the set $D_{\xi }$ of conditions deciding both values is dense. By F1 there is a filter meeting all $D_{\xi }$. Compatibility within the filter makes the decided values coherent, producing in the ground universe an injection $f:{\omega }_1\to S$ and $g:{\omega }_1\to \mathcal{I}$ with $f\mathrm{‘}\mathrm{‘}\xi \subseteq g(\xi )$. Put $Z=f\mathrm{‘}\mathrm{‘}{\omega }_1$. If $a\subseteq Z$ is countable, the set of its $f$-indices is bounded by some $\xi <{\omega }_1$, so $a\subseteq f\mathrm{‘}\mathrm{‘}\xi \subseteq g(\xi )\in \mathcal{I}$ and downward closure gives $a\in \mathcal{I}$. Thus $Z$ witnesses PID's first alternative. Together with the first sentence of step 3.1, this proves PID. AC is used exactly through A1 and the stated ZFC suppliers.

Let $P$ consist of pairs $(s,F)$ with $s\in [\omega {]}^{<\omega }$ and $F\in [\mathcal{A}{]}^{<\omega }$. Put $(t,G)\le (s,F)$ exactly when $s\subseteq t$, $F\subseteq G$, and $t\setminus s\subseteq \bigcap F$, taking $\bigcap \mathrm{\varnothing }=\omega$. For a fixed finite stem $s$, every finite collection of conditions with that stem has the common extension whose side set is the union of their side sets. Since there are countably many finite subsets of $\omega$, $P$ is sigma-centered and therefore ccc.

For each $a\in \mathcal{A}$, the set $E_a=\{(s,F):a\in F\}$ is dense, because adding $a$ to $F$ changes no stem. For each $n<\omega$, let $D_n=\{(s,F):(\mathrm{\exists }k\in s)k\ge n\}$. Given $(s,F)$ outside $D_n$, the strong finite intersection property makes $\bigcap F$ infinite, so choose $k\in \bigcap F$ with $k\ge n$ and extend the stem by $k$; hence $D_n$ is dense. The family of all $E_a$ and $D_n$ has cardinality at most ${\omega }_1$.

By F2 and F3, choose a filter $G\subseteq P$ meeting every set from step 2.1, and put $b=\bigcup \{s:(s,F)\in G\}$. Meeting all $D_n$ makes $b$ unbounded in $\omega$, hence infinite. Fix $a\in \mathcal{A}$ and choose $(s,F)\in G\cap E_a$. For any $(t,H)\in G$, directedness gives $(u,K)\in G$ below both; the order relative to $(s,F)$ gives $u\setminus s\subseteq \bigcap F\subseteq a$, and $t\subseteq u$. Thus $t\setminus a\subseteq s$, and after taking the union, $b\setminus a\subseteq s$ is finite. Therefore $b{\subseteq }^{\ast }a$ for every $a\in \mathcal{A}$, so $b$ is an infinite pseudointersection.

Since every at-most-${\omega }_1$ strong-finite-intersection family has such a pseudointersection, no family witnessing the definition of $\mathfrak{p}$ has cardinality at most ${\omega }_1$. By F1, $\mathfrak{p}>{\omega }_1$. Empty and finite $\mathcal{A}$ are included: the same forcing works, and for $\mathcal{A}=\mathrm{\varnothing }$ the constructed $b$ is simply infinite.

Assume for contradiction that some subspace $W\subseteq K$ is not Lindelöf. Regularity, Hausdorffness, and hereditary separability pass to subspaces, so replace $K$ by $W$. Choose an open cover with no countable subcover. Recursively for $\alpha <{\omega }_1$, select a cover member $U_{\alpha }$ and $x_{\alpha }\in U_{\alpha }$ outside ${\bigcup }_{\beta <\alpha }{U}_{\beta }$. Let $X=\{x_{\alpha }:\alpha <{\omega }_1\}$ and relabel $U_{x_{\alpha }}=U_{\alpha }$; then $U_x\cap X$ is countable for every $x\in X$. By regularity choose open $V_x$ with $x\in V_x\subseteq \overline{V_x}\subseteq U_x$. Define $\mathcal{I}=\{A\in [X{]}^{\le \omega }:(\mathrm{\forall }x\in X)\mathrm{\mid }A\cap \overline{V_x}\mathrm{\mid }<\omega \}$. It is an ideal containing all finite sets.

We first derive the needed domination fact from $\mathfrak{p}$. If $\mathcal{F}\subseteq {\omega }^{\omega }$ has size less than $\mathfrak{p}$, consider, on the countable set ${\omega }^{<\omega }$, the sets $A_f=\{s:(\mathrm{\forall }i<\mathrm{\mid }s\mathrm{\mid })f(i)\le s(i)\}$ for $f\in \mathcal{F}$ and $C_n=\{s:\mathrm{\mid }s\mathrm{\mid }\ge n\}$. Every finite intersection is infinite. A pseudointersection $B$ exists by the definition of $\mathfrak{p}$; thin it to distinct $s_n$ with $\mathrm{\mid }{s}_n\mathrm{\mid }>n$, and put $g(n)=s_n(n)$. Since $B{\subseteq }^{\ast }{A}_f$, $g$ eventually dominates every $f\in \mathcal{F}$. Now take $A_n\in \mathcal{I}$, replace them by their increasing finite unions, and enumerate each infinite $A_n$ as $\{a_{n,k}:k<\omega \}$. For each $x$, choose $f_x(n)$ past the finite set $A_n\cap \overline{V_x}$. As $\mathrm{\mid }X\mathrm{\mid }={\omega }_1<\mathfrak{p}$, choose one eventual dominator $g$ for all $f_x$, and set $A={\bigcup }_n\{a_{n,k}:k\ge g(n)\}$, ignoring finite $A_n$. Each $A_n{\subseteq }^{\ast }A$, while for fixed $x$ all sufficiently large rows avoid $\overline{V_x}$ and the finitely many remaining rows meet it finitely. Thus $A\in \mathcal{I}$, proving that $\mathcal{I}$ is a P-ideal.