Stick gives strong diagonal almost-disjoint guessing

Set $x_{\gamma }={\bigcup }_{\xi \in s_{\gamma }}{s}_{\xi }$. Each is countably infinite by F2. We prove that this single sequence, fixed independently of later targets, has the following stronger property: for every sequence $(H_{\alpha }{)}_{\alpha <\kappa }$ of countable sets with pairwise finite intersections, and every uncountable $X\subseteq \kappa$, some $x_{\gamma }\subseteq X$ has infinite remainder after removal of any finite union of the $H_{\alpha }$. For these fixed $H,X$, recursively choose $s_{{\xi }_i}\subseteq X$ disjoint from all earlier $s_{{\xi }_j}$ and all earlier selected $H_{{\zeta }_j}$, for $i<\kappa$; let ${\zeta }_i$ be the least $\alpha$ for which $H_{\alpha }\cap s_{{\xi }_i}$ is infinite, if one exists. At each stage the excluded union is countable by F2 and F4, so the remaining target is uncountable and stick supplies a choice; choose the least eligible stick index. F5 gives the recursion. The ${\xi }_i$ are distinct, as are the defined ${\zeta }_i$.

Apply stick to the uncountable set $\{{\xi }_i:i<\kappa \}$ and take $s_{\gamma }$ contained in it. Then $x_{\gamma }\subseteq X$. Suppose $x_{\gamma }\setminus {\bigcup }_{\alpha \in a}{H}_{\alpha }$ were finite for some finite $a$. For each of the infinitely many $i$ with ${\xi }_i\in s_{\gamma }$, the infinite set $s_{{\xi }_i}$ is almost contained in that finite union, so it meets some member infinitely and ${\zeta }_i$ is defined. Moreover $H_{{\zeta }_i}\cap s_{{\xi }_i}$ is infinite and almost contained in the same finite union; it meets some $H_{\alpha }$, $\alpha \in a$, infinitely. Pairwise finite intersections force ${\zeta }_i=\alpha \in a$. This puts infinitely many distinct ${\zeta }_i$ in the finite set $a$, an impossibility. Thus the strengthened property holds, including the empty finite union.

Fix a bijection $\pi :\omega \to \omega \times \omega$ by listing pairs in successive finite diagonals, and write its coordinates as ${\pi }_0,{\pi }_1$. Using AC and F4, choose surjections $q_{\alpha }:\omega \to \alpha$ for every $0<\alpha <\kappa$. Also choose increasing cofinal $\omega$-ladders for all $\alpha \in E$: from $q_{\alpha }$ choose successively a point above the preceding point and $q_{\alpha }(n)$, which is possible because $\alpha$ is limit. Recursively define countable $A_{\alpha }\subseteq \alpha$, starting with $A_0=\mathrm{\varnothing }$ and setting $A_{\eta +1}=\{\eta \}$. At $\alpha \in E$, for each $j<\omega$ put $Y_{\alpha ,j}=(x_{q_{\alpha }({\pi }_0(j))}\cap \alpha )\setminus {\bigcup }_{j^{\mathrm{\prime }}\le j}{A}_{q_{\alpha }(j^{\mathrm{\prime }})},J_{\alpha }=\{j:Y_{\alpha ,j}\text{ is infinite}\}\mathrm{.}$ In increasing order of $j\in J_{\alpha }$ select the least ${\xi }_{\alpha ,j}\in Y_{\alpha ,j}$ different from all earlier selections. Only finitely many selections precede stage $j$, so the choice exists. Put $R_{\alpha }^i=\{{\xi }_{\alpha ,j}:j\in J_{\alpha },{\pi }_1(j)=i\}$. If all these sets are cofinal in $\alpha$, call $\alpha$ good and set $A_{\alpha }^i=R_{\alpha }^i$, $A_{\alpha }={\bigcup }_i{R}_{\alpha }^i$. Otherwise let $A_{\alpha }$ be the fixed cofinal ladder and partition it into countably many disjoint infinite subsets $A_{\alpha }^i$, using the fibers of ${\pi }_0$ on its increasing enumeration. Each is cofinal. The data and least-choice rules make this an instance of F5.

Each final limit row is cofinal and disjoint. For $\beta <\alpha$, if $\alpha$ is a successor, $A_{\alpha }$ is a singleton; if it is a nongood limit, its increasing ladder meets $\beta$ in a finite set. If it is good, choose $j^{\mathrm{\prime }}$ with $q_{\alpha }(j^{\mathrm{\prime }})=\beta$. For every $j\in J_{\alpha }$ with $j\ge j^{\mathrm{\prime }}$, the definition of $Y_{\alpha ,j}$ excludes $A_{\beta }$. Thus $A_{\alpha }\cap A_{\beta }$ is contained in the finite set of selections at indices $j<j^{\mathrm{\prime }}$. Consequently $(A_{\alpha }{)}_{\alpha <\kappa }$ satisfies the hypothesis on $H$ in step 2.1.

Fix an uncountable $X\subseteq \kappa$. For each $ϵ<\kappa$, apply step 2.1 to the now completed sequence $H_{\alpha }=A_{\alpha }$ and to $X\setminus (ϵ+1)$, which is uncountable by F4. Choose the least ${\beta }_{ϵ}$ such that $x_{{\beta }_{ϵ}}\subseteq X\setminus (ϵ+1)$ has infinite remainder after every finite union of the $A_{\alpha }$. F3 supplies an ordinal $F(ϵ)<\kappa$ strictly above ${\beta }_{ϵ}$ and every member of $x_{{\beta }_{ϵ}}$. The set $D_X=\{\delta \in E:(\mathrm{\forall }ϵ<\delta )F(ϵ)<\delta \}$ is club. To prove unboundedness above any $b<\kappa$, recursively take increasing countable ordinals $a_n$ above $b$ such that $a_{n+1}>F(ϵ)$ for all $ϵ<a_n$; F2–F4 keep this possible. Their supremum $\delta <\kappa$ belongs to $E$ and closes under $F$. For closedness, if $\delta$ is a limit point of $D_X$ and $ϵ<\delta$, take $\eta \in D_X$ with $ϵ<\eta <\delta$; then $F(ϵ)<\eta <\delta$.

For $\delta \in D_X$, $i<\omega$ and $b<\delta$, choose $b<ϵ<\delta$. Then ${\beta }_{ϵ}<\delta$ and $x_{{\beta }_{ϵ}}\subseteq \delta$. Choose $k$ with $q_{\delta }(k)={\beta }_{ϵ}$ and the unique $j$ with $\pi (j)=(k,i)$. The set $Y_{\delta ,j}$ is exactly $x_{{\beta }_{ϵ}}$ minus a finite union of earlier $A$ sets, hence is infinite. Thus $j\in J_{\delta }$ and ${\xi }_{\delta ,j}\in R_{\delta }^i\cap X$ lies above $b$. Every $R_{\delta }^i$ meets $X$ cofinally, so $\delta$ is good and $\sup (A_{\delta }^i\cap X)=\delta$ for all $i$. In particular the recursive replacement rule has not removed these guesses.

Finally, given $(X_{\nu }{)}_{\nu <\kappa }$, obtain the clubs $D_{X_{\nu }}$ from step 5.1 and let $D$ be their diagonal intersection. The boundedness property F3 together with F4 says $\kappa$ is regular uncountable, so F6 applies. For $\delta \in D\cap E$ and $\nu <\delta$, $\delta \in D_{X_{\nu }}$; step 6.1 gives $\sup (A_{\delta }^i\cap X_{\nu })=\delta$ for every $i$. Intersect $D\cap E$ with any stationary $S\in \mathcal{P}$; this is stationary, because it meets every club after a finite club intersection (or directly because the intersection of two clubs is club). Cofinality, disjointness and finite cross-row intersections are step 4.1. This is precisely the strong diagonal property, whose finite-target consequence is F1, clause 4.