A tight strongly unbounded coloring gives finite-target AD guessing

We first record the elementary club tools used below. A1 supplies the countable choice needed by F3 and F4. By F4 and F5, $\kappa$ is regular uncountable and an uncountable subset of $\kappa$ is exactly an unbounded one. F7 therefore applies to countably many clubs, so a countable union of nonstationary sets is nonstationary: intersect clubs disjoint from its terms. A stationary set remains stationary after intersection with a club by the same finite-intersection fact. If $X\subseteq \kappa$ is unbounded, $\mathrm{acc}(X)=\{\delta \in E:\sup (X\cap \delta )=\delta \}$ is club: above any starting point take a strictly increasing sequence of points of $X$ and its countable supremum; closure follows by testing each bound below a limit point. In particular $E$ is club. Finally a regressive map $b:S\to \kappa$ on stationary $S\subseteq E$ has a stationary constant fiber. Otherwise choose a club $C_{\xi }$ disjoint from each fiber $b^{-1}\{\xi \}$ using A1; for $\alpha \in S\cap {\mathrm{△}}_{\xi <\kappa }{C}_{\xi }$, F8 gives $\alpha \in C_{b(\alpha )}$ because $b(\alpha )<\alpha$, contradicting the choice of that club.

Normalize the columns. An equal-column fiber is countable, since on an uncountable constant-column set each next-coordinate value set above a fixed prefix has at most one element, violating F2. The set of distinct columns has size ${\mathrm{\aleph }}_1$: choose each fiber's least old index; if there were only countably many fibers their union would be countable by step 1.1. Enumerate the representatives increasingly by $\kappa$; an uncountable subset of $\kappa$ has order type $\kappa$, since its initial segments are countable by F5. Reindex $c$ by these representatives. For every $T$, the old $[T{]}_c$ is uncountable exactly when uncountably many distinct columns have every prefix in $T$, exactly when the new $[T{]}_c$ is uncountable. Thus ${\mathcal{T}}_c$ and tightness persist, as does strong unboundedness on new indices. Henceforth the columns are distinct. Finite strings are countable, by grouping strings according to length plus sum of entries, with finite groups.

Every stationary $S\subseteq \kappa$ can be partitioned into $\kappa$ stationary pieces. By AC choose surjections $b_{\beta }:\omega \to \beta$ for every infinite $\beta <\kappa$. For $\xi <\kappa$ put $U_{\xi ,n}=\{\beta >\max (\xi ,\omega ):b_{\beta }(n)=\xi \}$. As $n$ varies these sets cover a tail of $\kappa$, so step 1.1 implies some $S\cap U_{\xi ,n}$ is stationary. Choose its least such $n=n(\xi )$. Some fixed $n$ is taken on an uncountable set of $\xi$, since a countable union of countable sets is countable. For this $n$ the corresponding stationary sets are pairwise disjoint, because a function has only one value at $n$. Reindex them by $\kappa$ and add all unused points of $S$ to the first part. Grouping the parts yields any specified nonzero number at most $\kappa$ of stationary pieces.

Construct a walk map $e$ locally. By AC choose for each $\beta \in E$ a strictly increasing cofinal sequence $C_{\beta }$ of order type $\omega$; choose an enumeration of $\beta$ and recursively pass above its next value to obtain it. Put $C_0=\mathrm{\varnothing }$ and $C_{\eta +1}=\{\eta \}$. For $\alpha <\gamma$, start at $\gamma$ and repeatedly pass from $\eta >\alpha$ to $\min (C_{\eta }\setminus \alpha )$. This point exists, lies at least at $\alpha$, and is strictly below $\eta$. The walk reaches $\alpha$ after finitely many steps, since an infinite decreasing ordinal sequence would have a least member followed by a smaller member. Let $W(\alpha ,\gamma )$ be its nonempty finite sequence of nodes before $\alpha$, and set $e(\alpha ,\gamma )=\max \{\mathrm{\mid }{C}_{\eta }\cap \alpha \mathrm{\mid }:\eta \in W(\alpha ,\gamma )\}$. Every count is finite: for a limit $\eta >\alpha$, only finitely many terms of its increasing cofinal $\omega$-sequence precede $\alpha$; a successor ladder is a singleton. Thus $e$ is natural-valued.

Fix $\beta \in E$ and $\gamma >\beta$. Let $P=W(\beta ,\gamma )$ and $j=e(\beta ,\gamma )$. The finite union of $C_{\eta }\cap \beta$ for $\eta \in P$ is bounded below $\beta$. Choose $ϵ<\beta$ at least as large as its members and with $\mathrm{\mid }{C}_{\beta }\cap ϵ\mathrm{\mid }\ge j$. If $ϵ<\alpha <\beta$, no $C_{\eta }$ at a node of $P$ meets $[\alpha ,\beta )$, so each walk step towards $\alpha$ agrees with its step towards $\beta$ until reaching $\beta$. Therefore $W(\alpha ,\gamma )$ is $P$ followed by $W(\alpha ,\beta )$. Also $C_{\eta }\cap \alpha =C_{\eta }\cap \beta$ on $P$, and $e(\alpha ,\beta )\ge \mathrm{\mid }{C}_{\beta }\cap \alpha \mathrm{\mid }\ge j$. Hence $e(\alpha ,\gamma )=\max (j,e(\alpha ,\beta ))=e(\alpha ,\beta )$ on that tail. Comparing both upper columns with the column at $\beta$ proves: for $\beta \le \gamma <\delta$, the maps $e(\cdot ,\gamma )$ and $e(\cdot ,\delta )$ agree eventually below $\beta$. Moreover $\mathrm{\mid }{C}_{\beta }\cap \alpha \mathrm{\mid }$ tends to infinity as $\alpha$ tends to $\beta$; the same equality shows that for every $\gamma \ge \beta$ and $m<\omega$, $\{\xi <\beta :e(\xi ,\gamma )\le m\}$ is bounded in $\beta$. These are the two walk properties needed below.

Choose a downward cofinal $\mathcal{U}\subseteq {\mathcal{T}}_c$ of size at most $\kappa$. Its nonempty finite lists have size at most $\kappa$: for a countable ordinal $\eta$, finite lists of indices below $\eta$ form a countable set; take the union over $\eta <\kappa$ and enumerate each countable block, using AC and $\mathrm{\mid }\kappa \times \omega \mathrm{\mid }={\mathrm{\aleph }}_1$. For this bound, well-order pairs first by their maximum coordinate and then lexicographically within each block. Every predecessor set is countable by F3 and F5, so its order type is below ${\omega }_1$ by F5. Sending each pair to that order type injects $\kappa \times \omega$ into $\kappa$; the reverse injection is $\eta \mapsto (\eta ,0)$. Split each $S\in \mathcal{P}$ into stationary pieces $S_{\sigma }$ indexed by these lists using step 2.2, with the notation retaining the original part $S$. Put $T^{\ast }=\{t:\mathrm{\mid }\{\beta :t\subseteq c_{\beta }\}\mathrm{\mid }={\mathrm{\aleph }}_1\}$. The union of the countable fibers of the remaining finite strings is countable by step 2.1, so choose $\rho <\kappa$ above all their indices. Thus every prefix of $c_{\beta }$ for $\beta \ge \rho$ belongs to $T^{\ast }$. This nonempty countable set is scheduled on successor ordinals by a map $f$: on successors $\omega \eta +n+1$ use the $n$th entry of a list of $T^{\ast }$ repeating each entry infinitely often. Every $t\in T^{\ast }$ occurs cofinally below every $\alpha \in E$. Indeed a terminal $\omega$-block gives infinitely many occurrences, and when there is no terminal block there are entire later blocks below $\alpha$. Recursively choose globally distinct indices ${\beta }_{\alpha ,j}\ge \rho$. If $\alpha \in S_{\sigma }$ for $\sigma =(T_0,\dots ,T_{r-1})$, let $r(\alpha )=r$ and choose ${\beta }_{\alpha ,j}\in [T_j{]}_c$; otherwise let $r(\alpha )=1$, choosing a column extending $f(\alpha )$ at a successor, and any column at zero. Each candidate set is uncountable and only countably many old indices are excluded, so the least eligible index exists. Let $d_{\alpha ,j}=c_{{\beta }_{\alpha ,j}}$; by step 2.1 all these functions are distinct.

Split $E$ into countably many stationary sets by step 2.2 and fix $z:\kappa \to \omega$ whose fiber at each $i$ contains the corresponding part. For $\xi <\alpha$ and $j<r(\alpha )$ define $h_{\alpha ,j}(\xi )=z(\min \{\gamma \in (\xi ,\alpha ]:\gamma =\alpha \text{ or }e(\gamma ,\alpha )\le \mathrm{\Delta }(d_{\xi ,0},d_{\alpha ,j})\})$. The second alternative is evaluated only when $\gamma <\alpha$; $\alpha$ is always a candidate. Distinctness in step 3.2 makes $\mathrm{\Delta }$ defined. Put $B_{\alpha ,j}^i=\{\xi <\alpha :h_{\alpha ,j}(\xi )=i,e(\xi ,\alpha )\le {\widehat{d}}_{\alpha ,j}(\mathrm{\Delta }(d_{\xi ,0},d_{\alpha ,j}))\}$ and $B_{\alpha }^i={\bigcup }_{j<r(\alpha )}{B}_{\alpha ,j}^i$. These definitions use only the natural-valued map of step 3.1.

If $\alpha \in E$, $\alpha \le \beta$, and $(\alpha ,j)\mathrm{\ne }(\beta ,j^{\mathrm{\prime }})$, then $B_{\alpha ,j}^i\cap B_{\beta ,j^{\mathrm{\prime }}}^{i^{\mathrm{\prime }}}$ is bounded in $\alpha$. To see this, put $n=\mathrm{\Delta }(d_{\alpha ,j},d_{\beta ,j^{\mathrm{\prime }}})$. If the intersection were cofinal, remove its bounded part where $e(\xi ,\alpha )\le {\widehat{d}}_{\alpha ,j}(n)$ using step 3.1. On the remaining cofinal set, the defining inequality and strict increase of the majorant force $\mathrm{\Delta }(d_{\xi ,0},d_{\alpha ,j})>n$. Thus $\mathrm{\Delta }(d_{\xi ,0},d_{\beta ,j^{\mathrm{\prime }}})=n$, so the other defining inequality gives $e(\xi ,\beta )\le {\widehat{d}}_{\beta ,j^{\mathrm{\prime }}}(n)$ on that cofinal set. This contradicts the bounded-sublevel property of step 3.1 at $\alpha$.

For $\alpha \in E$ and $i\mathrm{\ne }{i}^{\mathrm{\prime }}$, $B_{\alpha }^i\cap B_{\alpha }^{i^{\mathrm{\prime }}}$ is bounded in $\alpha$. Each intersection is a finite union over $j,j^{\mathrm{\prime }}$; the terms with $j\mathrm{\ne }{j}^{\mathrm{\prime }}$ are bounded by step 5.1, while a term with $j=j^{\mathrm{\prime }}$ is empty because $h_{\alpha ,j}$ cannot have two values. The same finite-union argument gives bounded intersections between $B_{\alpha }^i$ and $B_{\beta }^{i^{\mathrm{\prime }}}$ for $\alpha <\beta$ in $E$. Define $A_{\alpha }^i=B_{\alpha }^i\setminus {\bigcup }_{i^{\mathrm{\prime }}<i}{B}_{\alpha }^{i^{\mathrm{\prime }}}$. The members of each row are now disjoint, each loses only a bounded subset of $B_{\alpha }^i$, and the cross-row bounds persist. Cofinality is not yet asserted.

Fix uncountable targets $X_0,\dots ,X_{r-1}$, with $1\le r<\omega$, and a part $S\in \mathcal{P}$. Set $X_j^t=\{\xi \in X_j:t\subseteq d_{\xi ,0}\}$ and $V_j=\{t:X_j^t\text{ is uncountable}\}$. Removing the countable union of countable $X_j^t$ leaves uncountably many $\xi \in X_j$ all of whose column prefixes belong to $V_j$. Their distinct indices ${\beta }_{\xi ,0}$ belong to $[V_j{]}_c$, so $V_j\in {\mathcal{T}}_c$. Choose $T_j\in \mathcal{U}$ with $T_j\subseteq V_j$ and use the stationary piece $S_{\sigma }\subseteq S$ for $\sigma =(T_0,\dots ,T_{r-1})$. These are the columns and preliminary disjoint rows from steps 3.2 and 6.1.

Suppose the set of simultaneous guesses by these $A_{\alpha }^i$ is nonstationary. A club misses it. On its intersection with $S_{\sigma }$, some fixed pair $(i,j)\in \omega \times r$ fails on a stationary subset, by countable completeness from step 1.1. For such $\alpha$, $A_{\alpha }^i\cap X_j$ is bounded; step 6.1 implies $B_{\alpha ,j}^i\cap X_j$ is also bounded. Choose its supremum as a regressive bound, with value zero for an empty intersection. The pressing-down argument of step 1.1 yields a stationary $S_1\subseteq S_{\sigma }$ and fixed $b<\kappa$ with $B_{\alpha ,j}^i\cap X_j\subseteq b+1$ for every $\alpha \in S_1$.

Intersect the clubs $\mathrm{acc}(X_j^t)$ over those strings $t$ with uncountable $X_j^t$, obtaining a club $D$ by steps 1.1 and 2.1. Then $\mathrm{\Gamma }=D\cap E\cap z^{-1}\{i\}$ is stationary. Choose $\delta \in \mathrm{acc}(\mathrm{\Gamma })$ above $b$. For each $\alpha \in S_1$ above $\delta$, step 3.1 gives a threshold $a_{\alpha }<\delta$ such that $e(\xi ,\alpha )=e(\xi ,\delta )$ for $a_{\alpha }<\xi <\delta$. Since $\delta$ is countable, there is an uncountable $S_2\subseteq S_1\setminus (\delta +1)$ with a common threshold $a$. Choose $\gamma \in \mathrm{\Gamma }\cap \delta$ above $a,b$, and put $\nu =e(\gamma ,\delta )$. Thin $S_2$ to an uncountable $S_3$ on which $d_{\alpha ,j}\upharpoonright \nu$ is constant; there are only countably many strings of that length.

Apply strong unboundedness to the uncountable set of distinct indices $\{{\beta }_{\alpha ,j}:\alpha \in S_3\}$. Obtain a string $t$ of length $n$ with unbounded values $d_{\alpha ,j}(n)$ among its extensions in $S_3$. Necessarily $n\ge \nu$, since all earlier coordinates were fixed in step 9.1. At least one such column lies in $[T_j{]}_c$, so $t\in T_j\subseteq V_j$. Hence $X_j^t$ is uncountable and $\gamma \in D$ implies $X_j^t\cap \gamma$ cofinal in $\gamma$. The set $G=\{\xi <\gamma :e(\xi ,\delta )\le n\}$ is bounded below $\gamma$ by step 3.1. Choose $\xi \in X_j^t\cap \gamma$ above $a,b$ and every member of $G$. Choose $\alpha \in S_3$ with $t\subseteq d_{\alpha ,j}$ and $d_{\alpha ,j}(n)>\max (e(\xi ,\delta ),d_{\xi ,0}(n))$. Then $\mathrm{\Delta }(d_{\xi ,0},d_{\alpha ,j})=n$, and $e(\xi ,\alpha )=e(\xi ,\delta )<d_{\alpha ,j}(n)<{\widehat{d}}_{\alpha ,j}(n)$.

The minimum defining $h_{\alpha ,j}(\xi )$ is precisely $\gamma$. Indeed $e(\gamma ,\alpha )=e(\gamma ,\delta )=\nu \le n$, so $\gamma$ is a candidate. For $\xi <\eta <\gamma$ coherence gives $e(\eta ,\alpha )=e(\eta ,\delta )$, and $e(\eta ,\delta )\le n$ would put $\eta \in G$ above $\xi$, contrary to its choice. Also $\eta <\gamma <\alpha$, so the endpoint alternative is unavailable there. Thus $h_{\alpha ,j}(\xi )=z(\gamma )=i$. Together with step 10.1 this puts $\xi \in B_{\alpha ,j}^i\cap X_j$ above $b$, contradicting step 8.1. Therefore the simultaneous guesses are stationary for every finite list on every original part $S$. Every such successful row has all members cofinal, since the list is nonempty.

It remains to make every row cofinal, while preserving step 11.1. Call $\alpha \in E$ good if each $A_{\alpha }^i$ is cofinal. For a nongood $\alpha$, each prefix $d_{\alpha ,0}\upharpoonright n$ belongs to $T^{\ast }$ by step 3.2. The successor schedule gives cofinally many successors $\xi <\alpha$ with that prefix in $d_{\xi ,0}$, so $\mathrm{\Delta }(d_{\xi ,0},d_{\alpha ,0})\ge n$. Choose a strictly increasing sequence $({\xi }_n)$ cofinal in $\alpha$ with these differences tending to infinity, at stage $n$ passing above a fixed cofinal ladder's $n$th value as well as the preceding choice. Split its range into countably many disjoint infinite subsets and use them as the replacement row. All are cofinal, and their intersection with any smaller ordinal is finite. AC supplies these ladders and choices simultaneously.

A replaced upper row meets any lower row finitely; this handles also pairs of replaced rows. Consider a replaced lower row $\alpha$ and an unchanged upper row $\beta$. If its intersection with $A_{\beta }^i$ were cofinal in $\alpha$, the finite union defining $B_{\beta }^i$ would give one $j<r(\beta )$ with a cofinal intersection with $B_{\beta ,j}^i$. Put $n=\mathrm{\Delta }(d_{\alpha ,0},d_{\beta ,j})$. On a tail of the replacing ladder, step 12.1 gives $\mathrm{\Delta }(d_{\xi ,0},d_{\alpha ,0})>n$, hence $\mathrm{\Delta }(d_{\xi ,0},d_{\beta ,j})=n$. Membership in $B_{\beta ,j}^i$ then forces $e(\xi ,\beta )\le {\widehat{d}}_{\beta ,j}(n)$ on a cofinal subset of $\alpha$, contradicting step 3.1. Thus all cross-row intersections remain bounded.

Every final row is disjoint and cofinal by step 12.1, and intersections have the required bounds by steps 6.1 and 13.1. Every successful finite-target row of step 11.1 was good and therefore unchanged, so each stationary guessing set is preserved. This is exactly F1, on all of $E$ and every part of the given partition, proving the claim.