Moore's club extension lemma

Fix a sufficiently large regular $\mathrm{\Theta }$ and Skolem functions for $H(\mathrm{\Theta })$. The ordinals $\delta =M\cap {\omega }_1$ obtained from countable $M\prec H(\mathrm{\Theta })$ containing $A,B$ and all fixed walk data contain a club: Skolem hulls of successively larger countable ordinal sets give unboundedly many such cuts, while the union of an increasing $\omega$-chain of such hulls is elementary and has cut the supremum of their cuts. This is the standard club-of-cuts refinement of [F4], and is closed and unbounded in the sense of [F5]. Fix such $M$ and put $\delta =M\cap {\omega }_1$. Notice that every $x\in M\cap {\omega }_1$ is below $\delta$, while $\delta \mathrm{\notin }M$.

Suppose first that $R$ is equality, and fix $a\in A\setminus \delta$, $b\in B\setminus \delta$. By coherence in [F1], there is ${\gamma }_0<\delta$ above every $L(\delta ,b(j))$ and above all disagreements below $\delta$ among the finitely many pairs $e_{a(i)},e_{b(j)}$. Hence $e_{a(i)}(\xi )=e_{b(j)}(\xi )$ whenever ${\gamma }_0<\xi <\delta$. By the limit clause of [F2], choose $\gamma <\delta$ so that $\gamma <\xi <\delta$ implies ${\gamma }_0<\min L(\xi ,\delta )$.

We repeatedly use the following reflection observation. If $X\subseteq {\omega }_1$ belongs to $M$ and $\delta \in X$, then $X$ is uncountable: otherwise elementarity provides in $M$ an enumeration of $X$ by $\omega$, whence $X\subseteq M$ and $\delta \in M$, a contradiction. Thus $X$ is unbounded in ${\omega }_1$. Likewise, an uncountable $X\in M$ has $X\cap \delta$ unbounded in $\delta$, by elementarity applied to arbitrarily large members of $X$.

For the strict branch, let $E$ be the set of limit $\nu <{\omega }_1$ with the following property: for every $a^0\in A\setminus \nu$, ${\nu }_0<\nu$, $\epsilon <{\omega }_1$, $n<\omega$, and finite $K\subseteq {\omega }_1\setminus \nu$, some $a^1\in A\setminus \epsilon$ satisfies ${\nu }_0\le \mathrm{\Delta }(e_{a^0(i)},e_{a^1(i)})$ and $e_{a^1(i)}(\xi )>n$ for every $i<k$ and $\xi \in K$. This set is definable from $A$ and the fixed sequence, so $E\in M$.

Let $X$ consist of those ${\delta }^{+}<{\omega }_1$ for which some $a^{+}\in A\setminus {\delta }^{+}$ and $b^{+}\in B\setminus {\delta }^{+}$ simultaneously preserve each $e_{a(i)}$ and $e_{b(j)}$ below ${\gamma }_0$, agree coordinatewise on $({\gamma }_0,{\delta }^{+})$, satisfy ${\gamma }_0<\min L(\xi ,{\delta }^{+})$ and $\mu (\xi ,{\delta }^{+})(\min L(\xi ,{\delta }^{+}))=w_{{\delta }^{+}}=w_{\delta }$ for $\gamma <\xi <{\delta }^{+}$, and satisfy $\mu ({\delta }^{+},b^{+}(j))=\mu (\delta ,b(j))$ for every $j<l$. All parameters in this definition belong to $M$. The space $C(2^{\omega },\omega )$ is countable and belongs to $M$, so $w_{\delta }\in M$. Each $L(\delta ,b(j))$ and labelled trace $\mu (\delta ,b(j))$ is finite with ordinal entries below $\delta$ and labels in that countable space, hence is an element of $M$. Finally the restrictions of the $e$-functions below ${\gamma }_0$ are finite modifications, by [F1], of restrictions coded in $M$. Thus $X\in M$ without using either $\delta$ or $b$ itself as a parameter. Taking ${\delta }^{+}=\delta$, $a^{+}=a$, and $b^{+}=b$ shows $\delta \in X$; the trace and label requirements are [F2] and the first-label clause retained in [F3]. By step 2.1 choose ${\delta }^{+}\in X$ above $\delta$, with witnesses $a^{+},b^{+}$.

The cut $\delta$ belongs to $E$. Indeed, for given data at $\delta$, finite-to-one behavior lets us enlarge ${\nu }_0<\delta$ above every $\xi <\delta$ at which some $e_{a^0(i)}(\xi )\le n$. The restriction tuple $⟨e_{a^0(i)}\upharpoonright {\nu }_0:i<k⟩$ belongs to $M$ by coherence. The definable set of ordinals $\eta$ for which some $a^1\in A\setminus \eta$ has these restrictions and has all values above $n$ on $({\nu }_0,\eta )$ belongs to $M$ and contains $\delta$ with witness $a^0$. Step 2.1 makes it unbounded, so choose such $\eta$ above $\epsilon$ and $\max K$ (with no maximum needed when $K$ is empty). Its witness $a^1$ proves the defining demand. Hence $\delta \in E$, and step 2.1 also makes $E$ uncountable.

Put $L^{+}=L(\delta ,{\delta }^{+})$. It is nonempty because $\delta <{\delta }^{+}$. Its minimum exceeds ${\gamma }_0$, and [F2] splices it above the common old trace $L({\delta }^{+},b^{+}(j))=L(\delta ,b(j))$. Preservation below ${\gamma }_0$ gives the two strict $\mathrm{\Delta }$ bounds; agreement on $L^{+}$ gives clause 3. The equality of the old labelled traces gives clause 4, and the label at the new minimum is $w_{{\delta }^{+}}=w_{\delta }$, giving clause 5. Thus all five conclusions hold in the equality branch.

Fix $a\in A\setminus \delta$ and $b\in B\setminus \delta$. Choose ${\gamma }_0\in E\cap \delta$ above every old trace $L(\delta ,b(j))$, possible by steps 2.1 and 3.2, and then choose $\gamma <\delta$ by [F2] as in step 1.2. Reflecting exactly the finite restrictions, trace-tail and labelled-trace type used in step 3.1, but requiring the candidate cut to be a limit, gives a limit ${\delta }^{+}>\delta$ and $b^{+}\in B\setminus {\delta }^{+}$ such that the $e_b$-restrictions below ${\gamma }_0$ agree, ${\gamma }_0<\min L(\xi ,{\delta }^{+})$ with first label $w_{{\delta }^{+}}=w_{\delta }$ for $\gamma <\xi <{\delta }^{+}$, and $\mu ({\delta }^{+},b^{+}(j))=\mu (\delta ,b(j))$.

Put $L^{+}=L(\delta ,{\delta }^{+})$ and let $n$ be the maximum of the finitely many values $e_{b^{+}(j)}(\xi )$ for $j<l$ and $\xi \in L^{+}$. Apply the defining property of ${\gamma }_0\in E$ with $a^0=a$, a bound ${\nu }_0<{\gamma }_0$ above every old trace, $K=L^{+}$, and $\epsilon =\delta$. It gives $a^{+}\in A\setminus \delta$ preserving every $e_{a(i)}$ on the old trace and satisfying $e_{a^{+}(i)}(\xi )>n\ge e_{b^{+}(j)}(\xi )$ on $L^{+}$. The preservation of $b$ below ${\gamma }_0$, the splice calculation, and the label calculation from step 4.1 now verify clauses 1, 2, 4 and 5; the displayed strict inequality verifies clause 3.

The completed equality and strict branches cover the two and only two allowed relations. The club of cuts from step 1.1 is independent of the later choices of $a,b,R$, so it is the required $D$. No choice is used after selecting the Skolem functions and fixed data; those ZFC selections are precisely the declared AC dependency.