Fleissner's construction of a normal nonmetrizable Moore space from level data

Fix a well-ordering of the universe and enumerate the family $\mathcal{Z}:=\{Z\subseteq \mathrm{\Sigma }:\mathrm{card}Z\le \kappa \text{ and }Z\subseteq {\mathrm{\Sigma }}_k\text{ for some }k\in \omega \}$ as $\{Z_{\alpha }:\alpha <{\kappa }^{+}\}$, arranged so that $Z_0=\mathrm{\varnothing }$. Here $\mathrm{\Sigma }:={\bigcup }_{n\in \omega }{\mathrm{\Sigma }}_n$ with ${\mathrm{\Sigma }}_n:=\{f\upharpoonright n:f\in E^{\omega }\}$, and $[\sigma ]:=\{f\in F:\sigma \subseteq f\}$ for $F:=E^{\omega }$; write $a^{\ast }$ for the greatest ordinal in the range of a nonempty $a\in \mathrm{\Sigma }$, and put ${\mathrm{\Sigma }}^{\beta }:=\{\sigma \in \mathrm{\Sigma }:\sigma =\mathrm{\varnothing }\text{ or }{\sigma }^{\ast }<\beta \}$ and $\mathcal{Z}(\beta ):=\{Z_{\alpha }:\alpha <\beta \}$. Enumerating $\mathcal{Z}$ is legitimate: stationarity gives $\mathrm{\mid }E\mathrm{\mid }={\kappa }^{+}$, so [F1] with $\lambda ={\mathrm{\aleph }}_0$ gives $\mathrm{\mid }{E}^{\omega }\mathrm{\mid }=\mathrm{\mid }\mathrm{\Sigma }\mathrm{\mid }={\kappa }^{+}$; and [F1] with $\lambda =\kappa$ gives at most $({\kappa }^{+}{)}^{\kappa }={\kappa }^{+}$ subsets of size at most $\kappa$ on every finite level. Conversely, the singletons of ${\mathrm{\Sigma }}_1$ already give ${\kappa }^{+}$ members of $\mathcal{Z}$.

With $\kappa$, $({\kappa }_n)$ and the enumeration $\{Z_{\alpha }:\alpha <{\kappa }^{+}\}$ fixed as above, let $\mathrm{level}(Z)$ be the least $k$ with $Z\subseteq {\mathrm{\Sigma }}_k$ ($0$ for $Z=\mathrm{\varnothing }$), and for nonempty $\theta \in \mathrm{\Sigma }$ of length $r$ put $F_{\theta }:=\mathcal{Z}({\theta }^{\ast })\cap \{Z:\mathrm{level}(Z)<r\}$. This set has cardinality at most $\kappa$. Using [F2], fix an enumeration $e_{\theta }:{\mu }_{\theta }\to F_{\theta }$ with ${\mu }_{\theta }\le \kappa$, arranged with $e_{\theta }(0)=\mathrm{\varnothing }$; the latter is possible because $Z_0=\mathrm{\varnothing }\in F_{\theta }$. Let $P(\theta ,m):=e_{\theta }[\min ({\kappa }_m,{\mu }_{\theta })]$ and, for $\sigma \in \mathrm{\Sigma }$ of length $\mathrm{\ell }$, define $A(\sigma ,m):=\bigcup \{P(\theta ,m):\mathrm{\varnothing }\mathrm{\ne }\theta \subseteq \sigma \}$, with $A(\mathrm{\varnothing },m)=A(\sigma ,0)=\mathrm{\varnothing }$. Then ${\bigcup }_{m}A(\sigma ,m)=F_{\sigma }$: an element of $F_{\sigma }$ has an index $\xi <{\mu }_{\sigma }\le \kappa$ in $e_{\sigma }$, and cofinality of $({\kappa }_m)$ gives an $m$ with $\xi <{\kappa }_m$. Also, if $\sigma \subseteq {\sigma }^{\mathrm{\prime }}$ and $m\le m^{\mathrm{\prime }}$, then $A(\sigma ,m)\subseteq A({\sigma }^{\mathrm{\prime }},m^{\mathrm{\prime }})$. Finally, $\sigma$ has only finitely many nonempty prefixes, so $\mathrm{\mid }A(\sigma ,m)\mathrm{\mid }\le {\lambda }_{\mathrm{\ell },m}$, where ${\lambda }_{\mathrm{\ell },m}:={\kappa }_m$ if ${\kappa }_m$ is infinite and ${\lambda }_{\mathrm{\ell },m}:=(\mathrm{\ell }+1){\kappa }_m$ if ${\kappa }_m$ is finite. Thus the source's special $\kappa =\omega$ clause gives finite $A(\sigma ,m)$; it does not assert the false bound $\mathrm{\mid }A(\sigma ,m)\mathrm{\mid }\le {\kappa }_m$.

Construction requirement (well-definedness). The condition (12) below is imposed only at traces lying in $\mathrm{dom}(g)=\mathcal{Z}({\rho }^{\ast })$: for $\sigma \in {\mathrm{\Sigma }}_n$, a triple $⟨g,\rho ,\tau ⟩$ satisfies (12) at index $\sigma$ iff $Z\in A(\sigma ,n),Z\subseteq {\mathrm{\Sigma }}_m,Z\cap {\mathrm{\Sigma }}^{\rho (m)}\in \mathcal{Z}({\rho }^{\ast })⟹g(Z\cap {\mathrm{\Sigma }}^{\rho (m)})=1⟺\sigma \upharpoonright m\in Z\mathrm{.}$ No instance of (12) is used anywhere below at a trace outside $\mathcal{Z}({\rho }^{\ast })$, and the §6 instances are proved in Case 1 below, so this guard is a definition of the space, not an assumption. This is the documented discharge of the trace-domain obligation.

Lemma 3(a). If $T\subseteq \mathrm{\Sigma }$ satisfies $\bigcup \{[\sigma ]:\sigma \in T\}=F$, then for some $n$ the set $T\cap {\mathrm{\Sigma }}_n$ has a stafull subset, where $S\subseteq {\mathrm{\Sigma }}_n$ is stafull iff for all $\sigma \in S$ and $j<n$ the set $\{\tau (j):\sigma \upharpoonright j\subset \tau \in S\}$ is stationary in ${\kappa }^{+}$.

Let $Q_k$ be the set of triples $⟨g,\rho ,\tau ⟩$ such that $\rho ,\tau \in {\mathrm{\Sigma }}_k$, $\rho (i)<\tau (i)<\rho (i+1)<\tau (i+1)$ for all $i<k-1$, and $g$ is a function from $\mathcal{Z}({\rho }^{\ast })$ to $\{0,1\}$ (for $k=0$ read ${\rho }^{\ast }=0$), and let $Q:={\bigcup }_k{Q}_k$. For $\sigma \in {\mathrm{\Sigma }}_n$ let $C(\sigma )$ be the set of $⟨g,\rho ,\tau ⟩\in Q$ satisfying: $\sigma \subseteq \rho$ or $\sigma \subseteq \tau$; $\rho (0{)}_i=\tau (0{)}_i$ for all $i<n$; and the guarded (12) of step 1.3 at index $\sigma$. Put $B(\sigma ):=[\sigma ]\cup C(\sigma )$, and let $X:=F\cup Q$ with basis $\{B(\sigma ):\sigma \in \mathrm{\Sigma }\}\cup \{\{q\}:q\in Q\}$.

Suppose, for contradiction, that $T\cap {\mathrm{\Sigma }}_n$ has no stafull subset for every $n$; then every nonempty $S\subseteq T\cap {\mathrm{\Sigma }}_n$ is non-stafull, because a stafull subset of $S$ would be a stafull subset of $T\cap {\mathrm{\Sigma }}_n$. Call a finite $\rho \in \mathrm{\Sigma }$ acceptable iff for every $n>\mathrm{\mid }\rho \mathrm{\mid }$ and every nonempty $S\subseteq T\cap {\mathrm{\Sigma }}_n$ with $\sigma \supseteq \rho$ for all $\sigma \in S$, there are $\sigma \in S$ and $j$ with $\mathrm{\mid }\rho \mathrm{\mid }\le j<n$ such that $\{\tau (j):\sigma \upharpoonright j\subset \tau \in S\}$ is non-stationary. Then $\mathrm{\varnothing }$ is acceptable by non-stafullness of $S$ itself. We construct $f\in F$ whose every finite prefix is acceptable; then $f\upharpoonright 0=\mathrm{\varnothing }\mathrm{\notin }T$ (else $T\cap {\mathrm{\Sigma }}_0$ would be stafull) and $f\upharpoonright (i+1)\mathrm{\notin }T$ for all $i$, so $f\mathrm{\notin }\bigcup \{[\sigma ]:\sigma \in T\}$, contradicting the covering hypothesis.

Lemma 3(b) in the form used. If $S\subseteq {\mathrm{\Sigma }}_n$ is stafull and $h(\sigma )=\sigma (0{)}_i$ for a fixed $i<n$, then there is a stafull $S^{\mathrm{\prime }}\subseteq S$ on which $h$ is constant. Indeed $A_i:=\{\tau (0):\tau \in S\}\cap E$ is stationary, $h$ restricted to $A_i$ is regressive (ladder values are below their ordinal), so Fodor's lemma (F4) provides a stationary $A^{\mathrm{\prime }}\subseteq A_i$ on which $h$ is constant; then $S^{\mathrm{\prime }}:=\{\tau \in S:\tau (0)\in A^{\mathrm{\prime }}\}$ is stafull: for $j<n$ the branching set $\{\tau (j):\sigma \upharpoonright j\subset \tau \in S^{\mathrm{\prime }}\}$ equals the corresponding branching set of $S$ when $j\ge 1$ (the first coordinate is determined by $\sigma \upharpoonright j$) and equals $A^{\mathrm{\prime }}$ when $j=0$. Iterating for $i=0,\dots ,n-1$ gives a stafull $S^{\mathrm{\prime }}\subseteq S$ with $\sigma (0{)}_i={\sigma }^{\mathrm{\prime }}(0{)}_i$ for all $\sigma ,{\sigma }^{\mathrm{\prime }}\in S^{\mathrm{\prime }}$ and all $i<n$.

Lemma 3(c). If $S^{\mathrm{\prime }}\subseteq {\mathrm{\Sigma }}_n$ is stafull and $\beta <{\kappa }^{+}$, then there is $W=\{{\sigma }_{\alpha }:\alpha <\beta \}\subseteq S^{\mathrm{\prime }}$ with ${\sigma }_{\alpha }(i)<{\sigma }_{{\alpha }^{\mathrm{\prime }}}(i^{\mathrm{\prime }})$ iff $i<i^{\mathrm{\prime }}$ or ($i=i^{\mathrm{\prime }}$ and $\alpha <{\alpha }^{\mathrm{\prime }}$). Construct the coordinates level by level: having chosen levels $<i$ so that all level-$(i-1)$ values lie below a common bound $<{\kappa }^{+}$ and each prefix ${\sigma }_{\alpha }\upharpoonright i$ extends to a member of $S^{\mathrm{\prime }}$, stafullness of $S^{\mathrm{\prime }}$ makes each set $A_{\alpha }:=\{\tau (i):{\sigma }_{\alpha }\upharpoonright i\subset \tau \in S^{\mathrm{\prime }}\}$ stationary, hence unbounded above the bound; choose ${\sigma }_{\alpha }(i)\in A_{\alpha }$ strictly increasing in $\alpha$ and above the sup of the level-$(i-1)$ values, possible because that sup is $<{\kappa }^{+}$ by regularity (F1) since $\beta <{\kappa }^{+}$; the final level's choice lies in $S^{\mathrm{\prime }}$ by the definition of $A_{\alpha }$.

The family $\{B(\sigma )\}$ together with the singletons is a basis: for $x\in F$ and $x\in B(\sigma )\cap B({\sigma }^{\mathrm{\prime }})$ the sequences $\sigma ,{\sigma }^{\mathrm{\prime }}$ are both initial segments of $x$, hence comparable, and the longer one ${\sigma }^{\mathrm{\prime }\mathrm{\prime }}$ satisfies $B({\sigma }^{\mathrm{\prime }\mathrm{\prime }})\subseteq B(\sigma )\cap B({\sigma }^{\mathrm{\prime }})$ because $C({\sigma }^{\mathrm{\prime }\mathrm{\prime }})\subseteq C(\sigma )\cap C({\sigma }^{\mathrm{\prime }})$: the conditions (10) and (11) transfer downwards, and a guarded (12)-instance at ${\sigma }^{\mathrm{\prime }\mathrm{\prime }}$ restricts to the corresponding instance at $\sigma$ since $A(\sigma ,\mathrm{\mid }\sigma \mathrm{\mid })\subseteq A({\sigma }^{\mathrm{\prime }\mathrm{\prime }},\mathrm{\mid }{\sigma }^{\mathrm{\prime }\mathrm{\prime }}\mathrm{\mid })$ by step 1.2 and ${\sigma }^{\mathrm{\prime }\mathrm{\prime }}\upharpoonright m=\sigma \upharpoonright m$ for $m<\mathrm{\mid }\sigma \mathrm{\mid }$. For $x\in Q$ the singleton $\{x\}$ is a basis element contained in every basis element containing $x$.

$X$ is $T_1$, and $Q$ is open, hence $F=X\setminus Q$ is closed. To see the $T_1$ property, distinct branches in $F$ have incompatible finite initial segments, while every $q\in Q$ has the open singleton $\{q\}$. Conversely, for fixed $q=⟨g,\rho ,\tau ⟩\in Q_k$ and $f\in F$, choose $n>k$; then $q\mathrm{\notin }B(f\upharpoonright n)$ because condition (10) cannot make a length-$n$ sequence an initial segment of either length-$k$ coordinate. Thus every point other than $q$ has a neighbourhood avoiding $q$, so $\{q\}$ is also closed.

Successor step of the acceptability recursion. Let $\rho$ be acceptable with $\mathrm{\mid }\rho \mathrm{\mid }=i$. Call $S$ dangerous for $v$ if, for some $n>i+1$, $S\subseteq T\cap {\mathrm{\Sigma }}_n$ is nonempty, every $\sigma \in S$ extends $\rho \widehat{}v$, and $\{\tau (j):\sigma \upharpoonright j\subset \tau \in S\}$ is stationary for all $\sigma \in S$ and $i+1\le j<n$; let $B_1:=\{v\in E:$ some $S$ is dangerous for $v\}$, and $B_2:=\{v\in E:\rho \widehat{}v\in T\cap {\mathrm{\Sigma }}_{i+1}\}$. Then $B_1\cup B_2\mathrm{\ne }E$.

The branch neighbourhoods have the star property needed below. If $x\in F$ and open $D\ni x$, choose ${\sigma }_0\subseteq x$ with $B({\sigma }_0)\subseteq D$. For $n\ge \mathrm{\mid }{\sigma }_0\mathrm{\mid }$, the unique length-$n$ initial segment $\sigma =x\upharpoonright n$ satisfies $B(\sigma )\subseteq B({\sigma }_0)$ by step 3.1, so every length-$n$ basic member containing $x$ lies in $D$. At a point of $Q$ the singleton is a basic neighbourhood.

For disjoint closed $H,K\subseteq X$, it is enough to find disjoint open $U,V$ with $H\cap F\subseteq U$ and $K\cap F\subseteq V$. Indeed $U^{\mathrm{\prime }}:=(U\setminus K)\cup (H\cap Q),V^{\mathrm{\prime }}:=(V\setminus H)\cup (K\cap Q)$ are open, contain $H,K$ respectively, and are disjoint: $Q$ is discrete open, removing a closed set preserves openness, and each added isolated part lies in its own closed set and outside the other. Since $F$ is closed by step 3.2, the two traces are closed subsets of $F$.

Suppose $B_1\cup B_2=E$. If $B_1$ is stationary, choose for each $v\in B_1$ the least level $n_v$ of a dangerous $S_v$ and a least such $S_v$ in the fixed well-order; some $E^{\mathrm{\prime }}:=\{v\in B_1:n_v=n_0\}$ is stationary by the countable case split, and $V:={\bigcup }_{v\in E^{\mathrm{\prime }}}{S}_v\subseteq T\cap {\mathrm{\Sigma }}_{n_0}$ is nonempty with all members extending $\rho$. Acceptability of $\rho$ gives $\sigma \in V$ and $j\in [i,n_0)$ with $B(\sigma ,j,V)$ non-stationary. Pick $v\in E^{\mathrm{\prime }}$ with $\sigma \in S_v$. If $j\ge i+1$ then $B(\sigma ,j,V)\supseteq B(\sigma ,j,S_v)$ is stationary because $S_v$ is dangerous, a contradiction; if $j=i$ then $B(\sigma ,i,V)=\{\tau (i):\tau \in V\}\supseteq E^{\mathrm{\prime }}$, because $S_v\ni \sigma$ witnesses $v\in B(\sigma ,i,V)$ for every $v\in E^{\mathrm{\prime }}$, so $B(\sigma ,i,V)$ is stationary, again a contradiction. If $B_1$ is non-stationary then $E\setminus B_1$ is stationary and contained in $B_2$, so $B_2$ is stationary; applying acceptability of $\rho$ to the nonempty $S:=\{\rho \widehat{}v\in T\cap {\mathrm{\Sigma }}_{i+1}:v\in E\}$ and $n=i+1$ gives $B(\sigma ,i,S)$ non-stationary for some $\sigma \in S$, but $B(\sigma ,i,S)=\{\tau (i):\tau \in S\}\supseteq B_2$ is stationary. Both cases are contradictory, so $B_1\cup B_2\mathrm{\ne }E$.

Define ${\mathcal{G}}_n:=\{B(\sigma ):\sigma \in {\mathrm{\Sigma }}_n\}\cup \{\{q\}:q\in Q\}$. Each ${\mathcal{G}}_n$ is an open cover: a point $f\in F$ lies in $[f\upharpoonright n]\subseteq B(f\upharpoonright n)$, and every $q\in Q$ lies in its singleton. At $x\in F$, step 4.1 gives $\mathrm{St}(x,{\mathcal{G}}_n)\subseteq D$ at every sufficiently large level. If $x=q\in Q_k$, then for $n>k$ no $B(\sigma )$ with $\mathrm{\mid }\sigma \mathrm{\mid }=n$ contains $q$, because condition (10) would require $\sigma$ to be an initial segment of one of the length-$k$ sequences $\rho ,\tau$; thus $\mathrm{St}(q,{\mathcal{G}}_n)=\{q\}$. Moreover ${\bigcup }_n{\mathcal{G}}_n$ is a uniform base in the source's sense. A point $q\in Q_k$ belongs only to its singleton and to the finitely many $B(\sigma )$ whose $\sigma$ is an initial segment of its two length-$k$ coordinates, so it cannot lie in the intersection of an infinite subfamily. If an infinite subfamily $\mathcal{R}$ has $x\in \bigcap \mathcal{R}$, then $x\in F$ has a branch $\rho$, and the members of $\mathcal{R}$ are the sets $B(\rho \upharpoonright k)$ at arbitrarily large levels. Given open $D\ni x$, choose $n$ with $B(\rho \upharpoonright n)\subseteq D$; any member $B(\rho \upharpoonright k)$ of $\mathcal{R}$ with $k\ge n$ lies in $D$. Hence $\mathcal{R}$ is a neighbourhood base at $x$. By the Aleksandrov-Arhangel'skij equivalence of uniform bases with metacompact Moore spaces recorded in the source's Section 2, $X$ is a regular Moore space and is metacompact.

Choosing $v\in E\setminus (B_1\cup B_2)$ gives that $\rho \widehat{}v$ is acceptable — any nonempty $S\subseteq T\cap {\mathrm{\Sigma }}_n$ above $\rho \widehat{}v$ failing the acceptability witness would be dangerous for $v$ — and $\rho \widehat{}v\mathrm{\notin }T$. The recursion produces an infinite sequence $f\in E^{\omega }$ with every prefix acceptable and $f\upharpoonright (i+1)\mathrm{\notin }T$ for all $i$; since $f\in [\sigma ]$ means $\sigma =f\upharpoonright \mathrm{\mid }\sigma \mathrm{\mid }\in T$, the covering hypothesis is contradicted. Therefore some $T\cap {\mathrm{\Sigma }}_n$ has a stafull subset, proving Lemma 3(a).

For $n\in \omega$ and $Z\subseteq {\mathrm{\Sigma }}_n$ put $H_Z:=\bigcup \{[\sigma ]:\sigma \in Z\}$ and $K_Z:=F\setminus H_Z$. It suffices to separate $H_Z$ and $K_Z$ by disjoint open subsets of $X$ for every $n,Z$ (the source's Lemma 2). Here is the countable reduction. For disjoint closed $H,K\subseteq F$, let $H_n=\bigcup \{[\sigma ]:\mathrm{\mid }\sigma \mathrm{\mid }=n,[\sigma ]\cap H=\mathrm{\varnothing }\},K_n=\bigcup \{[\sigma ]:\mathrm{\mid }\sigma \mathrm{\mid }=n,[\sigma ]\cap K=\mathrm{\varnothing }\}\mathrm{.}$ Then $K\subseteq {\bigcup }_n{H}_n$ and $H\subseteq {\bigcup }_n{K}_n$, because the cylinders are a base for $F$. By the assumed $H_Z$--$K_Z$ separation, choose disjoint open $O_n,P_n$ containing $H_n,F\setminus H_n$, and disjoint open $R_n,S_n$ containing $K_n,F\setminus K_n$. Thus $\overline{O_n}\cap H=\mathrm{\varnothing }$ and $\overline{R_n}\cap K=\mathrm{\varnothing }$, since the corresponding $P_n,S_n$ are open and disjoint. The open sets $U={\bigcup }_n(R_n\setminus {\bigcup }_{i\le n}\overline{O_i}),V={\bigcup }_n(O_n\setminus {\bigcup }_{i\le n}\overline{R_i})$ contain $H,K$ respectively and are disjoint: if a point lay in the $n$th piece of $U$ and the $m$th piece of $V$, the case $n\le m$ contradicts removal of $\overline{R_n}$ from the latter, and $m\le n$ contradicts removal of $\overline{O_m}$ from the former. Step 4.2 then handles isolated points.

Non-metrizability. For $\delta \in E$ put $Y_{\delta }:=\{f\in F:f(0)=\delta \}$, a closed discrete family in $X$. If $X$ were metrizable it would be collectionwise normal (F6), so there would be pairwise disjoint open sets $U_{\delta }\supseteq Y_{\delta }$.

Fix $n\ge 1$ and $Z\subseteq {\mathrm{\Sigma }}_n$ with $H_Z$ as in step 6.1, and let $C:=\{\gamma <{\kappa }^{+}:\text{if }\beta <\gamma \text{ then }Z\cap {\mathrm{\Sigma }}^{\beta }=Z_{\alpha }\text{ for some }\alpha <\gamma \}$. Then $C$ is a club: it is closed because for a limit $\gamma$ of $C$-points and $\beta <\gamma$ some ${\gamma }^{\mathrm{\prime }}\in C$ has $\beta <{\gamma }^{\mathrm{\prime }}<\gamma$, so $Z\cap {\mathrm{\Sigma }}^{\beta }=Z_{\alpha }$ with $\alpha <{\gamma }^{\mathrm{\prime }}<\gamma$; and it is unbounded because the assignment $\gamma \mapsto \sup \{\alpha +1:\beta <\gamma ,Z\cap {\mathrm{\Sigma }}^{\beta }=Z_{\alpha }\}$ can be iterated countably many times, remains below ${\kappa }^{+}$ by regularity, and its limit lies in $C$. This uses $\mathrm{card}Z\le \kappa$, so that at most $\kappa$ distinct truncations occur, and $\mathrm{cf}({\kappa }^{+})>\omega$.

Assume such $\{U_{\delta }\}$ exist. Put $T:=\{\sigma \in \mathrm{\Sigma }:B(\sigma )\subseteq U_{\sigma (0)}\}$. Then $\bigcup \{[\sigma ]:\sigma \in T\}=F$: for $f\in F$ one has $f\in Y_{f(0)}\subseteq U_{f(0)}$, and since $f\mathrm{\notin }Q$ some basic neighbourhood $B(\sigma )\ni f$ lies in the open set $U_{f(0)}$; then $f\in [\sigma ]\subseteq F$, so $\sigma \subseteq f$ and $\sigma (0)=f(0)$, giving $B(\sigma )\subseteq U_{\sigma (0)}$ and $f\in [\sigma ]$ with $\sigma \in T$.

Let $\gamma (\beta )$ be the least element of $C$ greater than $\beta$, and for $\sigma \in {\mathrm{\Sigma }}_{n+3}$ choose $j(\sigma ):=1+\max (n+3,m_{\gamma (\sigma (n+1))}(\sigma (0)),m^{\ast }(\sigma )),$ where $m^{\ast }(\sigma )$ is the least $m$ with $Z\cap {\mathrm{\Sigma }}^{\sigma (n)}\in A(\sigma ,m)$ if $\gamma (\sigma (n))<\sigma (n+2)$, and $m^{\ast }(\sigma ):=0$ otherwise; this is a definable choice from the given data. Then $j(\sigma )>m_{\gamma (\sigma (n+1))}(\sigma (0))$ strictly, and whenever $\gamma (\sigma (n))<\sigma (n+2)$ the transfer of step 1.2 gives $Z\cap {\mathrm{\Sigma }}^{\sigma (n)}\in A(\sigma ,j(\sigma ))$ (and even in $A(\sigma ,j(\sigma )-1)$, which is what the applications at index ${\rho }^{\mathrm{\prime }}⊋\sigma$ of level $j(\sigma )$ need). Put $W(\sigma ):=\bigcup \{B(\rho ):\sigma \subseteq \rho \in {\mathrm{\Sigma }}_{j(\sigma )}\}$.

By Lemma 3(a) proved above there are $n\ge 1$ and a stafull $S\subseteq T\cap {\mathrm{\Sigma }}_n$. Apply step 2.3 $n$ times to get a stafull $S^{\mathrm{\prime }}\subseteq S$ with $\sigma (0{)}_i={\sigma }^{\mathrm{\prime }}(0{)}_i$ for all $\sigma ,{\sigma }^{\mathrm{\prime }}\in S^{\mathrm{\prime }}$ and $i<n$, and apply step 2.4 with $\beta :=\kappa$ to obtain $W=\{{\sigma }_{\alpha }:\alpha <\kappa \}\subseteq S^{\mathrm{\prime }}$ with the interleaving property. Put ${\lambda }_n:={\kappa }_n$ when $\kappa >\omega$ and ${\lambda }_n:=(n+1){\kappa }_n$ when $\kappa =\omega$. For $\sigma \in W$, step 1.2 gives $0<\mathrm{\mid }A(\sigma ,n)\mathrm{\mid }\le {\lambda }_n$: nonemptiness follows from $n\ge 1$, ${\kappa }_n\ge 1$, and $e_{\sigma }(0)=\mathrm{\varnothing }$. Enumerate it, repeating entries if necessary, as $\{Z(\sigma ,\delta ):\delta <{\lambda }_n\}$. Notice that ${\lambda }_n={\kappa }_n$ is infinite when $\kappa >\omega$, while ${\lambda }_n$ is finite when $\kappa =\omega$.

Claim (Case 1). Let $\sigma ,\nu \in {\mathrm{\Sigma }}_{n+3}$ and suppose $\sigma \upharpoonright n\in Z$, $\nu \upharpoonright n\mathrm{\notin }Z$, and $\gamma (\nu (n))<\sigma (n+2)$; assume $\sigma (0)<\nu (0)$. Then $W(\sigma )\cap W(\nu )=\mathrm{\varnothing }$.

For $\rho ,\tau \in W$ with $\rho (0)<\tau (0)$, the triple $⟨g,\rho ,\tau ⟩$ for any $g:\mathcal{Z}({\rho }^{\ast })\to \{0,1\}$ satisfies (7), (8), (10), (11) of step 2.1: $\rho ,\tau \in {\mathrm{\Sigma }}_n$; the interleaving of step 2.4 gives (8); $\rho \subseteq \rho$ gives (10); and (11) is exactly the constancy of $\sigma (0{)}_i$ from step 2.3. Since $B(\rho )\ni ⟨g,\rho ,\tau ⟩$ would put it into $U_{\rho (0)}\cap U_{\tau (0)}=\mathrm{\varnothing }$ (as $\rho (0)\mathrm{\ne }\tau (0)$), the guarded (12) must fail at index $\rho$ or at index $\tau$ for every $g$.

Let $⟨g,\rho ,\tau ⟩\in W(\sigma )\cap W(\nu )$: then there are ${\rho }^{\mathrm{\prime }}\supseteq \sigma$, ${\nu }^{\mathrm{\prime }}\supseteq \nu$ with $\mathrm{\mid }{\rho }^{\mathrm{\prime }}\mathrm{\mid }=j(\sigma )$, $\mathrm{\mid }{\nu }^{\mathrm{\prime }}\mathrm{\mid }=j(\nu )$ and $⟨g,\rho ,\tau ⟩\in C({\rho }^{\mathrm{\prime }})\cap C({\nu }^{\mathrm{\prime }})$, because $[\cdot ]\subseteq F$ and $F\cap Q=\mathrm{\varnothing }$. By condition (10) at both indices the cases $({\rho }^{\mathrm{\prime }}\subseteq \rho$ and ${\nu }^{\mathrm{\prime }}\subseteq \rho )$ and $({\rho }^{\mathrm{\prime }}\subseteq \tau$ and ${\nu }^{\mathrm{\prime }}\subseteq \tau )$ force $\sigma (0)=\nu (0)$, so after interchanging $\sigma$ and $\nu$ if necessary we have ${\rho }^{\mathrm{\prime }}\subseteq \rho$ and ${\nu }^{\mathrm{\prime }}\subseteq \tau$; hence $\rho (0)=\sigma (0)$, $\tau (0)=\nu (0)$, and $\rho (i)=\sigma (i)$ for $i<j(\sigma )$, $\tau (i)=\nu (i)$ for $i<j(\nu )$.

Claim (Case 2). If $\sigma (n+2)\le \gamma (\nu (n))$ under the hypotheses of step 9.1 (same membership assumptions), then $W(\sigma )\cap W(\nu )=\mathrm{\varnothing }$.

For a fixed pair $\rho ,\tau$ consider the two systems of constraints on $g$, with the exponent always the triple's first component $\rho$, and include only guarded instances whose trace lies in $\mathcal{Z}({\rho }^{\ast })$: the first system requires $g(Z\cap {\mathrm{\Sigma }}^{\rho (m)})=[\rho \upharpoonright m\in Z]$ for $Z=Z(\rho ,\delta )\in A(\rho ,n)$ of level $m$, and the second requires the analogous value $[\tau \upharpoonright m\in Z]$ for $Z=Z(\tau ,\eta )\in A(\tau ,n)$. Each guarded system is internally consistent: by step 2.3 and $(\rho \upharpoonright m{)}^{\ast }=\rho (m-1)<\rho (m)$, the required value depends only on the trace $Z\cap {\mathrm{\Sigma }}^{\rho (m)}$; the same holds for the $\tau$-system because $\tau (m-1)<\rho (m)$ by interleaving. No single total $g:\mathcal{Z}({\rho }^{\ast })\to \{0,1\}$ satisfies both guarded systems, by step 9.2. Hence a conflict exists: there are $\delta ,\eta <{\lambda }_n$ and $m<n$ whose common trace belongs to $\mathcal{Z}({\rho }^{\ast })$ and for which $Z(\rho ,\delta )\cap {\mathrm{\Sigma }}^{\rho (m)}=Z(\tau ,\eta )\cap {\mathrm{\Sigma }}^{\rho (m)}$ but $[\rho \upharpoonright m\in Z(\rho ,\delta )]\mathrm{\ne }[\tau \upharpoonright m\in Z(\tau ,\eta )]$. Otherwise the function assigning each constrained trace its required value and $0$ to every other member of $\mathcal{Z}({\rho }^{\ast })$ would satisfy both systems. This is (17) in trace form together with (18a) or (18b).

Condition (8) at the coordinates below $j(\sigma )\le \mathrm{\mid }\rho \mathrm{\mid }$ and $j(\nu )\le \mathrm{\mid }\tau \mathrm{\mid }$, together with step 10.1, gives $\sigma (n)<\nu (n)<\sigma (n+1)\le \sigma (n+2)$. Before applying guarded (12), verify its domain condition. Since $\gamma (\sigma (n))$ and $\gamma (\nu (n))$ are club points and $\sigma (n)<\nu (n)$, step 7.1 gives indices $\mathrm{index}(Z\cap {\mathrm{\Sigma }}^{\sigma (n)})<\gamma (\sigma (n))\le \gamma (\nu (n)),\mathrm{index}(Z\cap {\mathrm{\Sigma }}^{\nu (n)})<\gamma (\nu (n))\mathrm{.}$ The Case-1 hypothesis gives $\gamma (\nu (n))<\sigma (n+2)\le {\rho }^{\ast }$, so both traces belong to $\mathcal{Z}({\rho }^{\ast })$. Step 8.1 now puts $Z\cap {\mathrm{\Sigma }}^{\sigma (n)}$ in $A({\rho }^{\mathrm{\prime }},j(\sigma ))$ and $Z\cap {\mathrm{\Sigma }}^{\nu (n)}$ in $A({\nu }^{\mathrm{\prime }},j(\nu ))$. Their condition-(12) traces both reduce to $Z\cap {\mathrm{\Sigma }}^{\rho (n)}$ because $\rho (n)=\sigma (n)<\nu (n)$, while ${\rho }^{\mathrm{\prime }}\upharpoonright n\in Z$ and ${\nu }^{\mathrm{\prime }}\upharpoonright n\mathrm{\notin }Z$. Thus guarded (12) yields $1=g(Z\cap {\mathrm{\Sigma }}^{\rho (n)})=0$, a contradiction. Hence Case 1 holds.

Assume $⟨g,\rho ,\tau ⟩=:q\in W(\sigma )\cap W(\nu )$ and argue as in step 10.1 to get ${\rho }^{\mathrm{\prime }}\subseteq \rho$, ${\nu }^{\mathrm{\prime }}\subseteq \tau$; then (8) gives the chain $\nu (n)<\sigma (n+1)<\nu (n+1)<\sigma (n+2)\le \gamma (\nu (n))$, so both $\sigma (n+1)$ and $\nu (n+1)$ lie in the interval $[\nu (n),\gamma (\nu (n)))$, which contains no $C$-point by the minimality of $\gamma (\nu (n))$; applying the definition of $\gamma$ inside that gap gives $\gamma (\sigma (n+1))=\gamma (\nu (n+1))=\gamma (\nu (n))$. By step 8.1, $j(\sigma )>m_{\gamma (\sigma (n+1))}(\sigma (0)),j(\nu )>m_{\gamma (\nu (n+1))}(\nu (0))\mathrm{.}$ Put $m:=\max \{j(\sigma ),j(\nu )\}$ and $m^{\ast }:=\max \{m_{\gamma (\sigma (n+1))}(\sigma (0)),m_{\gamma (\nu (n+1))}(\nu (0))\}$; then $m^{\ast }<m$, and condition (11) at the indices ${\rho }^{\mathrm{\prime }}$, ${\nu }^{\mathrm{\prime }}$ gives $\rho (0{)}_i=\tau (0{)}_i$ for every $i<m$.

Define $U:=\bigcup \{W(\sigma ):\sigma \in {\mathrm{\Sigma }}_{n+3},\sigma \upharpoonright n\in Z\},V:=\bigcup \{W(\nu ):\nu \in {\mathrm{\Sigma }}_{n+3},\nu \upharpoonright n\mathrm{\notin }Z\}\mathrm{.}$ These sets are open and contain $H_Z,K_Z$ respectively: extend the length-$n$ initial segment of any branch to length $n+3$ and use $B(\sigma )\subseteq W(\sigma )$. If Cases 1 and 2 both hold, every cross-pair of their displayed constituents is disjoint, according as $\gamma (\nu (n))<\sigma (n+2)$ or $\sigma (n+2)\le \gamma (\nu (n))$ (interchange the two sequences first when their zeroth coordinates are reversed). Hence the two cases imply $U\cap V=\mathrm{\varnothing }$.

Colour each pair $\{\rho ,\tau \}\subseteq W$ with $\rho (0)<\tau (0)$ by the least conflict witness $(\delta ,\eta ,m,\mathrm{a}\mathrm{l}\mathrm{t})$ in a fixed well-ordering of ${\lambda }_n\times {\lambda }_n\times n\times 2$. If $\kappa =\omega$, this is a finite colour set and the infinite Ramsey theorem (F5) gives an infinite homogeneous set. If $\kappa >\omega$, then ${\lambda }_n={\kappa }_n$ is infinite and, since $2^{{\kappa }_n}<\kappa$, the Erdős–Rado theorem (F5) applied to a subset of $W$ of size $(2^{{\kappa }_n}{)}^{+}$ gives a homogeneous set of size ${\kappa }_n^{+}$. Either way there are $\rho ,\sigma ,\tau \in W$ with $\rho (0)<\sigma (0)<\tau (0)$ and one common alternative of (18), common indices $\delta ,\eta$, and a common $m$ such that (17) holds for each of the three pairs.