Club continuity produces strongly increasing subsequences

Put $\theta ={\kappa }^{++}$ and fix the family in F2. Define a continuous strictly increasing $\xi :\theta \to \lambda$. Start with $\xi (0)=0$. At a nonzero limit take the supremum of previous values. At stage $i+1$, for every $\eta \in E_{\kappa }^{\theta }$ put $h_{\eta ,i}(a)=\sup \{f_{\xi (j)}(a):j\in C_{\eta }\cap (i+1)\}$, with empty supremum zero. If some $\sigma <\lambda$ with $\sigma >\xi (i)$ satisfies $h_{\eta ,i}{<}_I{f}_{\sigma }$, record the least such $\sigma$; otherwise record $\xi (i)+1$. Choose $\xi (i+1)$ to be the least element of $U$ strictly above $\xi (i)$ and all recorded ordinals. There are at most $\theta <\lambda$ records, so F4 makes their supremum less than $\lambda$, and unboundedness of $U$ supplies that least point. Limit stages likewise remain below $\lambda$. F5 defines the recursion. Whenever a genuine bound was recorded, transitivity of ${<}_I$ gives $h_{\eta ,i}{<}_I{f}_{\xi (i+1)}$: outside the union of the two comparison-failure sets, the two strict ordinal inequalities compose.

Let $\delta ={\sup }_{i<\theta }\xi (i)$. F4 gives $\delta <\lambda$. The increasing cofinal enumeration shows $\mathrm{cf}(\delta )\le \theta$. If $B\subseteq \delta$ were cofinal of size less than $\theta$, send each $b\in B$ to the least $i$ with $b<\xi (i)$. These indices would be unbounded in $\theta$, contradicting its regularity from F6. Thus $\mathrm{cf}(\delta )=\theta$. The range $D$ of $\xi$ is club in $\delta$: continuity includes every nonzero limit point below $\delta$, and its supremum is $\delta$. By hypothesis choose $E_{\delta }$ and a bound $f_{\beta }$. F3 makes $D\cap E_{\delta }$ club in $\delta$; pulling back under the continuous enumeration gives a club $C=\{i<\theta :\xi (i)\in E_{\delta }\}$ in $\theta$. In detail unboundedness follows from that of $D\cap E_{\delta }$, and at a nonzero limit of indices in $C$, continuity of $\xi$ and closure of $E_{\delta }$ give membership in $C$. By F2 fix $\eta$ with $C_{\eta }\subseteq C$.

Every prefix supremum $h_{\eta ,i}$ now has a strict bound in the chain: it is pointwise at most ${\sup }_{\alpha \in E_{\delta }}{f}_{\alpha }{<}_I{f}_{\beta }$, and a chain member of index greater than both $\beta$ and $\xi (i)$ is also a strict bound. Thus all its bound questions in step 1.1 were positive. If $i<j$ belong to $C_{\eta }$, then $i+1\le j$ and $h_{\eta ,i}{<}_I{f}_{\xi (i+1)}{\le }_I{f}_{\xi (j)}$. The last comparison allows equality when $j=i+1$. Hence $h_{\eta ,i}{<}_I{f}_{\xi (j)}$ for all such pairs.

Enumerate $C_{\eta }$ continuously as $(c_{\rho }{)}_{\rho <\kappa }$. Its nonaccumulation points other than its first point are exactly $c_{\rho +1}$ for $\rho <\kappa$: at a nonzero limit the enumeration is continuous, whereas a successor has previous maximum $c_{\rho }$. Write $j_{\rho }=c_{\rho +1}$ and $t_{\rho }=f_{\xi (j_{\rho })}$. Define $Z_{\rho }=\{a:h_{\eta ,c_{\rho }}(a)\ge t_{\rho }(a)\}\in I$ by step 3.1. For $\rho <\sigma$, $j_{\rho }\le c_{\sigma }$, so $t_{\rho }\le h_{\eta ,c_{\sigma }}$ pointwise, and therefore $t_{\rho }(a)<t_{\sigma }(a)$ whenever $a\mathrm{\notin }{Z}_{\sigma }$. In particular $(t_{\rho }{)}_{\rho <\kappa }$ is strongly increasing, witnessed by $(Z_{\rho })$.

To obtain indices in $U$, set $u_{\rho }=f_{\xi (j_{\rho }+1)}$. Its index belongs to $U$ by step 1.1, and these indices strictly increase with $\rho$. Since $j_{\rho }<j_{\rho }+1\le j_{\rho +1}$, the chain gives $t_{\rho }{<}_I{u}_{\rho }{\le }_I{t}_{\rho +1}$. Put $P_{\rho }=\{a:t_{\rho }(a)\ge u_{\rho }(a)\}$ and $Q_{\rho }=\{a:u_{\rho }(a)>t_{\rho +1}(a)\}$, both in $I$, and set $W_{\rho }=Z_{\rho }\cup Z_{\rho +1}\cup P_{\rho }\cup Q_{\rho }$. For $\rho <\sigma$ and $a\mathrm{\notin }{W}_{\rho }\cup W_{\sigma }$, one has $u_{\rho }(a)\le t_{\rho +1}(a)\le t_{\sigma }(a)<u_{\sigma }(a)$. The middle inequality is equality if $\sigma =\rho +1$, and is strict by step 4.1 otherwise. Thus the $u_{\rho }$ are strongly increasing with witnesses $W_{\rho }\in I$. They form an order-type-$\kappa$ subsequence indexed in the arbitrary $U$, proving $(\ast {)}_{\kappa }$. Only finite unions of ideal sets were used; the countable finite-ideal specialization follows by taking those sets to be finite. QED.