Strongly increasing subsequences force a bounding projection

Write $p_{\alpha }$ for the projection of $f_{\alpha }$. It suffices to rule out the assumption that none strictly bounds the sequence. Under that assumption, for each $\alpha$ let $\eta$ be the least index for which $P=\{a:p_{\alpha }(a)\le f_{\eta }(a)\}$ is $I$-positive, since $f_{\eta }{<}_I{p}_{\alpha }$ fails. Take $b(\alpha )$ to be the least index greater than $\alpha$ and $\eta$. Outside the small failure set for $f_{\eta }{<}_I{f}_{b(\alpha )}$, the set $P$ witnesses $p_{\alpha }<f_{b(\alpha )}$, so $\{a:p_{\alpha }(a)<f_{b(\alpha )}(a)\}$ remains positive. Removing a small set cannot destroy positivity: otherwise its union with the removed set would put $P$ in $I$. Every later index has the same positive comparison, by composition with another strict comparison and the same finite-union argument. No completeness of $I$ is used.

Recursively choose strictly increasing indices $u_{\xi }<\lambda$ for $\xi <\lambda$ so that $u_{\xi }>b(u_{\zeta })$ whenever $\zeta <\xi$. At each stage the earlier indices and witness indices form a set of cardinality less than $\lambda$, so F2 bounds them below $\lambda$; choose the least larger index. F3 gives the sequence. Its range $U$ is unbounded, because a bounded subset of the cardinal $\lambda$ has cardinality less than $\lambda$ and cannot contain a strictly increasing sequence of length $\lambda$. By step 1.1, $\{a:p_{\alpha }(a)<f_{\beta }(a)\}$ is positive for all $\alpha <\beta$ in $U$. Apply $(\ast {)}_{\kappa }$ to obtain indices $(v_i{)}_{i<\kappa }$ in $U$ and strong-increase witnesses $Z_i\in I$.

Put $E_i=\{a:f_{v_i}(a)\ge s(a)\}\in I$. For each $i<\kappa$, the positive set $\{a:p_{v_i}(a)<f_{v_{i+1}}(a)\}$ remains positive after removing $Z_i\cup Z_{i+1}\cup E_i\cup E_{i+1}$. Here $i+1<\kappa$ because $\kappa$ is infinite regular. In particular the remaining set is nonempty; AC selects a coordinate $a_i$ in it. Some fixed $a$ is selected for $\kappa$ many indices. Indeed, if every fiber $\{i:a_i=a\}$ had size less than $\kappa$, it would be bounded in $\kappa$ by F2; there are $\tau <\kappa$ such bounds, so F2 would bound their union, contradicting that the fibers cover $\kappa$.

Let $T=\{i:a_i=a\}$ have size $\kappa$. For $i<j$ in $T$, the chosen coordinate gives $p_{v_i}(a)<f_{v_{i+1}}(a)$. If $i+1=j$ the next comparison is equality; otherwise strong increase gives $f_{v_{i+1}}(a)<f_{v_j}(a)$, since $a\mathrm{\notin }{Z}_{i+1}\cup Z_j$. Finally $a\mathrm{\notin }{E}_j$, so the genuine ceiling rule gives $f_{v_j}(a)\le p_{v_j}(a)\in S(a)$. Also $p_{v_i}(a)\in S(a)$. Hence $i\mapsto p_{v_i}(a)$ is a strictly increasing injection of $T$ into $S(a)$, contradicting $\mathrm{\mid }T\mathrm{\mid }=\kappa >\mathrm{\mid }S(a)\mathrm{\mid }$. The assumption in step 1.1 is impossible. For every admitted family $S$ some projection therefore strictly bounds the sequence, exactly the asserted property. The finite-ideal countable case satisfies all these hypotheses with $\tau ={\mathrm{\aleph }}_0$. QED.