Universal pcf sequences have strong increase and exact bounds

Choose a cofinality-$\lambda$ ultrafilter $D$, which exists by the definition of $\mathrm{pcf}(A)$. Put $L=A\cap (\lambda +1)$. If $L\mathrm{\notin }D$, its complement belongs to $D$: a maximal proper filter omitting $L$ has a member disjoint from $L$, since otherwise adjoining $L$ would generate a proper larger filter. On that complement every coordinate exceeds $\lambda$; F1 would give product cofinality greater than $\lambda$, a contradiction. Thus $L\in D$, and F3 implies $L\mathrm{\notin }I$. The least positive initial segment therefore exists with $\mu \le \lambda +1$. Also $\lambda \ge \min A>\mathrm{\mid }A\mathrm{\mid }$ by F1 and progressiveness. If $\mu =\theta +1$, minimality gives $A\cap \theta \in I$, and positivity of its union with $A\cap \{\theta \}$ forces $\theta \in A$ and $\{\theta \}\mathrm{\notin }I$. F2 gives $\theta \ge \lambda$, while $\theta +1\le \lambda +1$ gives $\theta \le \lambda$. Thus $\theta =\lambda$.

In this successor case let $N=A\cap \lambda$. Then $N\in I$. No member of $I$ contains a coordinate $a\ge \lambda$, by its singleton subset and F2; hence $I=\mathcal{P}(N)$. Define $f_{\xi }(a)=0$ on $N$ and $f_{\xi }(a)=\xi$ off $N$. Since $a\ge \lambda >\xi$ off $N$, these are product functions. For $\xi <\eta$, $f_{\xi }(a)<f_{\eta }(a)$ everywhere off $N$, so the whole sequence is strongly increasing with constant exceptional set $N$. Every unbounded $U\subseteq \lambda$ has an increasing sequence of indices of order type any infinite $\kappa \le \lambda$: recursively take its least member above previous indices, using regularity from F1 and boundedness from F7 at stages below $\lambda$, and F9 for the recursion. Restriction proves each required $(\ast {)}_{\kappa }$. Every cofinality-$\lambda$ ultrafilter contains $\{\lambda \}$: it avoids $N$ by F3 and cannot concentrate on the coordinates greater than $\lambda$ by F1; the finite intersection rule forces that singleton. At that coordinate the values $\xi$ enumerate $\lambda$, so the sequence is universal.

Otherwise $\mu$ is a limit ordinal and $A\cap \mu$ is unbounded in $\mu$. Indeed, a bound $\beta <\mu$ would give $A\cap \mu =A\cap (\beta +1)\in I$. An unbounded set of infinite cardinals has a cardinal supremum: for each $\gamma <\mu$ some cardinal $a$ satisfies $\gamma <a<\mu$, ruling out any bijection of $\mu$ with a smaller ordinal $\gamma$ by restriction to $a$. Thus $\mu$ is a limit cardinal. Also $\mathrm{cf}(\mu )\le \mathrm{\mid }A\cap \mu \mathrm{\mid }\le \mathrm{\mid }A\mathrm{\mid }<\min A<\mu$ by F7 and progressiveness. It is singular; since $\lambda$ is regular and $\mu \le \lambda +1$, it follows that $\mu <\lambda$. The support $A$ is infinite, and, as $\mu$ is a limit cardinal above $\mathrm{\mid }A\mathrm{\mid }$, $\mathrm{\mid }A{\mathrm{\mid }}^{+}<\mu$. For every infinite regular $\kappa <\mu$ its double successor and the ordinal ${\kappa }^{++}+1$ are below $\mu$. Minimality therefore gives $\{a\in A:a\le {\kappa }^{++}\}\in I$.

In the successor case of step 2.1, set $h(a)=a$ on $N$ and $h(a)=\lambda$ off $N$. These are positive limit ordinals and $h(a)\le a$. Every $f_{\xi }{<}_{I}h$. Given $g{<}_{I}h$, its failure set is a subset of $N$, so $g(a)<\lambda$ for every $a\mathrm{\notin }N$. There are at most $\mathrm{\mid }A\mathrm{\mid }<\lambda$ such values, so F7 bounds their successors by some $\xi <\lambda$; if their supremum is zero, take $\xi =1$. Then $g(a)<\xi =f_{\xi }(a)$ off $N$, proving exactness. This proves the complete conclusion in the successor case. In particular it covers finite $A$: the finitely many initial segments change only at successors, so a least positive one cannot first occur at a limit.

In the limit case of step 2.2, AC in A1 supplies the choice hypothesis of the following two existence results. Apply F4 to obtain a universal sequence $g$, and F5 using F3's directedness to obtain a strict chain $f$ with $g_{\xi }(a)<f_{\xi +1}(a)$ everywhere. For every uncountable regular $\kappa <\mu$, step 2.2 checks both eligibility inequalities and the small-coordinate condition, so this one chain has $(\ast {)}_{\kappa }$. It also has $(\ast {)}_{\omega }$: ${\omega }_1<\mu$ because $\mu$ is a limit cardinal greater than the infinite cardinal $\mathrm{\mid }A\mathrm{\mid }$, and ${\omega }_1$ is regular by F8; restrict each strong ${\omega }_1$ subsequence to its first $\omega$ terms. If $D$ has cofinality $\lambda$, F3 transfers every $I$ comparison to $D$. Given a product function $u$, universality of $g$ supplies $\xi$ with $u{\le }_D{g}_{\xi }{<}_D{f}_{\xi +1}$, proving universality of $f$.

Put $\tau =\mathrm{\mid }A\mathrm{\mid }$. By F8 ${\tau }^{+}$ is regular and uncountable; step 2.2 gives ${\tau }^{+}<\mu <\lambda$ and makes it eligible in F5. Thus F5's projection and exact-bound clauses supply an exact bound $v$ with positive limit values. Let $e(a)=a$, a pointwise bound for every product function. Leastness in F6 gives $v{\le }_{I}e$. Set $h(a)=\min \{v(a),a\}$; it is positive and limit-valued, lies below $a$ everywhere, and differs from $v$ only on the small set $\{a:v(a)>a\}$. Consequently each upper-bound comparison and each comparison $g{<}_{I}h$ is unchanged from $v$, so exactness is preserved. Together with the successor-case construction in step 3.1 and universality in step 3.2 this proves all the assertions. QED.