Pcf generators restrict, finitely cover, and carry scales

For nonempty $A_0\subseteq A$, progressiveness persists because $\mathrm{\mid }{A}_0\mathrm{\mid }\le \mathrm{\mid }A\mathrm{\mid }<\min A\le \min A_0$. Fix $\lambda \in \mathrm{pcf}(A_0)$ and put $B=B_{\lambda }[A]$. F6 gives $B\in J_{\le \lambda }[A]$, so F2 gives $B\cap A_0\in J_{\le \lambda }[A_0]$. If $D_0$ on $A_0$ has cofinality $\lambda$, define $D=\{Y\subseteq A:Y\cap A_0\in D_0\}$. F1 gives the same cofinality on $A$, so F6 gives $B\in D$, which means $B\cap A_0\in D_0$. F6's criterion on the progressive support $A_0$ now proves generation; its uniqueness proves equality modulo $J_{<\lambda }[A_0]$ with every chosen generator there.

Fix $\lambda \in \mathrm{pcf}(A)$, $I=J_{<\lambda }[A]$, $B=B_{\lambda }[A]$ and any universal sequence $f$. F6 says $B\mathrm{\notin }I$, so $I\upharpoonright B$ is proper and the restricted chain is strict. Suppose some $h\in \prod B$ is not strictly below any restricted $f_{\xi }$ modulo this ideal. Then $T_{\xi }=\{a\in B:f_{\xi }(a)\le h(a)\}$ is positive for each $\xi$. For $\xi <\eta$, outside the $I$-small failure set of $f_{\xi }<f_{\eta }$ we have $T_{\eta }\subseteq T_{\xi }$. Hence for any finitely many indices with largest index $\eta$, their $T$-intersection contains $T_{\eta }$ minus a finite union of small sets and is positive. Its intersection with $B\setminus E$ for any $E\in I$ is still nonempty. Thus the family of these $T$ sets and the dual ideal on $B$ has the finite intersection property and generates a proper filter. By F7 extend it to an ultrafilter on $B$, and extend by support to $D$ on $A$ using F1. This $D$ contains $B$, avoids $I$, and meets $J_{\le \lambda }$ through $B$; F3 gives cofinality $\lambda$. Extend $h$ by zero off $B$ to a product function $\bar{h}$. Every $T_{\xi }\in D$ yields $f_{\xi }{\le }_D\bar{h}$. Universality gives $\eta$ with $\bar{h}+1{\le }_D{f}_{\eta }$, a contradiction to those two inequalities on their $D$-large intersection. Here $\bar{h}+1$ is a product member because every coordinate is an infinite cardinal. Thus every $h$ is strictly dominated by a restricted term. F1 gives true cofinality $\lambda$, proving clause 3.

Put $M=\max \mathrm{pcf}(A)$, which exists by F4 and is infinite by F1. Every everywhere-cofinal family maps to a cofinal family in an ultraproduct of cofinality $M$, so has cardinality at least $M$. For the opposite inequality, use A1 and F5 to choose a universal sequence for every $\mu \in \mathrm{pcf}(A)$. Its terms are indexed by pairs $(\mu ,\xi )$ with $\mu \le M$ and $\xi <\mu$; there are at most $M$ such pairs by F8, since the set of cardinals at most $M$ injects into $M+1$, which has cardinality $M$. Let $H$ consist of all finite pointwise maxima of those terms, including the empty maximum $0$. For each finite length, F8 bounds the number of tuples by $M$; the countable union still has size at most $M$ by F8 and AC. All members of $H$ belong to the product and $H$ is closed under binary maxima. Given $g\in \prod A$, put $T_h=\{a:g(a)<h(a)\}$. The identity $T_{\max (h,k)}=T_h\cup T_k$ shows that if no $T_h=A$, the subsets of these sets form a proper ideal: its empty member comes from $h=0$, its union closure is the displayed identity, and it omits $A$. Extend its proper dual filter by F7 to $D$. For every $h\in H$, $A\setminus T_h\in D$, so $h{\le }_{D}g$. But $\mu =\mathrm{cf}(\prod A\mathrm{/}D)$ belongs to $\mathrm{pcf}(A)$ and its chosen universal sequence is contained in $H$; cofinality of that sequence dominates $g+1$ in $D$, contradicting its bound by $g$. Some $T_h=A$, proving strict everywhere domination. Weak cofinality has the same lower bound and strict cofinality the proved upper bound, so both equal $M$.

The empty set is covered by the empty family. Suppose a nonempty $X\subseteq A$ failed clause 2, and choose such an $X$ with least $\mu =\max \mathrm{pcf}(X)$; F4 applies by progressiveness as in step 1.1. By that step, $X\cap B_{\mu }[A]$ generates the larger ideal on $X$. Since $X\in J_{\le \mu }[X]$, its remainder $Y=X\setminus B_{\mu }[A]$ belongs to $J_{<\mu }[X]$. If $Y$ is empty, $B_{\mu }$ alone covers $X$. Otherwise F4 gives $\nu =\max \mathrm{pcf}(Y)<\mu$ by F2. Minimality of the counterexample gives a finite cover of $Y$ by generators indexed in $\mathrm{pcf}(Y)\subseteq \mathrm{pcf}(X)$, using F1. Adjoining $B_{\mu }$ covers $X$, again a contradiction. Thus the finite-cover assertion holds. If $X\in J_{<\lambda }[A]$, each index in this cover is below $\lambda$ by F2. Conversely for $\mu <\lambda$ with $\mu \in \mathrm{pcf}(A)$, $B_{\mu }\in J_{\le \mu }[A]\subseteq J_{<\lambda }[A]$ by F2 and F6; indices outside PCF contribute empty sets. Downward and finite-union closure prove the converse ideal characterization, including $\lambda =0$.

For any proper filter $F$, a set $S$ lies in every ultrafilter extension if and only if $S\in F$. One direction is inclusion. For the other, if $S\mathrm{\notin }F$, each $T\in F$ meets $A\setminus S$, since $T\subseteq S$ would force $S\in F$. Finite intersections in $F$ show that adjoining $A\setminus S$ has the finite intersection property; its generated filter is proper and F7 extends it to an ultrafilter omitting $S$. Now if a strict cofinal regular $\lambda$-chain exists modulo $F$, it remains strict and cofinal in every ultrafilter extension. F1 therefore gives cofinality $\lambda$ in all of them. Conversely suppose every extension has that cofinality. There is at least one extension by F7, so $\lambda \in \mathrm{pcf}(A)$. By F6 each extension contains $B_{\lambda }$, and by F3 each avoids $J_{<\lambda }$, hence contains each member of its dual. Applying the just-proved intersection property separately to these sets gives $B_{\lambda }\in F$ and $J_{<\lambda }^{\ast }\subseteq F$. Finally suppose these latter conditions hold. They imply $B_{\lambda }\mathrm{\ne }\mathrm{\varnothing }$, hence $\lambda \in \mathrm{pcf}(A)$ by our empty-generator convention. By F5 and step 1.2 take a scale on $B_{\lambda }$ modulo the restricted ideal and extend each term by zero off $B_{\lambda }$. For every comparison its successful set contains $B_{\lambda }\setminus E$ for some $E\in J_{<\lambda }[A]$, which belongs to $F$. For any product function, restrict it to $B_{\lambda }$ and use the scale there; the same set calculation proves cofinality modulo $F$. Strictness and F1 give true cofinality $\lambda$. This proves all three implications and covers cardinals outside PCF as well.

For an ultrafilter $D$, set $\lambda =\mathrm{cf}(\prod A\mathrm{/}D)$. F6 puts $B_{\lambda }\in D$ and F3 makes $D$ avoid $J_{<\lambda }$. If $\mu <\lambda$, then $B_{\mu }\in J_{<\lambda }$ by the ideal inclusion calculated in step 2.1 (or is empty), so $B_{\mu }\mathrm{\notin }D$. Thus the least generator index in $D$ exists and equals $\lambda$, which also proves the converse characterization of that least index. Clauses 1, 3 and 5 were proved in steps 1.1–1.3, clause 2 in step 2.1 and the filter equivalence in step 2.2. QED.