Pcf ideal directedness and ultrafilter cofinality cutoffs

If $A=\mathrm{\varnothing }$, the empty function is the sole product member; no ultrafilter exists by F1. If $I=J_{<\lambda }[A]$ is improper, every comparison failure set belongs to $I$, so the zero function weakly bounds every family. Assume henceforth that $A\mathrm{\ne }\mathrm{\varnothing }$ and $I$ is proper. For finite $A$, F1–F2 give $I=\mathcal{P}(S)$ with $S=\{a\in A:a<\lambda \}$. For $\mathrm{\mid }H\mathrm{\mid }<\lambda$, define $b(a)=\sup \{f(a)+1:f\in H\}$ on $A\setminus S$, and $b(a)=0$ on $S$. At each remaining coordinate, $\mathrm{\mid }H\mathrm{\mid }<\lambda \le a=\mathrm{cf}(a)$, so F6 gives $b(a)<a$. Thus $b$ strictly bounds $H$ modulo $I$. The empty $H$ has supremum zero. At $\lambda =0$ there are no families of size less than $\lambda$, so directedness is vacuous.

Let $A$ be infinite and $\tau =\mathrm{\mid }A\mathrm{\mid }$. Progressiveness gives $a>\tau$ for all $a\in A$. If $\lambda \le {\tau }^{+3}$, there are at most three cardinals in $A$ below $\lambda$. Their set $S$ belongs to $I$ by the finite PCF calculation in F1. The supremum formula of step 1.1 on $A\setminus S$ again bounds every $<\lambda$ family. If $\lambda >{\tau }^{+3}$, remove instead $S=A\cap ({\tau }^{+3}+1)$, also finite and $I$-small. Let $A^{\mathrm{\prime }}=A\setminus S$ and $I^{\mathrm{\prime }}=I\cap \mathcal{P}(A^{\mathrm{\prime }})=J_{<\lambda }[A^{\mathrm{\prime }}]$ by F2. Restriction and zero extension preserve and reflect comparisons modulo these ideals: a failure set differs from its restricted version only inside $S\in I$. Bounds on $A^{\mathrm{\prime }}$ therefore extend to bounds on $A$. We have $\min A^{\mathrm{\prime }}>{\tau }^{+3}$, $A^{\mathrm{\prime }}$ infinite and $\mathrm{\mid }{A}^{\mathrm{\prime }}\mathrm{\mid }\le \tau$. If $I^{\mathrm{\prime }}$ were improper, $A^{\mathrm{\prime }}=A\setminus S\in I$ would imply $A\in I$, impossible. It remains to bound families on this proper reduced support.

Keep $I^{\mathrm{\prime }}$ fixed and prove, by induction on cardinals $\rho <\lambda$, that every family $H\subseteq \prod A^{\mathrm{\prime }}$ of size $\rho$ has a bound. This induction is valid because a nonempty set of failed cardinal sizes below $\lambda$ would have a least member. If $\rho <\min A^{\mathrm{\prime }}$, the pointwise formula $b(a)={\sup }_{f\in H}(f(a)+1)$ is below every $a$ by F6 and is a strict bound. If $\rho$ is singular and all smaller sizes are bounded, enumerate $H=(h_{\xi }{)}_{\xi <\rho }$ by AC and fix cofinal indices $({\gamma }_j{)}_{j<\mathrm{cf}(\rho )}$ in $\rho$. Each subfamily with $\xi <{\gamma }_j$ has a bound by induction. AC chooses these bounds simultaneously; the family of bounds has size at most $\mathrm{cf}(\rho )<\rho$, so induction bounds it too. For each $h_{\xi }$ choose $j$ with $\xi <{\gamma }_j$; composing its two weak comparisons gives a common weak bound for $H$, and a coordinate successor gives a strict bound.

It remains to consider regular $\rho \ge \min A^{\mathrm{\prime }}>{\tau }^{+3}$. The induction assumption makes the product $\rho$-directed. Put $\kappa ={\tau }^{+}$, regular by F7 and uncountable because $\tau$ is infinite. Then ${\kappa }^{++}={\tau }^{+3}<\rho$, $\mathrm{\mid }{A}^{\mathrm{\prime }}\mathrm{\mid }\le \tau <\kappa$, and $\{a\in A^{\mathrm{\prime }}:a\le {\kappa }^{++}\}=\mathrm{\varnothing }\in I^{\mathrm{\prime }}$. Apply F3 to an enumeration of $H$, obtaining a strict $\rho$-chain $(f_{\xi })$ with each enumerated member pointwise below $f_{\xi +1}$ and with $(\ast {)}_{\kappa }$. Set ${\tau }^{\mathrm{\prime }}=\mathrm{\mid }{A}^{\mathrm{\prime }}\mathrm{\mid }$. Restrict those strong subsequences to their first $({\tau }^{\mathrm{\prime }}{)}^{+}$ terms; F7 makes that cardinal regular and $({\tau }^{\mathrm{\prime }}{)}^{+}\le {\tau }^{+}=\kappa <\rho$. F4 gives the $({\tau }^{\mathrm{\prime }}{)}^{+}$ projection property, and F5 gives a positive limit-valued exact bound $h$. Cap it pointwise at the coordinate identity, writing $h^{\mathrm{\prime }}(a)=\min \{h(a),a\}$. This is still an upper bound since both entries of the minimum are upper bounds. It is exact because $g{<}_{I^{\mathrm{\prime }}}{h}^{\mathrm{\prime }}$ implies $g{<}_{I^{\mathrm{\prime }}}h$, which is strictly below some $f_{\xi }$ by exactness. Also $h^{\mathrm{\prime }}$ is positive and limit-valued. Replace $h$ by $h^{\mathrm{\prime }}$, so now $h(a)\le a$ everywhere.

Put $B=\{a\in A^{\mathrm{\prime }}:h(a)=a\}$. If $B\mathrm{\notin }{I}^{\mathrm{\prime }}$, F2 gives an ultrafilter $D$ on $A^{\mathrm{\prime }}$ containing $B$, of cofinality at least $\lambda$, and containing $(I^{\mathrm{\prime }}{)}^{\ast }$. Let $J=\{X\subseteq A^{\mathrm{\prime }}:A^{\mathrm{\prime }}\setminus X\in D\}$, a larger proper ideal; comparison modulo $J$ is comparison modulo $D$. F5 transfers exactness of $h$ to $J$. In the linear order $\prod A^{\mathrm{\prime }}\mathrm{/}D$, the $\rho <\lambda$ chain cannot be cofinal, by its cofinality lower bound, so a point witnessing noncofinality strictly bounds the entire chain. Represent it by $t\in \prod A^{\mathrm{\prime }}$. On $B\in D$, $t(a)<a=h(a)$, hence $t{<}_{D}h$. Exactness modulo $D$ then gives $t{<}_D{f}_{\xi }$ for some $\xi$, contradicting that $t$ bounds every chain term. Therefore $B\in I^{\mathrm{\prime }}$. Reset $h$ to zero on $B$. The result belongs to $\prod A^{\mathrm{\prime }}$ because $h(a)<a$ off $B$, and it remains an upper bound modulo $I^{\mathrm{\prime }}$. Since each member of $H$ is below a chain term, it bounds $H$. This closes the regular case of the induction. Steps 1.1–3.1 now give directedness for every cutoff, including singular $\lambda$, after extension to $A$.

Write $\nu =\mathrm{cf}(\prod A\mathrm{/}D)$. If $X\in D\cap J_{<\lambda }[A]$, F2 gives $\nu <\lambda$. Conversely, suppose $D$ avoids $J_{<\lambda }[A]$ and $\nu <\lambda$. AC selects representatives of a cofinal family of $\nu$ quotient classes. The directedness just proved gives these representatives a strict upper bound modulo the proper ideal $J_{<\lambda }[A]$, hence modulo $D$: the complement of every ideal-small set belongs to $D$, since an ultrafilter contains either a set or its complement. A strict bound cannot bound a cofinal family in a linear order without a last element (F1). This contradiction proves $\nu \ge \lambda$ when the ideal is avoided, and establishes both directions of the first equivalence. Apply it also at ${\lambda }^{+}$: avoidance at $\lambda$ means $\nu \ge \lambda$, and meeting at ${\lambda }^{+}$ means $\nu <{\lambda }^{+}$. Since $\nu$ is a cardinal, these two inequalities hold exactly when $\nu =\lambda$. This proves both directions of the second equivalence. If $\lambda$ is finite or singular, $\nu =\lambda$ is impossible because F1 makes $\nu$ infinite regular, and the equivalent right side is correspondingly impossible. QED.