Pcf has no holes for progressive intervals

Begin with the directed assertion. Every singleton $\{a\}$ belongs to $I$: otherwise the family $(u_{\xi }{)}_{\xi <a}$ with $u_{\xi }(a)=\xi$ and $u_{\xi }(b)=0$ for $b\mathrm{\ne }a$ has size $a<\lambda$, yet no weak bound $v$ in the product, since $u_{v(a)+1}(a)>v(a)$ and this failure singleton is positive. The index $v(a)+1<a$ because $a$ is an infinite cardinal. Directedness contradicts this. Let $\beta$ be the least ordinal for which $A_0=A\cap \beta$ is positive; it exists since $A=A\cap (\sup A+1)$ is positive. It is not a successor: its preceding initial segment and the possible newly added singleton would both be small. Nor can $A_0$ be bounded below $\beta$, for then an earlier initial segment would equal it. Thus $\beta =\sup A_0=\mu$ is a limit cardinal, as a supremum of unbounded cardinal coordinates, and $A_0$ has no last member. Every proper initial segment of $A_0$ is small. By F7, $\mathrm{cf}(\mu )\le \mathrm{\mid }{A}_0\mathrm{\mid }<\min A_0<\mu$, so $\mu$ is singular. Put $I_0=I\cap \mathcal{P}(A_0)$. It is proper, and directedness restricts: extend any small family of functions by zero outside $A_0$, take an $I$-bound and restrict it. In particular $A_0$ is infinite and progressive and remains an interval.

Independently, the proposed interval identity holds if $A$ is finite, since F1 gives $\mathrm{pcf}(A)=A$ and the interval condition includes exactly all regular cardinals from its minimum to its maximum. For every nonempty progressive $A$, F1 and F4 also give the easy inclusion: its possible cofinalities are regular, at least $\min A$ and at most $M=\max \mathrm{pcf}(A)$.

Continue the directed assertion on $A_0$, with $\tau =\mathrm{\mid }{A}_0\mathrm{\mid }$. Since $\mu$ is a limit cardinal greater than $\tau$, ${\tau }^{+}<\mu \le \sup A<\lambda$. For each uncountable regular $\kappa <\mu$, ${\kappa }^{++}+1<\mu$, and hence $\{a\in A_0:a\le {\kappa }^{++}\}\in I_0$ by step 1.1. F5 applies to this restricted directed product with the constant zero prescribed family and supplies one strict $\lambda$-chain $f$ with every such $(\ast {)}_{\kappa }$. In particular ${\tau }^{+}$ is eligible and regular by F8. The exact-bound clause of F5 applies because $\lambda >{\tau }^{+}$, giving an exact bound $v$. Also $\kappa =\min A_0$ is uncountable, regular, greater than $\tau$ and below $\mu$, so F5 gives its projection property; F6 makes $\{a:\mathrm{cf}(v(a))<\min A_0\}$ small. First choose a positive limit representative using F6, and cap it at the identity: leastness gives $v{\le }_{I_0}(a\mapsto a)$, so $\min (v(a),a)$ is the same ideal class. Off a small set this has cofinality at least $\min A_0$; on the exceptional set replace its value by $a$. Denote the resulting bound by $h$. It remains exact, positive and limit-valued and satisfies $\min A_0\le c(a)=\mathrm{cf}(h(a))\le a$ for every $a$. The upper inequality follows because $h(a)\le a$ and a cofinal subset has size at most $\mathrm{\mid }h(a)\mathrm{\mid }\le a$. By F7 $c(a)$ is regular. The interval property, with endpoints $\min A_0$ and $a$, now gives $c(a)\in A_0$. This is the use of the interval hypothesis in the directed argument.

For each $\xi$, $f_{\xi }{<}_{I_0}h$ because $f_{\xi }{<}_{I_0}{f}_{\xi +1}{\le }_{I_0}h$. Reset $f_{\xi }$ to zero on its individual small failure set to obtain ${\tilde{f}}_{\xi }\in \prod h$. These changes preserve strict comparisons. Exactness makes this chain cofinal in $\prod h\mathrm{/}{I}_0$, so its true cofinality is $\lambda$ by F1. By F7 and A1 choose strictly increasing cofinal maps $e_a:c(a)\to h(a)$ for all $a$. F1 transfers true cofinality to ${\prod }_{a\in A_0}c(a)\mathrm{/}{I}_0$. Put $C=\mathrm{ran}c\subseteq A_0$ and $J=\{E\subseteq C:c^{-1}[E]\in I_0\}$. Preimages preserve finite unions and subsets, and $c^{-1}[C]=A_0\mathrm{\notin }{I}_0$, so $J$ is proper. Moreover $\mathrm{\mid }{A}_0\mathrm{\mid }<\min A_0\le \min C$, precisely the cardinal bound in F1's repetition transfer. Thus $\prod C\mathrm{/}J$ has true cofinality $\lambda$, and F1 gives $\lambda \in \mathrm{pcf}(C)\subseteq \mathrm{pcf}(A_0)\subseteq \mathrm{pcf}(A)$. This proves the directed assertion.

Suppose now $A$ has no largest member. Its supremum $\mu$ is a limit cardinal and $\mathrm{cf}(\mu )\le \mathrm{\mid }A\mathrm{\mid }<\min A<\mu$ by F7, so it is singular. Each regular cardinal $\rho$ with $\min A\le \rho <\mu$ lies in $A$: choose a member of $A$ above $\rho$ and apply the interval property. It therefore belongs to PCF by F1. For a regular $\rho$ with $\mu <\rho \le M$, the ideal $J_{<\rho }[A]$ is proper by F2, since $M\in \mathrm{pcf}(A)$ is not below $\rho$. F3 gives $\rho$-directedness, so step 3.1 applied with $\lambda =\rho$ gives $\rho \in \mathrm{pcf}(A)$. There is no regular cardinal equal to singular $\mu$. Together with step 1.2's opposite inclusion this proves the interval identity when there is no last member.

Finally let $A$ be infinite with a last member. The order type of $A$ is $\delta +n$ for a nonzero limit ordinal $\delta$ and a positive finite $n$. Indeed, repeatedly taking predecessors of a successor order type must reach a limit or zero after finitely many steps: failure would give an infinite descending sequence of ordinals, whose set of values has a least member followed by a smaller one. Reaching zero would make the original order type finite. Split off this finite terminal interval $F$ and let $A_0$ be the initial interval of order type $\delta$, with no maximum. Step 4.1 gives $\mathrm{pcf}(A_0)$ as the regular interval through $M_0=\max \mathrm{pcf}(A_0)$, and F1 gives $\mathrm{pcf}(A)=\mathrm{pcf}(A_0)\cup F$. There is no missing regular between the start of $A$ and $\max (M_0,\max F)$: a regular $\rho \le M_0$ is in the first interval; if $M_0<\rho \le \max F$, then $\rho >\sup A_0$ because $M_0\ge \sup A_0$ by F1, and the original interval condition puts $\rho \in A$, necessarily in $F$. The maximum of this union is $\max (M_0,\max F)$. This proves the identity in the remaining infinite case, while step 1.2 proves the finite case and step 3.1 proves the directed assertion. QED.