Progressive pcf has a maximum and continuous cutoff ideals

Put $P=\mathrm{pcf}(A)$, a nonempty set by F1, and $I={\bigcup }_{\lambda \in P}{J}_{<\lambda }[A]$. This is an ideal: it contains the empty set because $P\mathrm{\ne }\mathrm{\varnothing }$; it is downward closed by F2; and if two members lie in the ideals with cutoffs $\lambda ,\mu \in P$, both lie in the ideal with the larger of those two cutoffs, which also contains their union. It is proper because every constituent omits $A$. Its dual $I^{\ast }=\{A\setminus X:X\in I\}$ contains $A$, omits the empty set and is upward closed and closed under finite intersections, by complementing the ideal axioms. It is therefore a proper filter. F4 supplies an ultrafilter $D\supseteq I^{\ast }$. This $D$ avoids $I$, since containing both $X$ and its complement would put the empty set in $D$.

For each $\lambda \in P$ there is an ultrafilter of cofinality $\lambda$, and F3 supplies $X_{\lambda }\in J_{\le \lambda }[A]\setminus J_{<\lambda }[A]$. The corresponding nonempty witness sets lie in $\mathcal{P}(A)$; AC selects $X_{\lambda }$ simultaneously. If $\lambda <\mu$ in $P$, every member of $\mathrm{pcf}(X_{\lambda })$ is at most $\lambda <\mu$, so $X_{\lambda }\in J_{<\mu }[A]$, whereas $X_{\mu }\mathrm{\notin }{J}_{<\mu }[A]$. Thus $X_{\lambda }\mathrm{\ne }{X}_{\mu }$. This injection $P\to \mathcal{P}(A)$ gives $\mathrm{\mid }P\mathrm{\mid }\le \mathrm{\mid }\mathcal{P}(A)\mathrm{\mid }=2^{\mathrm{\mid }A\mathrm{\mid }}$ by F5: subsets correspond bijectively to their characteristic functions, and a bijection $\mathrm{\mid }A\mathrm{\mid }\to A$ transports these to binary functions on $\mathrm{\mid }A\mathrm{\mid }$.

Put $M=\mathrm{cf}(\prod A\mathrm{/}D)\in P$ by F1. For each $\lambda \in P$, step 1.1 shows that $D$ avoids $J_{<\lambda }$; hence $M\ge \lambda$ by F3. Since $M$ itself belongs to $P$, it is its maximum. In particular the maximum is infinite regular, not merely a supremum outside PCF. This proof applies to every nonempty progressive subset $X\subseteq A$, since $\mathrm{\mid }X\mathrm{\mid }\le \mathrm{\mid }A\mathrm{\mid }<\min A\le \min X$; the first inequality follows from its inclusion by F5.

Fix $\lambda >0$. If nonempty $X\in J_{<\lambda }[A]$, step 2.1 applied to $X$ gives the cardinal $\theta =\max \mathrm{pcf}(X)<\lambda$. Every possible cofinality of $X$ is at most $\theta$, hence $X\in J_{\le \theta }[A]$. For $X=\mathrm{\varnothing }$, use $\theta =0<\lambda$ and $\mathrm{pcf}(\mathrm{\varnothing })=\mathrm{\varnothing }$ from F1. Conversely, if $X\in J_{\le \theta }[A]$ with cardinal $\theta <\lambda$, every member of $\mathrm{pcf}(X)$ is at most $\theta$ and hence below $\lambda$; thus $X\in J_{<\lambda }[A]$. This proves both inclusions in the first identity for every nonzero cutoff, including one and singular cutoffs.

If $\lambda$ is a nonzero limit cardinal, it is infinite. For nonempty $X\in J_{<\lambda }$ take $\mu =\max \mathrm{pcf}(X)<\lambda$ as in step 3.1; since $\lambda$ is not a successor, ${\mu }^{+}<\lambda$, and $X\in J_{<{\mu }^{+}}$. For the empty $X$ choose cutoff zero, which is below $\lambda$ and whose ideal contains the empty set. The reverse inclusion follows from monotonicity of the cutoff ideals, proving the second identity. If $A=\mathrm{\varnothing }$, F1 gives empty PCF and every cutoff ideal equals $\{\mathrm{\varnothing }\}$; each positive-cutoff union has an index (zero), so equals that same ideal, and $0\le 2^0=1$ gives the cardinal bound. If $A$ is finite nonempty, F1 gives $P=A$, whose largest element is its maximum, in agreement with step 2.1. Finally at cutoff zero, F1 and singleton monotonicity show that no nonempty subset is null; hence $J_{<0}[A]=\{\mathrm{\varnothing }\}$, while the union indexed below zero has no members. This verifies the stated exceptions. QED.