Pcf cofinality ideals have single generators

First prove the criterion. If $K=I+B$, then $B\in K$ because $B\setminus B=\mathrm{\varnothing }\in I$. If $D$ has cofinality $\lambda$, F2 gives $X\in D\cap K$ and $D\cap I=\mathrm{\varnothing }$. Now $X\setminus B\in I$, so it is not in $D$, and its complement is in $D$. Intersecting with $X$ gives $X\cap B\in D$, hence $B\in D$. Conversely suppose $B\in K$ and every cofinality-$\lambda$ ultrafilter contains $B$. For $X\in K$, any ultrafilter containing $X\setminus B$ contains $X$, so its cofinality is at most $\lambda$ by F1. It cannot equal $\lambda$, since it then contains both $B$ and $X\setminus B$. Thus every such ultrafilter has cofinality below $\lambda$, and F1 gives $X\setminus B\in I$. This proves $K\subseteq I+B$. If $X\setminus B\in I$, then $X\subseteq (X\setminus B)\cup B\in K$ because $I\subseteq K$ and $K$ is an ideal. Thus $I+B\subseteq K$ as well. The complementary-pair rule used here follows directly from maximality of a proper filter: if adjoining a missing set preserved properness it would contradict maximality, so some old member is disjoint from it.

Take $f,h$ from F3 and put $B=\{a\in A:h(a)=a\}$. Every term satisfies $f_{\xi }{<}_{I}h$: $f_{\xi }{<}_I{f}_{\xi +1}{\le }_{I}h$, and $\xi +1<\lambda$ because $\lambda$ is an infinite cardinal by F5. For an ultrafilter $D$ avoiding $I$, its dual ideal $I_D=\{X\subseteq A:A\setminus X\in D\}$ contains $I$ and is proper. Thus the chain remains strict modulo $D$. Its exactness transfers by F4 if $A$ is infinite. For finite $A$ the same transfer follows directly: for $g{<}_{D}h$, reset $g$ to zero where it fails $g<h$. The reset function is everywhere below positive $h$, so exactness modulo $I$ makes it ${<}_I$ some $f_{\xi }$. Returning the old values changes it only on an $I_D$-small set, giving $g{<}_D{f}_{\xi }$. This reset argument works for infinite $A$ too, and explains the transfer without requiring singleton-smallness.

Let $D$ contain $B$. If it meets $I$, F2 gives cofinality below $\lambda$. Otherwise step 1.2 applies. For any $g\in \prod A$, $g(a)<a=h(a)$ on $B$, so $g{<}_{D}h$. Transferred exactness supplies $\xi$ with $g{<}_D{f}_{\xi }$. Therefore the strict chain $f$ is cofinal in $\prod A\mathrm{/}D$, and F5 makes its cofinality $\lambda$. Both cases give cofinality at most $\lambda$ for every $D$ containing $B$. The supported-ultrafilter test F1 therefore gives $B\in K$.

Let $D$ have cofinality $\lambda$. It avoids $I$ by F2. If $B\mathrm{\notin }D$, the complement belongs to $D$ by the rule proved in step 1.1. Define $u(a)=h(a)$ off $B$ and $u(a)=0$ on $B$. Off $B$ we have $h(a)<a$, and on $B$ we have $0<a$, so $u\in \prod A$. Step 1.2 gives $f_{\xi }{<}_{D}h{=}_{D}u$ for all $\xi$. But universality from F3 supplies $\eta$ with $u{\le }_D{f}_{\eta }$, contradicting the strict reverse inequality on the nonempty intersection of two $D$-large comparison sets. Hence $B\in D$.

Steps 2.1–2.2 meet the criterion of step 1.1, giving $K=I+B$. A cofinality-$\lambda$ ultrafilter exists by the hypothesis $\lambda \in \mathrm{pcf}(A)$; it contains $B$ and avoids $I$, so $B\mathrm{\notin }I$ and in particular $B\mathrm{\ne }\mathrm{\varnothing }$. If $C$ also generates $K$, then $B,C\in K$ and the two generation equalities give $B\setminus C\in I$ and $C\setminus B\in I$. Their union $B\mathrm{△}C$ lies in $I$. Conversely changing $B$ on an $I$-small set does not change the condition $X\setminus B\in I$, by taking the union with that small set in each direction. Finally $\mathrm{pcf}(A)$ is a set: ultrafilters on $A$ form a subset of $\mathcal{P}(\mathcal{P}(A))$, and Replacement collects their cofinalities. For each such $\lambda$ the subsets of $A$ generating its ideal form a nonempty set by the proof above. Apply A1 to this indexed family to select all $B_{\lambda }$. No compatibility condition between distinct selections was used or follows from this choice. QED.