Pcf cofinality ideals and cutoff conventions

Since $\mathrm{pcf}(\mathrm{\varnothing })=\mathrm{\varnothing }$, the empty set belongs to $J_{<\lambda }[A]$. If $E$ belongs and $H\subseteq E$, monotonicity gives $\mathrm{pcf}(H)\subseteq \mathrm{pcf}(E)\subseteq \lambda$. If $E,H$ both belong, finite-union preservation gives $\mathrm{pcf}(E\cup H)=\mathrm{pcf}(E)\cup \mathrm{pcf}(H)\subseteq \lambda$. Thus it is an ideal; the same argument with ${\lambda }^{+}$ gives $J_{\le \lambda }$. Larger cutoffs enlarge the ideals because the corresponding ordinal intervals are nested. For $E\subseteq B\subseteq A$, the statement $\mathrm{pcf}(E)\subseteq \lambda$ does not depend on the ambient set, proving the restriction identity in both directions.

By the singleton calculation, $\mathrm{pcf}(\{a\})=\{a\}$, so its membership is exactly $a<\lambda$, including failure at the endpoint $a=\lambda$. If $\lambda \in \mathrm{pcf}(A)$, then $\mathrm{pcf}(A)\subseteq ̸\lambda$, so $A\mathrm{\notin }{J}_{<\lambda }[A]$. If $A$ is empty, its only ideal here is $\{\mathrm{\varnothing }\}$, improper on the empty support. For nonempty $A$ and $\lambda =0$ or $1$, the singleton calculation and monotonicity show that only the empty subset is null, since all coordinates are infinite. For a finite $A$, the same calculation gives exactly $J_{<\lambda }[A]=\mathcal{P}(\{a\in A:a<\lambda \})$.

Suppose first that $E\in J_{<\lambda }[A]$ and $D$ is an ultrafilter on $A$ containing $E$. Restrict $D$ to $E$; by the support isomorphism its product cofinality is unchanged and belongs to $\mathrm{pcf}(E)$, hence is below $\lambda$. Conversely, if every such $D$ has cofinality below $\lambda$, any ultrafilter on $E$ extends to one on $A$ supported on $E$ with the same cofinality. Thus every member of $\mathrm{pcf}(E)$ is below $\lambda$, proving membership. For $E=\mathrm{\varnothing }$ both the universal ultrafilter assertions are vacuous and membership was proved in step 1.1.

A set $H\subseteq A$ belongs to every ultrafilter of cofinality at least $\lambda$ exactly when none of those ultrafilters contains $A\setminus H$, by the complementary-pair law in F2. By step 2.1 this is exactly $A\setminus H\in J_{<\lambda }[A]$, or $H\in J_{<\lambda }[A{]}^{\ast }$. This proves both inclusions in the dual identity. If there are no high-cofinality ultrafilters, step 2.1 with $E=A$ puts $A$ in the ideal, so the ideal and its dual are both $\mathcal{P}(A)$; the empty-intersection convention gives the same result. No witnesses are chosen in these computations beyond the AC-dependent facts already proved in F1. QED.