Tail suprema and normalized scales

Since $\theta <\lambda =\mathrm{cf}(\lambda )$, F2–F3 bound the index sequence's supremum below $\lambda$. It has no last index and is strictly increasing, so every ${\alpha }_{\zeta }<\delta$, $\delta$ is a limit and the sequence is cofinal there. Thus $\mathrm{cf}(\delta )\le \theta$ by F2. If a cofinal subset $C\subseteq \delta$ had size less than $\theta$, choose for each $c\in C$ the least ${\zeta }_c$ with $c<{\alpha }_{{\zeta }_c}$. Regularity of $\theta$ bounds these indices by some ${\zeta }^{\ast }<\theta$; then $c<{\alpha }_{{\zeta }^{\ast }}<\delta$ for every $c\in C$, contradicting cofinality. Therefore $\mathrm{cf}(\delta )=\theta >\omega$. A1 supplies the AC hypothesis of F3; the individual choices just used were least ordinals.

Fix $n\in T$. Since $n>k\ge m$, $\theta ={\mathrm{\aleph }}_m<{\mathrm{\aleph }}_n=\mathrm{cf}({\mathrm{\aleph }}_n)$. All $g_{\zeta }(n)$ lie below ${\mathrm{\aleph }}_n$ by their product type, so F2–F3 give $g(n)<{\mathrm{\aleph }}_n$. Strict increase with limit length $\theta$ makes $g(n)$ a positive limit, and its displayed sequence is cofinal, giving $\mathrm{cf}(g(n))\le \theta$. For any purported cofinal subset of $g(n)$ of size less than $\theta$, choose for each member the least sequence index above it. A common bound on those indices exists by regularity of $\theta$; the value at that bound would bound the purported cofinal subset strictly below $g(n)$, a contradiction. Hence $\mathrm{cf}(g(n))=\theta$ at every tail coordinate. Extending by zero on the finite prefix gives $\bar{g}\in {\prod }_{n\in B}{\mathrm{\aleph }}_n$.

For $\alpha <\delta$, step 1.1 gives $\zeta <\theta$ with $\alpha <{\alpha }_{\zeta }$. Strictness of the scale gives $f_{\alpha }{<}^{\ast }{f}_{{\alpha }_{\zeta }}$, and the supplied eventual equality on the tail and $g_{\zeta }\le g$ imply $f_{\alpha }{\le }^{\ast }\bar{g}$, since the prefix is finite. Thus $\bar{g}$ is an upper bound of the whole initial segment. Let $u$ be any other product upper bound and put $E=\{n\in T:u(n)<g(n)\}$. For each $n\in E$ choose the least ${\zeta }_n<\theta$ with $u(n)<g_{{\zeta }_n}(n)$, which exists by the supremum definition. If $E$ is empty then $g\le u$ on the tail already. Otherwise $E$ is countable and $\theta$ is uncountable regular, so F2–F3 give ${\zeta }^{\ast }<\theta$ at least all the ${\zeta }_n$. Pointwise monotonicity yields $u(n)<g_{{\zeta }^{\ast }}(n)$ for every $n\in E$. But $g_{{\zeta }^{\ast }}{=}^{\ast }{f}_{{\alpha }_{{\zeta }^{\ast }}}\upharpoonright T{\le }^{\ast }u\upharpoonright T$, since ${\alpha }_{{\zeta }^{\ast }}<\delta$ by step 1.1. Therefore $E$ is contained in the union of those two finite failure sets. In either case $E$ is finite, so $\bar{g}{\le }^{\ast }u$. This proves leastness.

By step 1.1 the index $\delta$ has uncountable cofinality, and step 2.1 supplies a least bound inside the full product. Thus F1 applies and makes $f_{\delta }$ another least bound. Leastness in each direction gives $\bar{g}{\le }^{\ast }{f}_{\delta }$ and $f_{\delta }{\le }^{\ast }\bar{g}$. Outside the union of their finite exceptional sets the values are equal, so $g{=}^{\ast }{f}_{\delta }\upharpoonright T$. Steps 1.1–1.2 give all remaining index, product and coordinate-cofinality claims. QED.