Easton head chain condition and tail closure

Under GCH every infinite cardinal $\mu$ satisfies $2^{\mu }={\mu }^{+}$, by [F4]. If $\lambda ={\mathrm{\aleph }}_0$, then $2^{<\lambda }={\sup }_{n<\omega }{2}^n={\mathrm{\aleph }}_0=\lambda$. If $\lambda ={\nu }^{+}$ for an infinite cardinal $\nu$, then GCH gives $2^{\nu }=\lambda$ and monotonicity gives $2^{\mu }\le \lambda$ for every $\mu <\lambda$, so again $2^{<\lambda }=\lambda$. If $\lambda$ is a limit cardinal above ${\mathrm{\aleph }}_0$, the successor cardinals ${\mu }^{+}$ for infinite $\mu <\lambda$ are cofinal in $\lambda$, while each is at most $\lambda$; thus $2^{<\lambda }={\sup }_{\mu <\lambda }{\mu }^{+}=\lambda$. These cases prove clause (a) for every infinite regular $\lambda$.

Now let $⟨p_{\xi }:\xi <\delta ⟩$ be a descending sequence in $P^{>\lambda }$ with $\delta <{\lambda }^{+}$, and put $p={\bigcup }_{\xi <\delta }{p}_{\xi }$. The conditions form a $\subseteq$-chain of functions, so $p$ is a function with values in $\{0,1\}$, and every triple in its domain has first coordinate $>\lambda$. Fix an infinite regular $\gamma >\lambda$ and let $A_{\xi }=\{(\kappa ,\alpha ,\beta )\in \mathrm{dom}(p_{\xi }):\kappa \le \gamma \}$. Each $\mathrm{\mid }{A}_{\xi }\mathrm{\mid }<\gamma$ by [F8], and $\mathrm{\mid }\delta \mathrm{\mid }\le \lambda <\gamma$. Regularity of $\gamma$ gives $\sigma :={\sup }_{\xi <\delta }\mathrm{\mid }{A}_{\xi }\mathrm{\mid }<\gamma$, so $\mathrm{\mid }{\bigcup }_{\xi <\delta }{A}_{\xi }\mathrm{\mid }\le \mathrm{\mid }\delta \mathrm{\mid }\cdot \sigma <\gamma$ (with the finite or empty cases immediate). For $\gamma \le \lambda$ the set in question is empty. Hence $p\in P^{>\lambda }$ and $p$ extends every $p_{\xi }$, so $P^{>\lambda }$ is ${\lambda }^{+}$-closed, clause (c), in the sense of [F3]; the same cardinal bound applies to a pairwise compatible family of at most $\lambda$ conditions, whose union is a function because the members agree on overlaps.

Consequently $2^{\mathrm{\mid }r\mathrm{\mid }}\le 2^{<\lambda }=\lambda$ for every set $r$ with $\mathrm{\mid }r\mathrm{\mid }<\lambda$, and $\lambda \otimes \lambda =\lambda$ by [F6]; with Choice [F9], this bounds a union of $\lambda$ many sets each of size at most $\lambda$ by $\lambda$.

We prove clause (b) by contraposition. Suppose $W\subseteq P^{\le \lambda }$ is an antichain with $\mathrm{\mid }W\mathrm{\mid }={\lambda }^{+}$. By the support condition of [F8] at the regular cardinal $\lambda$, every $p\in W$ has $\mathrm{\mid }\mathrm{dom}(p)\mathrm{\mid }<\lambda$. For any fixed domain there are at most $2^{<\lambda }=\lambda$ bit assignments. Thus there are ${\lambda }^{+}$ distinct domains, since otherwise Choice [F9] and $\lambda \otimes \lambda =\lambda$ would bound $\mathrm{\mid }W\mathrm{\mid }$ by $\lambda$. Select one condition for each distinct domain. Now [F1] applies with $\kappa =\lambda$, $\rho =2^{<\lambda }=\lambda$ of step 1.1 and yields $W^{\mathrm{\prime }}\subseteq W$ of size ${\lambda }^{+}$ and a root $r$ with $\mathrm{dom}(p)\cap \mathrm{dom}(q)=r$ for distinct $p,q\in W^{\mathrm{\prime }}$, in the sense of [F2].

The map $p\mapsto p\upharpoonright r$ takes at most $2^{\mathrm{\mid }r\mathrm{\mid }}\le \lambda$ values on $W^{\mathrm{\prime }}$ by step 2.1, and a union of $\lambda$ many classes each of size at most $\lambda$ has size at most $\lambda$; since $\mathrm{\mid }{W}^{\mathrm{\prime }}\mathrm{\mid }={\lambda }^{+}>\lambda$, two distinct $p,q\in W^{\mathrm{\prime }}$ have $p\upharpoonright r=q\upharpoonright r$.

For such $p,q$ the union $p\cup q$ is a function, because the domains meet exactly in $r$ and the two agree on $r$; all its triples still have first coordinate $\le \lambda$. Let $\gamma$ be an infinite regular cardinal with $\gamma \le \lambda$ and let $A_p=\{(\kappa ,\alpha ,\beta )\in \mathrm{dom}(p):\kappa \le \gamma \}$ and similarly $A_q$. By [F8] both have cardinality below $\gamma$, and $\mathrm{\mid }{A}_p\cup A_q\mathrm{\mid }\le \mathrm{\mid }{A}_p\mathrm{\mid }\oplus \mathrm{\mid }{A}_q\mathrm{\mid }<\gamma$ by [F6], the case of finite cardinalities being immediate since $\gamma$ is infinite. Hence $p\cup q$ satisfies the support condition at every regular $\gamma \le \lambda$, and the bound at $\lambda$ also implies the bound at every regular $\gamma >\lambda$. Thus it is a condition of $P^{\le \lambda }$ extending both $p$ and $q$. This contradicts the antichain property of $W$, so no antichain of $P^{\le \lambda }$ has cardinality ${\lambda }^{+}$: every antichain has cardinality below ${\lambda }^{+}$, which is clause (b) by [F3].

Every condition of $P(F)$ splits uniquely as $p=p^{\le \lambda }\cup p^{>\lambda }$ with each restriction supported on its respective side of $\lambda$; since the support condition at a regular $\gamma$ refers only to triples with $\kappa \le \gamma$, the two parts are conditions of $P^{\le \lambda }$ and $P^{>\lambda }$ respectively, every head-tail pair has disjoint domains and its union satisfies each support bound because the union of two sets of size below an infinite regular $\gamma$ still has size below $\gamma$, and the extension order is preserved in both directions. Hence $p\mapsto (p^{\le \lambda },p^{>\lambda })$ is an isomorphism $P(F)\cong P^{\le \lambda }\times P^{>\lambda }$ when $\mathrm{dom}(F)$ is a set, and for a class Easton function the same computation applies to each set-sized condition; $P^{\le \lambda }$ is then a set, since its conditions are partial functions on the set of triples with $\kappa \le \lambda$ of size below $\lambda$. This is clause (d) and completes the proof. [F8, given] ∎