Prikry Forcing and Gitik's Singular-Cardinal Model: Examples and Counterexamples

Every tail $A_{\delta }$ belongs to $U$: its complement is the union of fewer than $\kappa$ singletons, while $U$ is nonprincipal and $\kappa$-complete. Its minimum is $\delta +1$. Hence $p,q,r,p^{\ast }$ satisfy the upper-part inequality in F1; in particular, $\alpha <\alpha +1$, $\beta <\beta +1$, and $\gamma <\gamma +1$.

If a condition extended both $p$ and $r$, its stem would end-extend both one-entry stems. Its first entry would then have to be both $\alpha$ and $\gamma$, contrary to $\alpha <\gamma$. Thus $p\perp r$. Notice that shrinking either upper part cannot repair this disagreement at the first stem entry.

For $n<\omega$, write $D_n=\{(s,A):\mathrm{\mid }s\mathrm{\mid }\ge n\}$. From a stem of length $m<n$, choose successively $n-m$ increasing points of its upper part and then shrink above the last chosen point; F1 shows that the resulting condition lies in $D_n$. Thus $D_n$ is dense. For $\eta <\kappa$, let $E_{\eta }$ consist of conditions with nonempty stem and last entry above $\eta$. Given $(s,A)$, the measure-one set $A$ is unbounded, so choose $\xi \in A$ above both $\eta$ and every entry of $s$, append $\xi$, and shrink the upper part to $A\cap A_{\xi }$. This gives an extension in $E_{\eta }$, so $E_{\eta }$ is dense.

The stem $⟨\alpha ,\beta ⟩$ end-extends $⟨\alpha ⟩$, its new entry $\beta$ lies in $A_{\alpha }$, and $A_{\beta }\subseteq A_{\alpha }$. Thus $q\le p$. Their stems differ, so $q\mathrm{̸}{\le }^{\ast }p$. On the other hand $p^{\ast }\le p$, $A_{\beta }\subseteq A_{\alpha }$, and the stems of $p^{\ast }$ and $p$ agree, so $p^{\ast }{\le }^{\ast }p$.

Each $D_n$ and $E_{\eta }$ is a member of $M$, because it is defined there from the ground forcing and the displayed ground parameters. By F3, $G$ meets every one of them. Meeting all $D_n$ makes the compatible stems have union of domain $\omega$, and meeting all $E_{\eta }$ makes that union unbounded in $\kappa$. Since extensions only end-extend strictly increasing stems, the union is a strictly increasing cofinal $\omega$-sequence, exactly as F2 asserts.

For every direct extension $r{\le }^{\ast }p$ and every $\xi <\gamma$, let $C(r,\xi )$ be the nonempty set of direct extensions of $r$ deciding “$\check{\xi }\in \dot{x}$.” Nonemptiness is F1. Use F4 once to choose $d(r,\xi )\in C(r,\xi )$ simultaneously for all such pairs.

Define $p_{\xi }=(s,A_{\xi })$ for $\xi \le \gamma$. Start with $p_0=p$. Given $p_{\xi }$, put $p_{\xi +1}=d(p_{\xi },\xi )$. At a nonzero limit $\delta \le \gamma$, put $A_{\delta }={\bigcap }_{\xi <\delta }{A}_{\xi }$ and $p_{\delta }=(s,A_{\delta })$. Since $\delta \le \gamma <\kappa$, F3 keeps $A_{\delta }$ in $U$. Thus this is a direct-extension decreasing recursion, and $p_{\xi +1}$ decides the $\xi$th membership question.

Put $B={\bigcap }_{\xi \le \gamma }{A}_{\xi }$ and $q=(s,B)$. The family has cardinality below $\kappa$, so F3 gives $B\in U$ and $q{\le }^{\ast }{p}_{\xi }$ for every $\xi \le \gamma$. Define $x=\{\xi <\gamma :p_{\xi +1}⊩\check{\xi }\in \dot{x}\}\mathrm{.}$ For each $\xi <\gamma$, the stronger condition $q$ preserves the decision of $p_{\xi +1}$; hence it forces membership exactly for the ordinals in $x$. Together with $q\le p$ and $p⊩\dot{x}\subseteq \check{\gamma }$, extensionality gives $q⊩\dot{x}=\check{x}$, the conclusion in F2.

When $\gamma =3$, the recursion has $p_0,p_1,p_2,p_3$. If their three successive decisions are “$0\in \dot{x}$,” “$1\mathrm{\notin }\dot{x}$,” and “$2\in \dot{x}$,” then the defining calculation in step 3.1 gives $x=\{0,2\}$ and $B=A_0\cap A_1\cap A_2\cap A_3$. For $\gamma =0$, there are no decisions, $x=\mathrm{\varnothing }$, and the one-factor intersection returns $q=p$; for $\gamma =1$, there is exactly one deciding direct extension.

For each $\alpha <\kappa$, set $A_{\alpha }=\{\xi <\kappa :\alpha <\xi \}$. Its complement is the intersection-complement of the singletons $\{\xi \}$ for $\xi \le \alpha$: nonprincipality puts every $\kappa \setminus \{\xi \}$ in $U$, and F3 keeps their fewer-than-$\kappa$ intersection $A_{\alpha }$ in $U$.

Define $p_{\alpha }=(⟨\alpha ⟩,A_{\alpha })$. Since $\min A_{\alpha }=\alpha +1$, F2 makes $p_{\alpha }$ a condition. If $\alpha \mathrm{\ne }\beta$ and $r$ extended both $p_{\alpha }$ and $p_{\beta }$, the stem of $r$ would end-extend both one-entry stems, so its first entry would have to equal both $\alpha$ and $\beta$. This is impossible. Therefore $\{p_{\alpha }:\alpha <\kappa \}$ is a pairwise incompatible family, and the indexing is injective, so it is an antichain of cardinality exactly $\kappa$.

Because $\kappa$ is uncountable, the antichain in step 2.1 is uncountable. By F4 this violates ccc, furnishing the promised witness to the failure of the refuted statement.

There is no conflict with the weaker positive conclusion in F1. There are only $\kappa$ finite stems, and conditions with the same stem are compatible by F2. Thus among ${\kappa }^{+}$ conditions two share a stem and are compatible, so no antichain has size ${\kappa }^{+}$. The explicit antichain from step 2.1 has size $\kappa$, which is below that forbidden size.