Symmetric Collapse and Ultrafilter-Free Models: Examples and Counterexamples

From F1, $H_0=\mathcal{G}$, $H_1$ fixes forcing layer $0$, and $H_2$ fixes layers $0$ and $1$. Thus F2 says that $R_0$ uses no generic collapse layer, $R_1$ may use only layer $0$, and $R_2$ may use only layers $0$ and $1$. More explicitly, the fixed-value theorem built into F2 identifies their Boolean coefficients with the complete algebras of $P\upharpoonright 0$, $P\upharpoonright 1$, and $P\upharpoonright 2$, respectively.

Instantiating F3 and F4 gives the three concrete composites $g_0=e_0\circ f_1:\omega \twoheadrightarrow R_0,g_1=e_1\circ f_2:\omega \twoheadrightarrow R_1,g_2=e_2\circ f_3:\omega \twoheadrightarrow R_2\mathrm{.}$ Here $e_m$ codes the initial-layer Boolean names, while the next unused canonical collapse $f_{m+1}$ makes its ordinal domain countable in $N$; F5 verifies this composition in general. This is a finite list of specified maps, so forming the triple $(g_0,g_1,g_2)$ requires no Choice.

F6 says every real lies in some later $R_m$. Suppose, however, that $N$ contained a sequence $⟨h_m:m<\omega ⟩$ with each $h_m:\omega \twoheadrightarrow R_m$. Then $q(m,k)=h_m(k)$ maps $\omega \times \omega$ onto ${\bigcup }_m{R}_m={\mathbb{R}}^N$; repeated or equal layers do not affect surjectivity. Composing with the explicit Cantor pairing bijection between $\omega$ and $\omega \times \omega$ would make the reals countable, contradicting F7. Therefore the individual maps illustrated in step 2.1 cannot be assembled for all layers inside $N$. The obstruction is precisely simultaneous countable choice, not failure of any fixed layer enumeration.

Let $T$ be the consistent theory supplied by F1. It contains ZF and asserts that a sequence $⟨R_m:m<\omega ⟩$ consists pointwise of countable sets and satisfies $\mathbb{R}={\bigcup }_{m<\omega }{R}_m$. Since $T$ contains ZF, it also proves from F2 that $\mathbb{R}$ is not countable.

Suppose for contradiction that ZF proved the statement refuted. Then $T$ would inherit that theorem. Applying it to the specific sequence in step 1.1 would make $\mathbb{R}$ countable, contradicting the same step's ZF proof that $\mathbb{R}$ is uncountable. Thus $T$ would be inconsistent, contrary to F1, and the claimed ZF theorem is not provable under the stated consistency hypothesis.

Let $T$ be the consistent theory supplied by F1. It contains ZF and the exact equality $\mathrm{cf}({\omega }_1)=\omega$. By F3, the ZF part of $T$ proves $\omega \mathrm{\ne }{\omega }_1$. The ordinals here are the target model's own $\omega$ and ${\omega }_1$; no ground-model ordinal is being substituted.

Suppose for contradiction that ZF proved the statement refuted. By F2, $T$ would then prove $\mathrm{cf}({\omega }_1)={\omega }_1$. Together with step 1.1 it would prove $\omega ={\omega }_1$, contradicting the ZF theorem recorded there. This would make $T$ inconsistent, contrary to F1. Hence, under $\mathrm{Con}(\mathrm{Z}\mathrm{F})$, regularity of ${\omega }_1$ is not provable in ZF.

Let $A=\{(n+1,k):k\ge k_0\}$ and let ${\pi }_A$ toggle the value of a condition exactly on $A$. By the Given hypothesis, $A\cap \mathrm{dom}(p)=\mathrm{\varnothing }$, so no value of $p$ is changed and ${\pi }_{A}p=p$. Coordinates other than $n+1$ are fixed pointwise.

Write $S=S_{n+1}$. For $k<k_0$, the flip does not act and $k\in {\pi }_{A}S$ exactly when $k\in S$. For $k\ge k_0$, it toggles the generic bit and $k\in {\pi }_{A}S$ exactly when $k\mathrm{\notin }S$. Hence ${\pi }_{A}S\mathrm{△}(\omega \setminus S)=\{k:k<k_0\}=k_0\mathrm{.}$ When $k_0=0$ this is exact complementation; when $k_0=1$ the only possible discrepancy is bit $0$. For every $k_0$, the discrepancy is finite, while the flipped set is an infinite tail. No selection or Choice principle is used.

No finite set $F$ belongs to $U$. Otherwise, since $U$ is not principal, no singleton $\{n\}$ belongs to $U$; F2 then puts every $\omega \setminus \{n\}$ in $U$. Intersecting these complements for the finitely many $n\in F$ puts $\omega \setminus F$ in $U$, so propriety is contradicted by $F\cap (\omega \setminus F)=\mathrm{\varnothing }$. This includes $F=\mathrm{\varnothing }$, which is excluded directly by propriety. By F2, every cofinite set therefore belongs to $U$.

Put $D=\omega \setminus (X\mathrm{△}Y)$. By F3 this is exactly the set on which $X$ and $Y$ agree, and it is cofinite by the Given hypothesis; hence $D\in U$ by step 1.1. If $X\in U$, then $X\cap D\in U$, while agreement gives $X\cap D\subseteq Y$, so upward closure gives $Y\in U$. Exchanging $X$ and $Y$ proves the reverse implication. Thus the displayed equivalence holds, including $X=Y$, $X=\mathrm{\varnothing }$, and $X=\omega$.

A flip of only finitely many bits replaces $X$ by some $Y$ with finite $X\mathrm{△}Y$, so step 2.1 preserves its membership status in $U$ and supplies no contradiction. Feferman's automorphism instead sends the selected real to a finite modification of $\omega \setminus X$: invariance then transfers its membership to the complement, which conflicts with F2. The distinction is between a finite flip and an infinite tail flip, not between two descriptions of the same automorphism.

Suppose $c\in N$ chooses one member of $f(k)$ for every $k$ in an infinite $K\subseteq \omega$. By F2, a unique definition of $c$ uses only $f$, ordinals, and reals $s_1,\dots ,s_t\in S\setminus \{f\}$ coming from finitely many coordinate indices $m_1,\dots ,m_t$. Let $k$ be the least member of $K\setminus \{m_1,\dots ,m_t\}$; this canonical fresh choice uses no Choice principle. Interchanging the two labels if necessary, suppose $c(f(k))=\delta (a_k)$.

By F4 choose a finite $p\in G$ forcing both the unique defining formula for $c$ and $c(f(k))=\delta (a_k)$. Let $b=0$ if $p$ mentions no bit of row $k$, and otherwise let $b$ exceed every mentioned bit. Flip all $(k,j)$ with $j\ge b$. By F3 this fixes $p$, every ordinal, and each $s_i$. It sends $a_k$ to a finite modification of $\omega \setminus a_k$, so it interchanges $\delta (a_k)$ and $\delta (\omega \setminus a_k)$. It fixes $f(k)$ as an unordered pair and fixes all other values of $f$, hence fixes $f$.

Apply F5 to the assertion forced in step 2.1. Because its condition and every defining parameter are fixed, the same $p$ forces that the same uniquely defined $c$ satisfies $c(f(k))=\delta (\omega \setminus a_k)$. Since $p\in G$, both value equations hold in the extension, but F1 says their right-hand sides are distinct. This contradicts functionality of $c$. Thus no choice function exists on an infinite subfamily. A choice on the empty subfamily, on one pair, or on any other fixed finite subfamily is not excluded; the obstruction is exactly the infinite partial-choice claim. The example uses no assertion about ultrafilters on arbitrary sets.