Solovay's Model and Regularity of All Sets of Reals — Examples

The complete restriction map gives $V[G]=V[G_{\xi }][G^{\xi }]$, where $G^{\xi }$ is generic for the quotient supported on $[\xi ,\kappa )\times \omega$. The actual initial generic, not merely $t$, is present at this stage.

Absorption recodes $G_{\xi }$ together with the quotient into an $H$ generic for a fresh $\mathrm{Lv}(\kappa {)}^{V[t]}$, yielding $V[G]=V[t][H]$. A formula $\phi (t,\vec{\alpha })$ omitting $H$ is therefore decided by tail homogeneity; a formula mentioning $G_{\xi }$ need not be fixed.

Choose Borel $B\in N$ representing $b$ modulo null. If $x$ is $N$-random, its ultrafilter on the measure algebra contains $b$ exactly when $x\in B$; the random-forcing truth lemma gives $N[x]\models \psi (x)⟺x\in B$.

Homogeneity makes the $R_x$-Boolean value of $\phi (x,a)$ either $0$ or $1$. Consequently the actual tail-generic extension satisfies $\phi (x,a)$ exactly when the top of $R_x$ forces it, that is, exactly when $N[x]\models \psi (x)$. Step 1.1 therefore gives $V[G]\models \phi (x,a)⟺x\in B$. For a nongeneric $x$ no equivalence is asserted; those exceptions are removed only after placing them in the coded null set. This is a tail-forcing calculation, not upward absoluteness of arbitrary $\phi$.

Below every $q\le p$ there are two conditions forcing incompatible finite prefixes of $\dot{\tau }$. Otherwise all prefixes forceable below some $q$ would be compatible. For every $k$, finite iteration of F2 would then give a unique $u_k\in 2^k$ forceable below $q$. Definability of forcing forms $(u_k{)}_{k<\omega }$ in $N$, and $q$ forces $\dot{\tau }={\bigcup }_k{u}_k\in N$, contradicting the newness hypothesis in F1.

Put $p_{\mathrm{\varnothing }}\le p$ in $D_0$. Use step 1.1 to choose two extensions forcing incompatible prefixes. Successively refine the two ordered pairs into $E_0$; openness preserves the first requirement while the reverse ordered pair is handled. Call the resulting conditions $p_0,p_1$, and strengthen them to decide incompatible prefixes $u_0,u_1$ of length at least $1$.

Below each of $p_0,p_1$, apply step 1.1 to choose two successors. Prefix decisions inherited from the parents separate successors from different parents, and the new splits separate siblings. Successively refine the four nodes through $D_0,D_1$ and all $12$ ordered pairs through $E_0,E_1$, then use F2 to decide extensions $u_{ij}$ of length at least $2$. There are only finitely many requirements, and downward closure preserves every earlier one.

Repeat at level three with eight nodes: meet $D_0,D_1,D_2$ at every node, meet $E_0,E_1,E_2$ for all $8\cdot 7=56$ ordered pairs of distinct nodes, and decide pairwise incompatible prefixes $u_{ijk}$ of length at least $3$.

Thus agreement of branches through level $m$ forces agreement of their interpreted reals through the already decided length-$m$ prefix, while their first split forces distinct interpretations. The continuity modulus is: input agreement through level $m$ implies output agreement through $m$ digits. Continuing the same finite procedure meets every enumerated dense set and realizes the endpoint asserted by F1.

Let $\beta ={\sup }_n({\alpha }_n+1)$ and define $s(\pi (n,k))=s_n(k)$. Then $s:\omega \to \beta$ is in $S$. The fixed inverse of $\pi$ recovers $s_n(k)=s(\pi (n,k))$ uniformly.

Encode the finite formula number and finite ordinal tuple for the $n$th definition in slots $\pi (n,0),\pi (n,1),\dots$, shifting the values of $s_n$ to later tagged slots. Finite tags are ordinals below a common bound. Thus one sequence recovers every formula, ordinal tuple and $S$-parameter, exactly as permitted by F1. Empty tuples use their length tag zero. This is an explicit coding calculation, not an application of the separate omega-closure theorem.

For a Vitali selector, rational translates are disjoint: measure zero makes their countable cover null, and positive measure makes finitely many translates exceed a containing interval.

For a Bernstein set, it and its complement contain no perfect subset; at least one is uncountable, contradicting PSP.

For a Hamel basis, one coefficient kernel is a proper measurable subgroup: positive measure makes it all of $\mathbb{R}$, while measure zero makes its rational-coset cover null.

For an additive map, a positive-measure bounded level set exists; Steinhaus makes the map bounded near zero and hence continuous and linear.

These are exactly the four named cases and use, respectively, translation invariance, PSP, subgroup rigidity, and Cauchy regularity; only countable ideal closure uses DC.

The comparison follows in all four cases.

F5 supplies Countable Choice inside $M$. By F2 and F4 all terms are measurable and $0<V=\lambda (K)<\mathrm{\infty }$. Finite additivity gives $V={\sum }_{i<m}\lambda (A_i)$. F3 gives ${\sum }_{i<m}\lambda (g_i{A}_i)={\sum }_{i<m}\lambda (A_i)=V$.

But disjoint congruent copies give $\lambda (K_0\bigsqcup K_1)=\lambda (K_0)+\lambda (K_1)=V+V=2V$. The same target set was the reassembly in step 1.1, so $V=2V$, hence $V=0$, contradicting $0<V<\mathrm{\infty }$ and verifying F1's exclusion by the promised calculation.

By F1, every $\alpha <\kappa$ has in $V[G]$ a real coding a surjection $\omega \twoheadrightarrow \alpha$; F2 puts each code in both $M$ and $L(\mathbb{R})$, so every such $\alpha$ is internally countable. Conversely, an internal surjection $\omega \twoheadrightarrow \kappa$ would belong to $V[G]$, contradicting F1 there. Thus the designated construction ordinal is ${\omega }_1$ in both inner models and is not inaccessible.

F3 has logical form $\mathrm{Con}(T_{\mathrm{i}\mathrm{n}\mathrm{a}\mathrm{c}\mathrm{c}})\to \mathrm{Con}(T_{\mathrm{r}\mathrm{e}\mathrm{g}})$. It neither reverses this arrow nor inserts “there is an inaccessible” into $T_{\mathrm{r}\mathrm{e}\mathrm{g}}$. The construction also does not prove that no other ordinal can be inaccessible in a chosen target model; that stronger assertion is not needed.

Hence both the proposed survival of the designated $\kappa$ and the inference from relative consistency to an internal inaccessible are invalid.