# Solovay's model, some correspondences and undefined notions

I'm studying the Solovay's model in which every set of reals is Lebesgue measurable (under the assumption of the existence of an inacessible cardinal).

I'm following Jech's book "Set Theory", a master's dissertation of my university and the original article. I have found some problems:

1. I find easier to work with $2^\omega$ (the set of functions from $\omega$ to 2) rather than $\mathbb{R}$ when building Borel codes and forcing with random reals. Let $Fn(\omega,2) = \{s \subset \omega \times 2: |s| < \omega\ and\ s\ is\ a\ function\}$ and $u_s = \{f \in 2^\omega: s \subset f\}$ and let $\tau$ be the topology of $2^\omega$ generated by the basic opens $u_s$. Let $Bor (2^\omega)$ be the borelianos of $2^\omega$ (i.e., the least $\sigma-algebra$ generated by the topology $\tau$). Now let $m: P(2) \rightarrow \{0, 1, 2\}$ be a measure as follows: $$m(\emptyset) = m(\{0\}) = m({1}) = \frac{1}{2} m(2) = 1$$ There is only one $\mu : Bor(2^\omega) \rightarrow [0,1]$ which is regular and satisfies $$\mu(u_s) = \Pi_{\alpha \in dom (s)} m(s(\{\alpha\}))$$ where $s \in Fn(\omega,2)$. If $M,N$ are transitive models of ZFC, $M \subset N$ and $A \in Bor(2^\omega)^M$ then $\mu(A)^M = \mu(A^*)^N$ where $A^*$ is the image IN $N$ of a borel code of $A$.

My first question is: Once the correspondence $A \rightarrow A^*$ preserves the measure on $2^\omega$ for transitive models of ZFC, how can I guarantee that the Lebesgue measure on the reals is preserved as well?

Furthermore, let $A_1, A_2 \in Bor(2^\omega)$ and say they are equivalent iff $\mu(A_1 \Delta A_2)=0$. Now let $[A]$ the corresponding equivalence class and $B_m = \{[A]: A \in Bor(2^\omega)\ and\ \mu(A) > 0\}$. Define a order $\leq$ on $B_m$ by $[A] \leq [B]$ iff $\mu(A - B) = 0$. Claim: $B_m$ has the countable chain condition.

Again, how can I guarantee that the corresponding $B_m$ for Lebesgue measure has the countable chain condition?

In general, my first problem is how to justify that the results for $2^\omega$ follows for the reals with the Lebesgue measure and the usual topology as well.

2.The notion of "$M[x]$", the least transitive model of ZFC which contains $x$. Let $M$ be a model of ZFC and $S$ a set of reals. $S$ is said Soloval over $M$ iff there is a formula $\phi$ with parameters in $M$ such that for all real $x$, $$s \in S \leftrightarrow M[x] \models \phi (x)$$ How can I guarantee that this least model exists? Or this definition makes sense WHEN this model exists?

If I have a $G$ a filter $B_m^M$-generic filter over $M$. We can build a random real $h$, and if we have a random real $h$ we can build a generic model $G$ which contains $h$ and is the least model for ZFC which contains $h$. This is a motivation for defining $M[G] = M[h]$

However I find again the use of this undefied notions: Let $Lev(\kappa, \omega_1)$ be the Levy collapse of the inacessible cardinal $\kappa$ to $\omega_1$ and let $L$ be the complete boolean algebra related to $Lev(\kappa, \omega_1$. (Factor Lemma) Let $G$ an $L$-generic filter over $M$ and $X \in M[G]$ with $M[G] \models X\ is\ a\ countable\ set\ of\ ordinals$. Then there is $H$ an $L$-generic over $M[X]$ such that $M[X][H] = M[G]$

Again, how can I guarantee that this model always exists? Or the factor lemma asserts that there is such a model $M[X]$ and also $M[X][H] = M[G]$?

Maybe those are naive/dumb questions, but I really can't understand specially the second doubt since we have to carefully build the generic G and assert a name to it, etc.

Thank you!

• I find Kanamori's treatment of this model to be very very clear. Commented Oct 4, 2013 at 6:21
• (Also, I find the formatting of your question to be very difficult; and finding the actual questions requires more effort than it should.) Commented Oct 4, 2013 at 6:27

These are good questions, but addressing them properly is delicate and takes some time. Here, I only run through sketches. One needs to be somewhat careful. For example, to address the first question, the key difficulty is making precise what you mean. (I do not mean that this is a fault of your presentation, rather it is an inherent mathematical difficulty of the question.)

Given $M\subset N$ and a set or reals $A$ in $M$, it is not necessarily the case that there is a set of reals $A^*$ in $N$ that "corresponds" to $A$ in any reasonable sense. Of course, in some cases, there is such an $A^*$. For example, $A$ could be a definable set in some sufficiently absolute sense. (Borel sets being given by codes are a good example of what I mean by this. For example, the codes are sufficiently robust that if a model thinks of a real $r$ that it belongs to the set given by a code $t$, then indeed $r$ is in the set coded by $t$ in any outer model as well.) Nowadays, we formalize this in general via the theory of universally Baire sets, which suffices, at least when we deal between models $M$ and their forcing extensions. Here, a set is universally Baire iff there are proper class trees $T,S$ such that their projections are complementary sets of reals in any forcing extension where they are computed, and $T$ projects to $A$. One can check that if $T,S$ and $T',S'$ have this property then, in any forcing extension, the projections of $T$ and $T'$ coincide, so this notion is independent of the specific pair of trees we use as witnesses; moreover, given any real $r$ in some forcing extension, if $r$ is in the projection of $T$ in that extension, then in fact $r$ is in the projection of $T$ in any extension where $r$ belongs.

Borel sets and in fact $\Sigma^1_1$ sets are universally Baire (provably in $\mathsf{ZF}+\mathsf{DC}$), but in the presence of reasonable large cardinal assumptions the notion goes much further. The point is that if $A$ is universally Baire, we have now an unambiguous interpretation of what we mean by $A$, whether we work in $M$ or in an extension of $M$, so we can now address the question.

Regardless of whether we use this formalization or another approach, what we have is that, as long as $A$ has a reasonably robust interpretation, given a Borel code in $M$ for a set $B$, if $B\subseteq A$ in $M$, then the same holds in any outer model where $A$ makes sense. Similarly if instead $B\supseteq A$. Now recall that Lebesgue measure is regular (provably in $\mathsf{ZF}+\mathsf{DC}$), so (as long as $A$ is measurable) the measure of $A$ coincides with the supremum of the measures of the compact sets contained in $A$, and with the infimum of the measures of open sets containing $A$. Since these two numbers are the same, then the measure of $A$ is also independent of the model where it is computed.

The reason that these two numbers coincide is that if $B\subseteq A\subseteq B'$ and $B,B'$ are Borel, then we know that $\mu(B)\le\mu(B')$. Hence, the supremum of the measures of compact subsets of $A$ is at most the infimum of the measures of open supersets of $A$. But already in the ground model we have that these two numbers coincide. In any outer model $N$, the Borel sets of the ground model suffice to verify that these two numbers coincide as well (precisely, because the measure of Borel sets is absolute).

Essentially the same argument shows that the countable chain condition holds for the algebra of Lebesgue measurable sets. The point is that any Lebesgue measurable set has the same measure as a Borel subset of it (the countable union of an appropriate sequence of compact subsets), and this set can be chosen explicitly, so there are no issues with limited amounts of choice here. But then if the algebra of Lebesgue measurable sets had an uncountable antichain, the corresponding Borel subsets would form an antichain as well, and this would be a contradiction.

Of course, the moral of all this is that the key difficulty here is not the transference of results from Borel sets to Lebesgue measurable sets once good "coding devices" are in place but rather, the actual existence of these coding devices. I mentioned universally Baire sets, but their theory was developed more recently than Solovay's results, and anyway it requires some large cardinal assumptions to take us beyond $\Sigma^1_1$ sets. This is why Solovay works with sets definable from a sequence of ordinals, using the sequence and the definition as the coding device, and why he needs results about nice properties of the Levy collapse, so that he can argue that these coding devices preserve enough properties between models so the arguments above can be carried through.

You also ask about $M[x]$. This is unambiguous and is always defined (but the definition seems harder to locate than one would expect): It is $\bigcup_{\alpha\in\mathsf{ORD}} L(\{x\}\cup (V_\alpha\cap M))$ where, if $M$ is a transitive set rather than a class containing all the ordinals, "$\mathsf{ORD}$" is really $\Omega=\mathsf{ORD}^M$, and "$L(T)$" is really $L_\Omega(T)$. For any $x$, this is a model of $\mathsf{ZF}$ (if $M$ is a set model, we restrict to sets $x$ that belong to some forcing extension of $M$). Choice may fail in general, but it holds if $x$ is a set of ordinals. Any real can be seen as a set of ordinals (in fact, natural numbers), so $M[x]$ is a model of choice.

There is a general result that ensures that if $M$ is a model of $\mathsf{ZFC}$, $N=M[G]$ is a forcing extension of $M$, $x\in N$, and $M[x]$ is a model of choice, then $M[x]$ is in fact a forcing extension of $M$. Moreover, if $\mathbb P$ is the complete Boolean algebra associated with the forcing poset that $G$ is generic for over $M$, then we can factor $\mathbb P=\mathbb P_1*\dot{\mathbb Q}$ and $M[x]=M[G']$ for some $G'$ generic over $M$ for $\mathbb P_1$ (and, of course, $M[G]=M[G'][H]$ where $H$ is generic for $\dot{\mathbb Q}_{G'}$ over $M[G']$).

Solovay is a fantastic and very careful writer, but the paper you are reading is one of the first nontrivial papers on forcing, so the language in it may be a bit less friendly than in more recent papers. Besides Jech's book, let me suggest that you read the relevant sections in either Kanamori's excellent book The higher infinite, where this material is also discussed in detail, or in Bartoszyński -Judah Set theory: On the structure of the real line.

• Regarding the definition of $M[x]$, if $P\in M$ is a "complicated enough" notion of forcing, and $G$ is a generic filter, I don't see why $M[G]$ as defined by you, and $M[G]$ as the interpretation of names are the same model. Commented Oct 4, 2013 at 7:44
• The generic extension $M[G]$ contains $M$ and $G$ and is a model of set theory, so it contains $L(\{G\}\cup(M\cap V_\alpha))$ for all $\alpha$. Conversely, the just defined $M[G]$ contains $G$, and it also contains $M$ because it contains $M\cap V_\alpha$ for all $\alpha$. Since it contains $M$, it contains $M^{\mathbb P}$. If $\tau$ is a $\mathbb P$-name, any model of set theory containing $\tau$ and $G$ can decode $\tau_G$. But then our $M[G]$ contains all interpretations of $\mathbb P$-names. So the two versions coincide. Note that the "complexity" of $\mathbb P$ ends up not being an issue. Commented Oct 4, 2013 at 14:15
• I'm a bit traumatic with this notation because Grigorieff uses it in his famous "Intermediate Models" paper, but there he uses $L[x]$ for what we would use $L(x)$; and $M[x]$ as what you are using. It ends up being very confusing altogether, but I guess it sort of makes sense now. I also see why the "complexity" of $\Bbb P$ isn't an issue if we assume $\sf AC$ in $M$. We can just encode it as an order on some $\kappa$, so the generic is a set of ordinals again, and the decoding function comes from $M$ so it's fine. Commented Oct 4, 2013 at 14:19
• Yes, the notation is unfortunate; $M[G]$ should actually be $M(G)$ -- regardless of whether the round- and square-bracket conventions coincide in this case. Commented Oct 4, 2013 at 18:16
• Hello @AndrésE.Caicedo. I know that this question is pretty old, but I am curious: Is this definition of $M[x]$ written in some paper/book?
– user79594
Commented May 13, 2017 at 22:05