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Let's take a Vitali set in a model of ZFC, then map its elements to the corresponding reals in the Solovay model and consider them as a set. We get a Vitali set in the Solovay model while it shouldn't exist there. Where is the flaw in that argument?

Would that collection of reals simply not constitute a set in the Solovay model?

Or is there a problem with the mapping?

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  • $\begingroup$ My guess would be that the problem is somehow with the mapping... $\endgroup$
    – Glinka
    Commented Apr 19 at 15:15
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    $\begingroup$ I think the problem is that in the Solovay model you are unable to prove that the mapped set is not measurable (in fact, in the Solovay model the set WILL BE measurable). As an analogy, the (geometric) $2$-sphere as a model for a certain non-Euclidean geometry is such that there is a certain (provable) theorem involving parallel lines, but if you map this $2$-sphere to the model of ordinary Euclidean $3$-space, then the theorem can no longer be proved (in fact, in the Euclidean model the statement WILL BE false). $\endgroup$
    – Dave L. Renfro
    Commented Apr 19 at 16:07
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    $\begingroup$ @DaveL.Renfro But the proof of non-measurability of a Vitali set doesn't seem to rely on the Axiom of Choice. And the construction of real numbers doesn't rely on AC. So can't we construct two identical sets of real numbers in both models, then only invoke AC to find a Vitali set in the ZFC model, then map the elements, and finally for the mapped set prove non-measurability inside the Solovay model? $\endgroup$
    – Maxim
    Commented Apr 19 at 16:46
  • $\begingroup$ I don't know these things in detail, but I do know that one needs to be VERY careful in dealing with non-ZFC models of set theory, since quite often standard results one has learned and standard methods one has used for many years might not hold in the model. $\endgroup$
    – Dave L. Renfro
    Commented Apr 19 at 17:11

2 Answers 2

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The Solovay model $M$ is built by adding a lot of new (generic) sets to the ZFC model $K$ that one starts with. Among those new sets, there are some bijections between $\mathbb N$ (which is the same in $K$ and $M$) and any Vitali set $V$ of $K$. Thus, although $V$ is nonmeasurable in the original model $K$, it is countable and thus measurable (with measure zero) in $M$.

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    $\begingroup$ In fact, the whole real line (and lots of even bigger sets) of the ground model $K$ are countable in $M$. The first uncountable cardinal of $M$ is an inaccessible cardinal of $K$. $\endgroup$
    – Andreas Blass
    Commented Apr 19 at 17:13
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    $\begingroup$ That's mind-bending! If the whole real line is countable in $M$, what happens to the Cantor's diagonal argument there? Wouldn't it be able to find some missing real numbers? $\endgroup$
    – Maxim
    Commented Apr 19 at 17:42
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    $\begingroup$ @Maxim The whole real line of $M$ contains not only the real numbers of $K$ (which form only a countable-in- $M$ set), but also a lot of new (generic) real numbers. Cantor's diagonal argument remains perfectly valid in $M$, since it's a model of ZF. If you apply the diagonal argument (in $M$) to an enumeration of the real numbers of $K$, it will produce one of the new real numbers of $M$. $\endgroup$
    – Andreas Blass
    Commented Apr 19 at 17:53
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    $\begingroup$ I can't wrap my head around the fact that The set of real numbers is a different set in another model. Aren't the constructs like sequences of rationals or decimal digit representations independent of the axiom of choice? $\endgroup$
    – Glinka
    Commented Apr 19 at 18:18
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    $\begingroup$ @Maxim The issue is that the sequence of digits is not an element of $K$, so you cannot use it to specify a real number in $K$. This becomes less mind-bending if you consider that $K$ could be a countable model of set theory, which only contains countably many sequences of digits (from our external perspective) $\endgroup$
    – Alex Kruckman
    Commented Apr 19 at 18:43
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So, you are talking about the Solovay model and the real numbers. Let's look at $L$, which is an inner model of the universe where $\sf ZFC$ holds. It is the smallest one, and so $L$ is included in the Solovay model.

How come that $\sf ZFC$ holds in it and fails in the Solovay model, and specifically, that $\Bbb R$ can no longer be well-ordered there? Well, we added new reals.

But what is a real number? It can be seen as a subset of $\Bbb N$. If you will, in one coding or another. The natural numbers are not changed between $L$, the Solovay model, and the between "the universe". It's always the same natural numbers.

So, if we take some $x\subseteq\Bbb N$ which is in the Solovay model but not in $L$, and we map each $n\in x$ to itself, how come that $x\notin L$?

Well, exactly the same way. The Solovay model perhaps managed to capture all the real numbers of the universe, but it fails to capture all the subsets of the real numbers. It does catch the Borel sets, and more, and enough of those that we can consider sets of reals which are outside the Solovay model a kind of "irrelevant" in a sense, but they are still example of sets of real numbers which are not in the Solovay model itself.

If this is a problematic concept, ask yourself, if every set must be in the Solovay model, in what sense the Solovay model is different from the "full universe of $\sf ZFC$"? And if they are not, then how can the Axiom of Choice fail in one of those and hold in the other? Indeed, different models of set theory (at least those with the same ordinals) must disagree on the power set operation of some set which lies in both models. Otherwise, they are just equal. So, the Solovay model is simply missing out on the non-measurable sets.

Your next question, then, may very well be, what happened to the Vitali set that $L$ had? Well, the reality is that by adding new real numbers the Vitali set of $L$ is actually countable in the Solovay model. But more generally, we can very well add new real numbers—even without changing the cardinality involved—and turn a set of reals into a null set.

So the sets that existed in $L$ that were not measurable are null in the Solovay model; the sets that exist "in the universe" and are not measurable are simply not inside the Solovay model. So there is no real problem.

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  • $\begingroup$ Thank you for a detailed explanation! So would it be correct to say that a non-measurable set mapped from a model of $\sf ZFC$ into the Solovay model can either be a null set (if mapped from a smaller model, e.g. $L$) or not exist (if mapped from a larger model, e.g. the full universe of $\sf ZFC$)? $\endgroup$
    – Maxim
    Commented Apr 19 at 21:34
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    $\begingroup$ The whole question is kinda like asking about why words in English are not words in French, since the letters in English are also letters in French, to be honest. Yes, we can talk about a mapping from a set of reals into the Solovay model, but it's, in a sense, just a function. We might as well as about it being a function from the reals into $\{0,1\}$. I mean, sure, but what about that? What you want to ask about is rather what can we say about sets that are in the Solovay model. Are they measurable? Does the measure agree with the "real" Lebesgue measure, etc. $\endgroup$
    – Asaf Karagila ♦
    Commented Apr 19 at 23:31
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    $\begingroup$ I think that the bigger issue here is to understand how different models of set theory relate to each other, and how the "standard mathematics" behave with respect to these different models. This is the sort of thing that takes time and mathematical maturity to wrap your head around... $\endgroup$
    – Asaf Karagila ♦
    Commented Apr 19 at 23:33

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