# Do the real numbers “know” that they are countable in a larger model?

(This was first posted to math.stackexchange but had no answers there after several days):

Let ${\mathbb R}$ be the set of real numbers in whatever is your favorite model of $ZFC$. Then (by Levy collapse) there is some larger model in which ${\mathbb R}$ is countable.

I wonder whether there are any mathematically interesting facts about ${\mathbb R}$ that are, on the one hand, entirely internal to the original model, but on the other hand, best proven (or perhaps best discovered) by reference to the countability in the larger model.

In other words, I am looking for arguments of the following form: "Choose an enumeration of ${\mathbb R}$ in some larger model $M'$. Then ..... and therefore $X$'', where $X$ is a statement about the real numbers that would both make sense and be interesting to a person who had never heard of model theory. Is there a reason to believe that no such arguments are likely to exist?

• To the question in the title: No. They do not know that. – Asaf Karagila Mar 3 '15 at 5:45

## 2 Answers

In the theory of Borel equivalence relations, one can define something called a pinned equivalence relation. Roughly speaking, a Borel equivalence relation $E$ on a Polish space $X$ is unpinned if there is a forcing which adds $E$-classes consisting only of new elements of $X$. (For a more precise definition, see http://users.math.cas.cz/~zapletal/f2note.pdf. Kanovei's book also has a good discussion, though I don't have my copy on me currently to give a more specific reference than that.) So a Borel equivalence relation is pinned if any new elements of $X$ added by forcing are $E$-equivalent to old elements of $X$ (again, roughly speaking).

Many Borel equivalence relations are pinned. For example, every countable Borel equivalence relation is pinned. Also notably, the relation $E_{l^\infty}$ on $\mathbb{R}^\mathbb{N}$, which makes two sequences of reals equivalent if their $l^\infty$ distance is finite, is pinned. It's also not too hard to show that being pinned is closed downward under Borel reducibility, i.e. if $E$ Borel reduces to $F$ and $F$ is pinned, then $E$ is pinned.

On the other hand, let $E_{ctble}$ be the equivalence relation on $\mathbb{R}^\mathbb{N}$ given by $(x_n) E_{ctble} (y_n)$ iff $(x_n)$ and $(y_n)$ enumerate the same set of reals. One can see that $E_{ctble}$ is unpinned by doing a collapse forcing. If you collapse the continuum to be countable, then you get elements of $\mathbb{R}^\mathbb{N}$ in the new model which enumerate all of the old reals. Clearly these sequences are not $E_{ctble}$-equivalent to any of the old elements of $\mathbb{R}^\mathbb{N}$!

This means that $E_{ctble}$ does not Borel reduce to $E_{l^\infty}$. I would say that this is a fact about (equivalence relations on) $\mathbb{R}^\mathbb{N}$ which is completely internal to the model, but as far as I know, right now the only way we know to prove it is through this forcing argument.

Probably the interesting examples involve large cardinals.

Instead of just collapsing $\mathbb R$ to be countable, force with $Col(\omega,<\kappa)$, where $\kappa$ is an inaccessible cardinal. This is the direct limit of repeatedly making the reals countable. So for each ordinal $\alpha \leq \kappa$, there is a new version of the reals $\mathbb R_\alpha$, and for $\alpha < \kappa$ these all become countable in the end. This feature is essential to the landmark theorem of Solovay that in the final model, every definable set of reals--definable meaning is in $L(\mathbb R)$--is Lebesgue measurable, equal to an open set modulo a meager set, and if uncountable contains a perfect subset. The proof is presented in Chapter 26 of Set Theory by Thomas Jech. It is a bit involved, but a key step that relates to your question is the following. Suppose $\mathbb R$ is the version of the reals in the final model, and $r \in \mathbb R$. Then in the final model, the intersection of all measure one Borel sets coded in $L(r)$ is measure one in $L(\mathbb R)$. This is because in the final model there are only countably many of these.

Now with larger cardinals this enables us to actually say things about "our own universe." If there is a Woodin cardinal $\delta$, then there is a forcing called the stationary tower that turns $\delta$ into $\aleph_1$ and has some special features. Say we start in the model $V$ and $G$ is the generic filter for the stationary tower. Then:

(1) There is a submodel $M$ of $V[G]$ that is closed under countable subsequences (thus contains all the new reals), and an elementary embedding $j : V \to M$.

(2) There is another forcing extension $V[H]$ with the same reals as $V[G]$, where $H$ is generic for $Col(\omega,<\delta)$.

Thus Solovay's theorem may be applied to say that every set of reals in $L(\mathbb R)^{V[G]}$ has all the regularity properties. By the elementarity of $j$, this holds for $L(\mathbb R)^V$ as well.

There are set theorists on MO who are far more expert than me on these topics, so probably even more fascinating results can be given.

• I still wouldn't say that $\Bbb R$ "knows" it's countable somewhere. – Asaf Karagila Mar 3 '15 at 7:13
• I think this is a problem only for the title, not the actual question. – Monroe Eskew Mar 3 '15 at 7:16
• We don't quite "choose an enumeration of the reals" in a larger model either. – Asaf Karagila Mar 3 '15 at 7:17
• Sure we do, when we argue using the fact the intermediate $\mathbb R$'s are countable, order to show almost all reals are Random or Cohen or something. – Monroe Eskew Mar 3 '15 at 7:19
• I don't really understand your comment because we still get to make the conclusion about properties of the ground model reals in the presence of a Woodin cardinal. And the countability of "old" sets of reals is key to this. – Monroe Eskew Mar 4 '15 at 15:42