# Conflating reals and sets of countable ordinals "nicely"

It is consistent with ZFC that $2^{\aleph_1}=2^{\aleph_0}$. This can be gotten easily via forcing; more interestingly, it is a direct consequence of forcing axioms (which also set this value at $\aleph_2$). However, just because a bijection exists between the subsets of $\omega$ and the subsets of $\omega_1$ doesn't mean that such a bijection is "nice" in any sense.

My question is, roughly:

Can we ever have a "nice" injection from $\mathcal{P}(\omega_1)$ to $\mathcal{P}(\omega)$, or at least a "nice" surjection from $\mathcal{P}(\omega)$ to $\mathcal{P}(\omega_1)$?

Exactly what "nice" here means is a bit up for grabs; my instinct at first is to lean on generalized descriptive set theory and ask for something like "Borel on $\omega_1^{\omega_1}$," but I'm not at all sure that's a good thing to do.

Another take might be along the lines of the following:

Is it consistent with ZFC that there is a parameter-freely-definable surjection from $\mathcal{P}(\omega)$ to $\mathcal{P}(\omega_1)$?

We can also ask a "local" version of this question:

Is it consistent with ZFC that for every $s\subset\omega_1$ there is a $r\subset\omega$ with $s\in L[r]$?

While this wouldn't establish the existence of a "nice" correspondence, it would at least establish a more meaningful equivalence than follows immediately from them having the same cardinality. (Note, however, that this would imply that $\omega_1=\omega_1^r$ for some real $r$; since this is contradicted by large cardinals, this seems like it might be the wrong question to ask.)

• Don't you get this by forcing MA over $L$ and using almost disjoint forcing? You need an $\omega_1$-sequence of almost disjoint subsets of $\omega$. You can find one in $L$ definable without parameters. Aug 15 '17 at 5:24
• @AndrésE.Caicedo I'm not very familiar with almost disjoint forcing (although plenty of people have told me I should learn it); can you elaborate a bit? Aug 15 '17 at 5:25
• Sure, I'll post some details as an answer (in 2 hours or so). Aug 15 '17 at 11:41
• The answer to the middle question is yes, because any set can be made definable in a forcing extension by coding it into the GCH pattern. You can do this well above $\omega_1$. So fix any surjection (in a suitable model) and then make it definable. Aug 15 '17 at 13:56
• Joel's argument highlights a requirement you may want, namely, that your definitions are local in a suitable sense. Aug 15 '17 at 15:29

The technique of almost disjoint forcing was introduced in

MR0289291 (44 #6482). Jensen, R. B.; Solovay, R. M. Some applications of almost disjoint sets. In Mathematical Logic and Foundations of Set Theory (Proc. Internat. Colloq., Jerusalem, 1968), pp. 84–104, North-Holland, Amsterdam, 1970.

Fix an almost disjoint family $X=(x_\alpha:\alpha<\omega_1)$ of subsets of $\omega$, so $x_\alpha\cap x_\beta$ is finite for all $\alpha<\beta<\omega_1$. Given $s\subseteq\omega_1$, the poset $\mathbb P_{X,s}$ adds a real $r$ such that $$s=\{\alpha<\omega_1:r\cap x_\alpha\mbox{ is infinite}\}.$$ This shows that $s\in L[X,r]$; if $X\in L$ then in fact $s\in L[r]$. The poset $\mathbb P_{X,s}$ is ccc, and it follows from standard arguments that if $\mathsf{MA}_{\omega_1}$ holds and $\omega_1=\omega_1^L$, then $\mathcal P(\omega_1)=\bigcup_{r\subseteq\omega}\mathcal P(\omega_1)^{L[r]}$. Moreover, by taking as $X$ the first almost disjoint family in the standard enumeration of $L$, we see that $X$ is definable without parameters, and this gives us a simple parameter-free injection of $\mathcal P(\omega_1)$ into $\mathcal P(\omega)$ as long as there is, for instance, a parameter-free definable well-ordering of $\mathcal P(\omega)$. This is consistent with $\mathsf{MA}_{\omega_1}+\omega_1=\omega_1^L$.

(In fact, it is a consequence of $\mathsf{BPFA}+\omega_1=\omega_1^L$, see my paper with Friedman listed below and available at my page. For the consistency result just for $\mathsf{MA}$, this is a result of Friedman, see Theorem 8.51 in his class forcing book.)

In general, in the context of forcing axioms, to get a nice well-ordering of $\mathcal P(\omega_1)$, it suffices to find a nice well-ordering of $\mathcal P(\omega)$ and apply almost disjoint forcing. If we further arrange that the almost disjoint family we begin with is sufficiently nice, this reflects as well in the complexity and/or parameters of the well-ordering of $\mathcal P(\omega_1)$. The same holds for a nice injection of $\mathcal P(\omega_1)$ into $\mathcal P(\omega)$ and similar variants.

MR2895389 (2012m:03123). Caicedo, Andrés Eduardo; Friedman, Sy-David. $\mathsf{BPFA}$ and projective well-orderings of the reals. J. Symbolic Logic 76 (2011), no. 4, 1126–1136.

I like the description of the poset $\mathbb P_{X,s}$ in Jech's book: A condition is a function $f\!:d_f\to2$ whose domain $d_f$ is a subset of $\omega$ such that $d_f\cap x_\alpha$ is finite for all $\alpha\in s$ and $\{n\in d_f: f(n)=1\}$ is finite. The order is inclusion.

Note that if $p,q$ are incompatible conditions, then in particular $\{n:p(n)=1\}\ne\{n:q(n)=1\}$. This shows that $\mathbb P_{X,s}$ is actually Knaster.

For any $p$ and any $\alpha\in\omega_1\smallsetminus s$, we can extend $p$ to a condition $q$ whose domain contains $x_\alpha$ and satisfies $q(n)=0$ for all $n\in x_\alpha\smallsetminus d_p$ (since the $x_\beta$ are almost disjoint).

Similarly, for any $\alpha\in s$ and any $k\in\omega$, we can extend any condition to ensure that its domain meets $x_\alpha$ in a set of size at least $k$.

From this, standard density arguments give us the result.

By the way, it is consistent that $\mathsf{MM}$ holds and there is a parameter-free well-ordering of $\mathcal P(\omega_1)$, which naturally gives us a nice injection of $\mathcal P(\omega_1)$ into $\mathcal P(\omega)$. (I am using "nice" loosely here, but the definition of the well-ordering, and therefore of the injection, is actually not too bad.) See

MR2474445 (2009k:03085). Larson, Paul B. Martin's maximum and definability in $H(\aleph_2)$. Ann. Pure Appl. Logic 156 (2008), no. 1, 110–122.

• I should mention that it is still open whether $\mathsf{MM}$ implies the existence of such a well-ordering. Also, of course you do not need to bring large cardinals into the picture, but since you mentioned forcing axioms, I figured you would be interested in highlighting results in their context. Aug 15 '17 at 15:08