# Which forcing technique implies "every set is countable from some perspective"? Which notion of "the same set" is used between models?

https://plato.stanford.edu/entries/paradox-skolem/ contains this claim:

Further, the multiverse conception leads naturally to the kinds of conclusions traditional Skolemites tended to favor. Let $$a$$ be a set in some model $$\mathbf M$$ (where $$\mathbf M$$ lives somewhere in the multiverse). Then $$\mathbf M$$ has a forcing extension, $$\mathbf M[\mathbf G]$$, in which $$a$$ is only countable. This provides a natural gloss on the Skolemite claim that “every set is countable from some perspective.” Similarly, the Skolemite's bias in favor of countability (see section 3.2) can be explained by the fact that, if $$a$$ is countable in one model $$\mathbf M$$, then it stays countable in all extensions of that model. In contrast, uncountable sets can always be made countable by passing to an appropriate forcing extension. For more on the multiverse, see Hamkins 2011 and Hamkins 2012. For some criticisms, see Koellner 2013 (under Other Internet Resources).

The claim that every model of set theory can somehow be extended to make any particular set countable sounds very implausible and surprising to me. I haven't been able to figure out what the exact forcing method is that they use to make this claim.

Whatever method is used, I'd like to understand its implications for this simple example: Suppose $$M$$ is a transitive model of ZFC (using the notion of "model" that implies $$M$$ is a set), and $$a \in M$$ such that $$M \models \mathrm{uncountable}(a)$$. Furthermore suppose $$a$$ is uncountable. Is there a model $$M'$$ of ZFC such that "the same set" as $$a$$ is uncountable in $$M'$$?

• Is $$a$$ an object of $$M'$$ and $$M' \models \mathrm{uncountable}(a)$$, or is whoever is making the claim that "every set is countable from some perspective" using a different notion of "the same"?
• What is the notion of how $$M'$$ extends $$M$$? Is $$M \subseteq M'$$? Can $$M'$$ be made to be a standard model?

## Where I got stuck:

The notion of forcing I'm most familiar with is the one described at the beginning of chapter 14 of Jech's Set Theory (http://iitp.ru/upload/userpage/300/jech_03.pdf). But for this one, a generic set $$G$$ is only guaranteed to exist if $$M$$ is countable (which by assumption, it isn't).

I've heard of forcing with Boolean algebras but don't understand the details. In particular, it seems like some equivalence relation is used to collapse the Boolean-valued model into a model of ZFC, and so I'd like to know if "every set is countable from some perspective" just boils down to "every set can be made countable if you quotient out its elements by a sufficiently strong equivalence relation" (if it does boil down to that, I'm still interested in the details).

Hamkins' 2011 paper (https://arxiv.org/abs/1108.4223) looks unhelpful; it doesn't contain much mathematical detail and appears to be full of sleights of hand of things being built out of classes and "let's pretend that". (I expect every claim involving class-sized "models" of ZFC, and all metamathematics, to be translatable into a claims about set-sized models such as my example $$M$$; if that's somehow not possible, I'd like to know why.)

• By Cantor diagonalization, it is straightforward to show that for every list of infinite binary sequences, there is another infinite binary sequence that is not on the list. On the other hand, you will likely have trouble making precise the sentence "every set is countable from some perspective." The complexity of making this sentence precise suggests that this sentence is the one that has errors in its proof and/or errors in its interpretation. Commented May 20 at 23:34
• @Michael what's the relevance of the thing about Cantor diagonalization? Regarding errors, yeah I'm suspecting that, but I want to know if I'm missing something. And there seem to be multiple people involved (Hamkins, the SEP author, one guy who critiqued Hamkins' philosophy but doesn't seem to object to this claim) treating the meaning of this argument as obvious, so I'm hoping someone here might know it too. Commented May 21 at 4:35
• Cantor diagonalization is a very easy and precise argument that there are uncountably infinite sets. The set $B=\{(a_1, a_2, a_3, ...) : a_i \in \{0,1\} \quad \forall i \in \mathbb{N}\}$, which is the set of all infinite binary sequences, is uncountably infinite. Any attempt to map the natural numbers $\mathbb{N}$ to elements of this set necessarily excludes something. That is, no function $f:\mathbb{N}\rightarrow B$ is surjective. You might learn about flaws in your philosophy of countable sets by trying to apply that philosophy to the set $B$. Commented May 21 at 12:02
• Take any function $f:\mathbb{N}\rightarrow B$. This function maps each $n \in \mathbb{N}$ to a binary sequence $f(n)=(f_1(n), f_2(n), f_3(n), ...)$. However, there is no $n \in \mathbb{N}$ that maps to the binary sequence $(1-f_1(1), 1-f_2(2), 1-f_3(3), 1-f_4(4), ...)$ because $f_n(n)\neq 1-f_n(n)$. Commented May 21 at 12:11
• @Michael I think the OP knows the diagonal argument. The question is about forcing. In fact, it is a theorem that for any set $X$ in any model $M$ of ZFC, there is a forcing extension $M'$ of $M$ such that $X$ is countable in $M'$. This is not incompatible with the diagonal argument! Commented May 21 at 12:27

## 1 Answer

The forcing result alluded to is standard/straightforward. Given $$a\in V,$$ let $$P$$ be the set of all finite partial functions $$\omega\to a$$ (according to $$V$$). The generic will be a surjective function $$\omega \to a$$ and we'll have $$V[G]\models \mbox{a is countable}.$$

The thorny part is, as you have touched on, understanding what the forcing extension is, in cases where no generic filter actually exists. To answer your second question in

What is the notion of how $$M'$$ extends $$M$$? Is $$M \subseteq M'$$? Can $$M'$$ be made to be a standard model?

clearly not in the conventional sense. For instance, if $$M$$ is $$V_\alpha$$ for some ordinal $$\alpha$$ then $$M$$ has no extensions at all of the same height, so no forcing extensions in the usual sense from a transitive model approach. In fact, for your specific question, it's even a little more than that (probably a large part of the source of counterintuitiveness)... if $$M$$ is transitive and uncountable and $$a\in M$$ is uncountable in the universe (not just according to $$M$$) then no bona fide extension of $$M$$ to a larger transitive model will have $$a$$ countable... there's no bijection between $$a$$ and $$\omega$$ for the extension to subsume. So even if the forcing extension succeeds in the standard way for some forcing posets in $$M$$, the above collapsing poset won't be one of them.

What we do have is described in theorems 2 and 3 of the Hamkins paper. (Since you didn't initially find it helpful, perhaps you would prefer Hamkins and Seabold, whose introductory sections cover much of the same ground with more detail and less philosophy.) For any model $$V$$, there is an elementary embedding $$j:V\to \overline V$$ into a not-necessarily-well-founded model (this is called the Boolean ultrapower / Boolean ultrapower embedding), that has an extension $$\overline V\subseteq\overline V[G]$$ that for all intents and purposes is the forcing extension of $$\overline V.$$ This entire construction can be carried out in $$V$$ (in the same sense that e.g. we can construct $$V$$'s constructible universe in $$V$$, with the minor differences that in this case we may need to use parameters and that the membership relation for $$\overline V$$ and $$\overline V[G]$$ is nonstandard).

So what we do have, at a minimum, for any $$a\in V$$, is a non-well-founded model where $$j(a)$$ is countable, where $$j(a)$$ is the image of $$a$$ under an elementary embedding (and thus similar to $$a$$ in a strong sense). Theorem 2 is a more syntactic version that maybe be more appealing, though it conflates $$V$$ and the predicate for the ground model, so hides the passage to an elementary extension a bit (the presentation in Hamkins/Seabold is better IMO).

In practice, we typically don't worry about countable models or whether the generic actually exists... since there's a forcing poset that forces it, we just say there's an extension of the ground model / universe where $$a$$ is countable. I think it's fair to say the above theorems form a convincing explanation of why that practice is legitimate, not only in a technical sense (for which there are several well-known approaches), but in the sense that it's very close to being literally true. But it's only "very close", and your mileage may vary on on the "sameness" of $$a$$ and $$j(a)$$ on the philosophical side of things.

• +1 Another "very close" attempt would be to leave $V$ unchanged and treat $V[G]$ as a Boolean-valued model. It seems to me that it does better with "the same" (because $\check a$ is built the same way as $a$) but does worse with "model" (because it isn't $2$-valued). Commented May 21 at 15:17
• @AndreasBlass That's a good point. (And for the OP and others interested, the $V[G]$ of the answer can be obtained as a quotient of said Boolean-valued model by any ultrafilter on the corresponding Boolean algebra. Where the tight isomorphism between $a$ and $\check a$ breaks is that when interpreted by the ultrafilter, the elements of $\check a^U$ are not just the interpretations of (check names of) elements of $a$, but also include mixtures thereof.) Commented May 21 at 17:54