Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

In Devlin's "The Joy of Sets" the author introduces the Boolean valued model $V^{{\mathcal B}}$ based on a given Boolean algebra ${\mathcal B}$ and describes how to assign a truth value $\|\phi\|\in{\mathcal B}$ to any formula $\phi$ of set theory. In particular, one has the following recursion rules for quantifiers: $\|\exists u\phi(u)\|=\bigvee_{u\in V^{{\mathcal B}}} \|\phi(u)\|$ and $\|\forall u\phi(u)\|=\bigwedge_{u\in V^{{\mathcal B}}}\|\phi(u)\|$

I don't understand how this can be turned into a proper definition of $\phi$ since the indexing ranges over the proper class $V^{{\mathcal B}}$. Also, what kind of object is $\|-\|$ supposed to be in the end?

I thought about defining $\|\phi\|$ by restricting quantification to larger and larger $V^{{\mathcal B}}_{\alpha}$, but it is not clear to me why this should stabilize.

Also, I am confused since for ${{\mathcal B}}=\{0,1\}$ it seems that the existence of such a $\phi$ would contradict Tarski's undefinability theorem.

I'm grateful for any clarification!

share|cite|improve this question
up vote 5 down vote accepted

The definition of $\|\varphi\|$ is separate for each formula (that is, the function $\varphi\mapsto\|\varphi\|$ is not definable, and why should it be? $\varphi$ is not an element of $V$ nor $V^\cal B$). This is a recursive construction in the meta-theory, taking a formula in the language of set theory, returning the formula for $\|\varphi\|$.

You're correct when you refer to Tarski's theorem. But this is exactly the case with the reflection theorem that states that $$V\models\varphi\implies\exists\alpha: V_\alpha\models\varphi$$

This theorem is a meta-theorem, and it cannot be internalized (uniformly) for the same reason that $\|\varphi\|$ cannot be internalized (uniformly), and for the same reason that $p\Vdash\varphi$ cannot be internalized (uniformly) when considering the internal definition using pre-dense sets below $p$ (rather than the definition using generic filters).

share|cite|improve this answer
These points confused me the first time I learned about forcing – it seems that most authors regard the theory/meta-theory distinction as being obvious! – Zhen Lin May 18 '14 at 7:38
Well, I can understand why. If I would have been told that when I learned forcing the first time, I'd have been totally confused. But then again, it took almost two years for me to internalize forcing, and then another year to internalize these points. So maybe I was doing it wrong. ;-) – Asaf Karagila May 18 '14 at 7:46
Thank you, Asaf & Zhen! @Asaf: Just to make sure I understand things correctly; how would, for example, the formula $\|u\in v\|$ look like explicitly? Do we prove first, for fixed $\alpha\in\text{Ord}$, the existence and uniqueness of $\|\in\|_\alpha: V^{{\mathcal B}}_\alpha\times V^{{\mathcal B}}_\alpha\to{\mathcal B}$ and $\|=\|_\alpha: V^{{\mathcal B}}_\alpha\times V^{{\mathcal B}}_\alpha\to{\mathcal B}$ satisfying the desired extensionality, and then take as $\|u\in y\|$ the formula $\exists\alpha\in\text{Ord}: x=\|\in\|_\alpha(u,v)$, and similar for $\|u=y\|$? – Hanno May 18 '14 at 11:12
@AsafKaragila How should the reflection meta-theorem be read? Clearly, the left-to-right direction is just the reflection theorem. But I'm finding it difficult to see why the right-to-left direction is true (even as a meta-theorem). – GME May 18 '14 at 11:52
@Hanno: If you know the usual proofs on how to internalize induction, using finite sequences to say that the induction went "this far" and so on, then the process is somewhat similar here as well. It's just that writing explicit formulas for these things is quite a horrible task. And the whole point of these mechanisms of induction is to allow us to ensure the existence of suitable formulas without worrying about what they look like at the end. – Asaf Karagila May 19 '14 at 8:08

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.