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See Noah Schweber's post on MathOverflow: https://mathoverflow.net/q/78083.

He writes:

Let $V$ be a model of $ZFC_2$. Then I claim CH holds in $V$ if and only if $CH$ is actually true. The proof of this is as follows. First, $\omega^V$ has order type $\omega$: clearly it has a subset of order type $\omega$, and by the second-order version of the powerset axiom, $P^V(\omega^V)=P(\omega^V)$, so if $\omega^V$ had the wrong order type $V$ would "see" the error. A fortiori, we can deduce that $\omega^V$ is countable.

Can someone expand on this argument? Specifically I don't understand what relation is implicit in the talk about order types of $\omega^V$ and $P(\omega)^V$, and how the equation $P^V(\omega^V)=P(\omega^V)$ is used in the argument. To be sure: this argument takes place in some first-order set theory (at least strong enough to define $\models_2$ and CH)?

Also, my axiomatization of ZFC2 (see Shapiro's Foundations without Foundationalism) does not include a second-order powerset axiom, only second order replacement: $\forall f \forall x \exists y \forall z (z \in y \leftrightarrow \exists w (w \in x \land z = fw))$. What is the second-order powerset axiom and how does it follow in second-order logic from second-order replacement + the other axioms of ZF?

Jori
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2 Answers2

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I'm going to write "$W$" for our assumed set model of $\mathsf{ZFC_2}$ and leave "$V$" (as usual) for the actual universe of sets. Also, I'll conflate $a\in W$ with $\{b\in W: W\models b\in a\}$; this will be especially benign since the first thing we'll show is that $W$ is well-founded and hence isomorphic to some transitive set.

First, let's outline what we need to do.

Think of "CH-in-W" as the statement

$CH_W:\quad$ There is a bijection in W between W's version of $\omega_1$ and $W$'s version of $2^{\omega}$.

(Or more snappily, $W\models (\omega_1)^W\equiv (2^\omega)^W$.)

Meanwhile, "CH-in-reality" (or "CH-in-$V$" if you prefer) is the statement

$CH:\quad$ There is a bijection between $\omega_1$ and $2^{\omega}$.

We want to show (in $V$) that $CH$ is equivalent to $CH_W$. The issue is that we need the relevant objects and collections to "match up:"

  • $W$ computes $\omega_1$ and $2^{\omega}$ correctly: there are in $V$ bijections $i_1:\omega_1^W\equiv\omega_1$ and $i_2:(2^{\omega})^W\equiv 2^\omega$. An important step towards proving this is to show that $W$ is well-founded, after which we can conflate $W$ with the transitive set it's isomorphic to via Mostowski collapse.

  • $W$ computes equicardinality correctly: for $a,b\in W$ such that there is (in $V$) a bijection between $a$ and $b$, there is also a bijection between $a$ and $b$ in $W$. (Note that the converse holds trivially.)

OK, now let me address the second-order powerset bit.

The definition of $\mathsf{ZFC_2}$ that I was taught included the "second-order powerset" axiom; this is basically the statement that true powersets exist in our model. Specifically, it's:

For every $a$ there is some $b$ such that

  • every element of $b$ is a subset of $a$, and

  • for every $X\subseteq a$ there is some $c\in b$ such that $X=c$.

Here capital letters denote second-order variables, and "$X=c$" is an abbreviation for "$\forall d(d\in X\leftrightarrow d\in c)$." Basically, thinking in terms of a structure $W$, we have that $W\models\mathsf{Pow_2}$ iff for every element of $W$ the true powerset of that element also exists in $W$.

This is however redundant: it follows from second-order Replacement. Given $W\models\mathsf{ZFC_2}$, $a\in W$, and $X\subseteq a$, consider the function $F_X:a\rightarrow a$ sending each $x\in X$ to itself and sending each $y\not\in X$ to some fixed $x\in X$. Now apply the first-order powerset axiom inside $W$.

So if you like, you can think of $\mathsf{Pow_2}$ as a lemma rather than a separate axiom.

We are now all set to prove $CH_W\iff CH$. This is a kind of "bootstrapping" argument: we iteratively demonstrate more and more levels of correctness.

  • First, by $\mathsf{Pow}_2$ we get that whenever $a,b\in W$ we have $a\equiv b\iff W\models a\equiv b$ (think about the set $a\times b$).

  • Any linear order in $W$ which $W$ thinks is well-ordered is actually well-ordered. This is because any descending sequence would be an element of $W$ by $\mathsf{Pow_2}$. This implies that $W$ is well-founded (think about the ordinals), and in particular that $\omega^W=\omega$.

  • This tells us that anything $W$ thinks is countable is actually countable and vice versa: using $\mathsf{Pow_2}$ we have "countability in $W$" = "in bijection with $\omega^W$ in reality" = "in bijection with $\omega$ in reality" (via the previous bulletpoint). But this is equivalent to $\omega_1^W=\omega_1$.

  • Similarly, $\omega=\omega^W$ plus $\mathsf{Pow}_2$ implies $(2^\omega)^W=2^\omega$.

So we get the desired equivalences needed. Continuing this argument we also get e.g. that $W\models 2^{\omega_{17}}=\omega_{18}$ iff in fact $2^{\omega_{17}}=\omega_{18}$; we need to go quite a ways before we get to statements which any $W\models{\mathsf{ZFC_2}}$ could be wrong about.

Finally, as you say this all takes place inside a theory strong enough to talk about second-order logic over set-sized structures. The usual first-order theory $\mathsf{ZFC}$ is indeed sufficient for this task. Specifically, just like first-order logic over a set-sized structure $A$ is dealt with at the level of $\mathcal{P}(A)$, second-order logic over a set-sized structure $A$ is dealt with at the level of $\mathcal{P}(\mathcal{P}(A))$. So $$W\models\mathsf{ZFC_2}\implies(CH\iff CH_W)$$ is formalized in the language of set theory as a statement of the form $$\forall w\mbox{[stuff about $\mathcal{P}(\mathcal{P}(W))$]}\implies(CH\iff \mbox{[stuff about $\mathcal{P}(W)$]}).$$

Note that the above means that what we're really proving is that every set-sized model of $\mathsf{ZFC_2}$ is correct about $CH$. This is because of a limitation of $\mathsf{ZFC}$: it can't even talk about the satisfaction of first-order theories in proper-class-sized structures, let alone second-order theories in such (unless those theories are of bounded quantifier complexity - which ours aren't). If we want to treat class-sized structures we need to pass to a hyperclass theory (just like we need to pass to a class theory in order to talk about first-order semantics of class-sized structures).

Community
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Noah Schweber
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  • I can see that "W ⊨ Pow2 iff for every element of W the true powerset of that element also exists in W" if we assume that W is transitive and its associated membership relation is $\in$. By Mostowski this follows if W is well-founded, which in turn follows easily from second-order replacement and infinity. So then Pow2 does not replace second-order replacement in your axiomatization but it is just a conservative addition, is that what you imply? – Jori Apr 28 '20 at 10:48
  • Thanks a lot by the way for you answer and the time it must have took to write it. I really really appreciate it! For people eager to learn about foundations of mathematics people like you are a goldmine. Even though I study logic at a university (ILLC, Amsterdam) it is still hard to find experts on many mathematical logic topics! – Jori Apr 28 '20 at 10:49
  • OK, I've mostly understood your argument now. Several questions: you say that $W$ is well-founded implies $\omega^W = \omega$, I'm not sure if I see why. I can suggest that: $W \models (\omega^W \models \text{PA})$ implies by the absoluteness of $\models$ (since WLOG $W \models \text{ZF-P}$ is transitive with the real $\in$) that $\omega^W \models \text{PA}$, and so $\omega^W = \omega$ because there are no non-standard well-founded models of $\text{PA}$? Or ... ? – Jori Apr 28 '20 at 15:36
  • What do you mean by "first-order logic over a set-sized structure A is dealt with at the level of P(A)"? Do you refer to the recursion used to define $\models$ involves functions from free variables in a formula to the domain? In that case is it not more accurate to talk about finite subsets of A instead of the complete P(A)? – Jori Apr 28 '20 at 15:42
  • Doesn't your result imply that for any inaccessible $\alpha$, CH is true iff $V_\alpha \models \text{CH}$? Intuitively it appears to me that this would make deciding CH much easier: there is only a single concrete model to check. Presumably this intuition is false: what goes wrong? – Jori Apr 28 '20 at 15:46
  • @Jori "Doesn't your result imply that for any inaccessible $\alpha$, CH is true iff $V_\alpha\models CH$?" In fact there's no need to go to an inaccessible - CH is already correctly determined in $V_{\omega+3}$. "there is only a single concrete model to check" Except that that model really isn't concrete in any simple sense - this is a big takeaway from the development of forcing, for example. – Noah Schweber Apr 28 '20 at 17:52
  • @Jori "What do you mean by "first-order logic over a set-sized structure A is dealt with at the level of P(A)"? ... In that case is it not more accurate to talk about finite subsets of A instead of the complete P(A)?" No. We need to be able to talk about arbitrary functions on $A$ - basically, the putative Skolem functions for the sentences in question - in order to make sense of $\mathcal{A}\models\varphi$ for all sentences $\varphi$ (we can do better if we restrict attention to a specific level in the quantifier hierarchy). – Noah Schweber Apr 28 '20 at 17:53
  • "Why $\omega_1^W\le\omega_1$?" If $W$ thinks $\alpha$ is countable, then $W$ contains a bijection $b$ between $\alpha$ and its $\omega$. But by a prior observation $b$ is in fact a bijection between $\alpha$ and true $\omega$. So $\alpha$ is actually countable ("existential" facts are "upwards absolute"). – Noah Schweber Apr 28 '20 at 17:55
  • If $\omega^W\not=\omega$, then $\omega^W$ is ill-founded. We know right off the bat that $\omega^W$ is a linear order with no top element in which nonzero elements has a predecessor (these facts are appropriately absolute); the only well-founded such order is $\omega$. – Noah Schweber Apr 28 '20 at 17:56
  • Oh, yes I see: if $\omega_1^W > \omega_1$ then since $W$ is well-founded we can assume that $\omega_1 \in W$ (up to isomorphism, or in actuality if we assume that $W$ is transitive), because the ordinals of a well-founded model are isomorphic to an initial segment of the actual ordinals. – Jori Apr 29 '20 at 12:32
  • Is it hard to see that $V_{\omega+3}$ decides CH? What do you mean with "this is a big takeaway from the development of forcing"? That we cannot prove in ZFC whether $V_{\omega+3} \models CH$ or not? For me as a beginner it feels like $\text{HF} = V_\omega$ is fairly natural and then $V_{\omega+3}$ is only a few powersets away. But probably the difficulty is exactly in those few powersets of infinite collections... ? – Jori Apr 29 '20 at 12:36
  • @Jori "Is it hard to see that $V_{\omega+3}$ decides CH?" No - in $V_{\omega+3}$, although you don't have $\omega_1$ itself you do have the linear order $\Omega_1$ of well-orderings of $\omega$ modulo isomorphism which is isomorphic to $\omega_1$, you have $\mathbb{R}$, and (if such exists) you have a bijection between $\Omega_1$ and $\mathbb{R}$. "$V_{\omega+3}$ is only a few powersets away" One of the big takeaways of set theory is that the powerset operation is incredibly complicated. This is a deep observation, one key aspect being the development of forcing. – Noah Schweber Apr 29 '20 at 14:07
  • @Jori Skolem functions are maps $A^n\rightarrow A$ for finite $n$. If $A$ is infinite (finite structures are boring here, so I'm ignoring them) we have $\mathcal{P}(A)\equiv \bigcup_{n\in\omega}A^{(A^n)}$. So there is indeed an equivalence between quantifying over $\mathcal{P}(A)$ and quantifying over potential Skolem functions on $A$. – Noah Schweber Apr 29 '20 at 14:09
  • Oh, yes, how silly of me, that was much, much easier than I realized (but orderings of $\omega$ have rank $\omega+1$, so any order of those should have rank $\omega + 4$?). – Jori Apr 29 '20 at 17:07
  • I'm still not sure whether I understand your comment about the Skolem functions. I went to Kunen to see again exactly how $\models$ is defined. But it seems to me that the most complicated quantifier says "there exists a $d \subset K$" where $K$ is the collection of all pairs $(\varphi, \sigma)$ such that $\varphi$ is a formula and $\sigma$ an assignment for $\varphi$ in $A$. – Jori Apr 29 '20 at 17:12
  • @Jori You can use flat pairing functions to keep ranks lower than one might expect. In fact, it turns out that CH is already decided in $V_{\omega+2}$! See the calculation in the first part of my answer here. That said, if you want to avoid the tedium around paying attention to the specific choice of pairing function, in this context there's nothing really lost by working in (say) $V_{\omega+4}$ instead. – Noah Schweber Apr 29 '20 at 17:13
  • @Jori Re: your second question, first notice that you're overloading "$A$" there. More importantly, think about how you would check that $A\models\forall x\exists y P(x,y)$. This is witnessed by a function sending each $x\in A$ to an appropriate $y\in A$. There's no single assignment of the finitely many variables in $\varphi$ to the elements of $A$ which is enough to check this; intuitivley the point is that our "opponent" has infinitely many possible candidates for $x$ that we need to respond to. (But this is getting too long for a comment thread; maybe ask a separate question about it.) – Noah Schweber Apr 29 '20 at 17:15
  • Although note that per Tarski it can't possibly be the case that we can define $A\models$ inside $A$ in general. In general we have to be forced, somehow, to look a bit "above" $A$ to tell whether $A\models\varphi$ for arbitrary $\varphi$. – Noah Schweber Apr 29 '20 at 17:17
  • Thanks again, your answers have been very insightful. I will open a new question to ask about the 'complexity' of $\models$. One thing though: I just realized that this result follows already directly from the theorem that any model of ZFC2 is an inaccessible rank model, since as you said the truth of CH is already reflected in $V_{\omega+2}$. I believe this is called Zermelo's categoricity theorem, and I recall the proof was quite easy (but I'm not completely sure). – Jori May 01 '20 at 16:31
  • Sorry, about the overloading of $A$, I was fixing it but you were almost immediate in your answer. Does my comment make more sense now? – Jori May 01 '20 at 16:32
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Second-order power set axiom would denote the following statement: if $a$ is a set and $P(x)$ is any second-order predicate which satisfies $\forall x (P(x)\to x\in a)$, then there is $b$ such that $b=\{x\mid P(x)\}$. This follows from the full second-order separation: take $b:=\{x\in a\mid P(x)\}$. Since the second-order replacement proves second-order separation, we are done.

$V$ has its 'own' power set operation, since it is a model of $\mathsf{ZF}_2$ (hence that of $\mathsf{ZF}$.) Noah Schweber's argument shows the internal power set $P^V(\omega^V)$ of $\omega^V$ in $V$ coincides with the true power set $P(\omega^V)$ of $\omega^V$: since the second-order power set axiom (under the full semantics, of course!) catches arbitrary subsets of $\omega^V$.

If $\omega^V$ is not standard (i.e. $\omega^V\neq \omega$,) then $\omega$ is a proper subset of $\omega^V$. Since $V$ contains all subsets of $\omega^V$, we have $\omega\in V$. We can see that $\omega$ is the least inductive set in $V$, but it contradicts with the definition of $\omega^V$ (the least inductive set in $V$.) This shows $\omega=\omega^V$.

As you mentioned, this argument works over a set theory that can formulate $\models_2$. Especially, this argument works over $\mathsf{ZF}$. To note, however, the existence of models of $\mathsf{ZF_2}$ is not provable from $\mathsf{ZF}$ alone. Its existence is equivalent to the existence of an inaccessible cardinal. Our argument would be nothing more than meaningless if we have no models of $\mathsf{ZF_2}$, although the preceding argument holds.

Hanul Jeon
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