An Alternative Proof of Scott’s Theorem

A rather famous result of Dana Scott’s is that the existence of a measurable cardinal implies that V\neq L. The original proof is a nice argument utilizing the machinery of elementary embeddings. In this post, I want to give an alternate proof that I came across while reading Paul Larson’s “The Stationary Tower”. Of course, I claimed that I would use this blog to work through that particular text, so a post regarding it is probably overdue. Additionally, this proof utilizes the interaction between measures and elementary submodels. As usual, none of these results are mine, and the proofs mainly follow the ones given in “The Stationary Tower”. Our main tool is the following lemma.

Lemma: Let M be a transitive model of ZFC-Replacement, and let \lambda= M\cap ON. Let \kappa be a cardinal in M, and let \mu\subseteq\mathcal{P}(\kappa) be such that

M\models\mu is a \kappa-complete, uniform, normal measure on \kappa“.

Let Z be an elementary substructure of M such that \mu\in Z, and Z\cap \mathcal{P}\in ([\mathcal{P}(\kappa])^{<\kappa})^M. Assume that either cf(\lambda)>\kappa, or Z\cap \lambda is cofinal in \lambda. For each \gamma\in \kappa, set

Z[\gamma]=\{ f(\gamma) | f:\kappa\rightarrow M \wedge f\in Z \}.

In this case, the following hold.

1) For each \gamma\in \kappa, Z[\gamma] is an elementary submodel of M.

2) \bigcap\{ A\subseteq \kappa : A\in\mu\cap Z\}\neq\emptyset.

3) For each \gamma\in\bigcap\{ A\subseteq \kappa : A\in\mu\cap Z\}, Z[\gamma]\cap\gamma= Z\cap\gamma= Z\cap \kappa.

Before proving the lemma, let’s take a look at the lemma itself. We will be using this in the case that \kappa is measurable, and \lambda limit large enough such that cf(\lambda) > \kappa, and V_\lambda\models “\mu is a \kappa-complete, uniform, normal measure on \kappa“. Here, we see the hypotheses of the lemma are satisfied if we let Z be a countable elementary submodel containing \mu. Next, note that Z\subseteq Z[\gamma] since the function that takes every element of \kappa to x for x\in Z is definable from parameters in Z (\kappa\in Z since \mu\in Z). Finally, note that each Z[\gamma] contains \gamma as an element. With this in mind, let’s prove the lemma.

Proof of Lemma:

Claim 1): We simply need to show that Z[\gamma] satisfies the Tarski-Vaught criterion. So, let x_1,\ldots x_n\in Z[\gamma] be such that

M\models \exists y \phi(y,x_1,\ldots, x_n).

Let f_1,\ldots f_n be functions in Z with domain X such that f_i(\gamma)=x_i for each i. For each \alpha<\kappa, if

M\models \exists y \phi(y,f_1(\alpha),\ldots, f_n(\alpha)),

then let y_\alpha be of minimal rank such that

M\models \phi(y_\alpha,f_1(\alpha),\ldots, f_n(\alpha)).

If we suppose that cf(\lambda)>\kappa, then the set \{rank(y_\alpha) \} is bounded in \lambda. Thus there is some \eta in M such that for every \alpha<\kappa, if

M\models \exists y \phi(y,f_1(\alpha),\ldots, f_n(\alpha)),

then there is a \in M with rank less than \eta such that

M\models \phi(y,f_1(\alpha),\ldots, f_n(\alpha)),

and thus there is some \eta with this property in Z. If Z\cap ON is cofinal in \lambda, then for each rank(y_\alpha), we can find some \xi_\alpha\in Z such that \xi_\alpha>rank(y_\alpha). But then, sup\{ \xi_\alpha \}\in Z, and so in either case, there is some ordinal \eta in Z satisfying the above property. Therefore, in Z there exists a well ordering of all of the elements of M which have rank smaller than \eta. Thus there exists a function g: \kappa\to M in Z such that, for \alpha<\kappa, if

M\models \exists y \phi(y,f_1(\alpha),\ldots, f_n(\alpha)),


M\models \phi(g(\alpha),f_1(\alpha),\ldots, f_n(\alpha)).

But then, g(\gamma)\in Z[\gamma], and by elementarity the existence of such a g finishes the proof of this claim.

Claim 2: Note that |Z\cap \mathcal{P}|^M<\kappa by assumption, and since M thinks that \mu is \kappa-complete, it follows that \bigcap\{A \subseteq \kappa : A\in Z\cap \mu\}\neq\emptyset.

Claim 3: We first show the second equality. We begin by noting that if \alpha\in Z\cap \kappa, then since Z thinks \mu is a uniform ultrafilter, it follows that \kappa\setminus \alpha\in Z\cap \mu. Thus, if \gamma\in \bigcap\{ A \subseteq \kappa : A\in Z\cap \mu\}, we see that in particular \gamma \in \bigcap\{\kappa\setminus\alpha : \alpha\in\kappa\cap Z\}. Thus, Z\cap\gamma=Z\cap \kappa.

To see the first equality, recall that Z\subseteq Z[\gamma], and in particular, Z\cap \gamma\subseteq Z[\gamma]\cap\gamma. So, fix a function f:\kappa\to M, and suppose that f(\gamma)<\gamma. Then, by comprehension, we see that \{\alpha<\kappa : f(\alpha)<\alpha\}\in Z. In particular, we have that \gamma \in \{\alpha <\kappa : f(\alpha) <\alpha\}, and so this set must be in \mu\cap Z. By normality, there is some \beta\in Z\cap\kappa such that \{ \alpha <\kappa : f(\alpha)=\beta\}\in \mu\cap Z. Thus f(\gamma)=\beta, and by elementarity since \beta is definable in Z, it follows that \beta\in Z, which completes the proof.

We will now use the above relationship between measurable cardinals and elementary submodels to prove Scott’s Theorem.

Theorem (Scott): If there exists a measurable cardinal, then V\neq L.

Proof: We begin by letting \kappa be a measurable cardinal, and \lambda a limit ordinal such that cf(\lambda)>\kappa. Now, let X be a countable elementary submodel of V_\lambda. By the above lemma, there exists an elementary submodel Y of V_\lambda such that Y\cap \kappa is a proper end extension of X\cap\kappa. Simply put, since X is countable, there is some \gamma that isn’t in X but is in the intersection of all the full measure subsets of \kappa that X sees. So, we apply the above lemma to X and see that X[\gamma] is the desired set. In fact, since X is countable, we see that Y=X[\gamma] is countable, and thus we may apply the lemma again to yield another countable elementary substructure with the above property. In particular, we may apply the lemma inductively through the countable ordinals. This gives us that, for any function f:\omega_1\to \omega_1, there exists a countable elementary submodel of V_\lambda such that ot(X\cap\kappa)>f(X\cap\omega_1).

This contradicts V=L as follows. We suppose otherwise, assume that V=L, and \kappa is measurable. For each \alpha<\omega_1, we let f(\alpha) be the least \beta such that \alpha is countable in L_\beta. Let \lambda be the least cardinal such that cf(\lambda)>\kappa, and (V_\lambda)^L=L_\lambda. Let X be a countable elementary submodel of V_\lambda such that ot(X\cap \lambda)>f(X\cap\omega_1). Let X\cap\omega_1=\alpha. By condensation, the transitive collapse of X is L_{ot(X\cap\lambda)}, and by assumption, \alpha must be countable in L_{ot(X\cap\lambda}. But, this is a contradiction since X\cap\omega_1=\alpha implies that \alpha is uncountable in L_{ot(X\cap\lambda)}.

It is worth noting that, from the above proof, we can isolate a bit more about the relationship between L, countable elementary submodels, and measurables.

Lemma: If there exists a measurable cardinal \kappa, and \lambda is a cardinal such that cf(\lambda)>\kappa, then for any function f:\omega_1\rightarrow\omega_1, there exists a countable elementary submodel X of V_\lambda such that ot(X\cap \kappa)>f(X\cap\omega_1).

Lemma: If V=L, then there exists a function f:\omega_1\rightarrow \omega_1 such that for any countable elementary submodel X of V_\lambda, where \lambda is the first uncountable strong limit cardinal, ot(X\cap\lambda)<f(X\cap\omega_1).


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