# 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

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

then

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

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).