A rather famous result of Dana Scott’s is that the existence of a measurable cardinal implies that . 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 be a transitive model of ZFC-Replacement, and let . Let be a cardinal in , and let be such that

“ is a -complete, uniform, normal measure on “.

Let be an elementary substructure of such that , and . Assume that either , or is cofinal in . For each , set

.

In this case, the following hold.

1) For each , is an elementary submodel of .

2) .

3) For each , .

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

**Proof of Lemma**:

**Claim 1)**: We simply need to show that satisfies the Tarski-Vaught criterion. So, let be such that

.

Let be functions in with domain such that for each . For each , if

,

then let be of minimal rank such that

.

If we suppose that , then the set is bounded in . Thus there is some in such that for every , if

,

then there is a with rank less than such that

,

and thus there is some with this property in . If is cofinal in , then for each , we can find some such that . But then, , and so in either case, there is some ordinal in satisfying the above property. Therefore, in there exists a well ordering of all of the elements of which have rank smaller than . Thus there exists a function in such that, for , if

,

then

.

But then, , and by elementarity the existence of such a finishes the proof of this claim.

**Claim 2**: Note that by assumption, and since thinks that is -complete, it follows that .

**Claim 3**: We first show the second equality. We begin by noting that if , then since thinks is a uniform ultrafilter, it follows that . Thus, if , we see that in particular . Thus, .

To see the first equality, recall that , and in particular, . So, fix a function , and suppose that . Then, by comprehension, we see that . In particular, we have that , and so this set must be in . By normality, there is some such that . Thus , and by elementarity since is definable in , it follows that , 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 .

**Proof**: We begin by letting be a measurable cardinal, and a limit ordinal such that . Now, let be a countable elementary submodel of . By the above lemma, there exists an elementary submodel of such that is a proper end extension of . Simply put, since is countable, there is some that isn’t in but is in the intersection of all the full measure subsets of that sees. So, we apply the above lemma to and see that is the desired set. In fact, since is countable, we see that 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 , there exists a countable elementary submodel of such that .

This contradicts as follows. We suppose otherwise, assume that , and is measurable. For each , we let be the least such that is countable in . Let be the least cardinal such that , and . Let be a countable elementary submodel of such that . Let . By condensation, the transitive collapse of is , and by assumption, must be countable in . But, this is a contradiction since implies that is uncountable in .

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

**Lemma**: If there exists a measurable cardinal , and is a cardinal such that , then for any function , there exists a countable elementary submodel of such that .

**Lemma**: If V=L, then there exists a function such that for any countable elementary submodel of , where is the first uncountable strong limit cardinal, .