## Finishing up Claim 3.3

September 21, 2011 at 16:35 | Posted in Uncategorized | Leave a commentSo why does work? Let us define

we should show

It should come as no surprise that we’re going to accomplish this using our prior postings on elementary submodels. In particular, we need to jump back to our posting of May 24, where we proved a version of Shelah’s Claim 3.3A. The notation and hypotheses are slightly different in our current setting, and since last May is long time ago, I’ll just formulate another version of things and then prove it.

Lemma 1Suppose is a singular cardinal and is as usual. Suppose we have objects , , and such that

- is an elementary submodel of ,
- is a cardinal less than and in ,
- , and
- for every and , there is an such that .
Then

*Proof:* Suppose this fails, and define

and

It should be clear that is well-defined and . It is also obvious that . We will now show that .

This follows immediately if , as then will be unbounded in . If , then

and so ordinals of the form for are cofinal in . But each of these ordinals is also in (as is in together with all members of ), and this tells us that must contain an ordinal bigger than .

Thus, no matter what the cofinality of , we know

Pausing for a moment, we note that the salient property of guarantees

This is quite powerful, given that we know . For example, this immediately implies that cannot be a successor ordinal, for example, as its predecessor would violate the above condition. Similarly, must have cofinality greater than — if not, then we can use the fact that to derive a contradiction.

We’ll finish by showing that also leads to a contradiction. Given the definition of , we know that any element of belongs to a subset of of cardinality at most that is definable in from finitely many members of . In particular, there is a set such that

- ,
- , and
- is definable in from finitely many members of .

(We discussed this in more detail back in May.)

The definability of guarantees that it is an element of as well, and therefore as it can be defined as the minimal element of .

Thus is an element of . Let us fix a strictly increasing function mapping onto a cofinal subset of . Let us define

Note that . We have assumed , so this means

Thus our assumptions give us an with .

But is an element of as well, hence so is the ordinal . We’ve now got a problem: since is an increasing function from into , it follows that

contradicting the fact that .

Now that we’ve established the lemma, let’s look back at what we need to verify in order to apply it. Looking at the bullet points in the statement of our lemma, all we need to worry about is the fourth one. Suppose and . Choose large enough so that

Looking back at the construction, we then have

and our choice of guarantees the existence of a function with

Thus, the hypotheses of the lemma have been satisfied by the construction from the previous post, and we are finished.

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