## Main Proof (part 3)

March 8, 2011 at 13:43 | Posted in Uncategorized | Leave a commentWe will now define certain increasing sequences and as in the last post, along with another sequence .

Start by setting and . Given and , we let be the least element of above with the property that

and define

What is the point? Note that this arrangement ensures that

for any . Since , it follows that we can choose such that “ majorizes beyond ”, i.e.,

whenever .

Now define

and

We know the following facts from previous posts:

- ,
- , and
- .

Thus, the theorem will be done provided we establish

Here we use the “Elementary submodel argument” from February 12, which tells us that we have what we want provided

The above is satisfied trivially for as ; thus, we need only verify things for which are greater than . Furthermore, we know , so

in any case.

Assume now by way of contradiction that we have such that

- ,
- , and
- .

Choose so large that

- , and
- .

Since and , it follows from the definition of that

But

as both and are in the latter model. Thus, on the one hand we have

and on the other,

Clearly, this is absurd.

## Leave a Comment »

Create a free website or blog at WordPress.com.

Entries and comments feeds.

## Leave a Reply