# Grigorieff forcing collapsing the continuum

This is a short technical post, more a note-to-self so that I know where to look this up if I ever need it again. It is also somewhat of a correction of something I said during my talk in Toronto in June.

### Grigorieff forcing

If you don't remember, here's the quick and dirty (i.e. traditional) way to define Grigorieff (or Gregorieff depending on your choice of latinization) forcing: it consists of partial functions on

Grigorieff ForcingGiven a ultrafilteron , let Partially order such functions by

iff , i.e., has more information.

You can think of Grigorieff conditions as perfect binary trees with complete branching on an ultrafilter set and "parallel movement" elsewhere. But I said quick and dirty is enough here, so let's not worry too much.

Grigorieff forcing is famous for being the forcing that Shelah used for the first model without P-points. One of the reasons this is possible is as follows.

Theorem (Shelah)Ifis a P-point, then is proper and -bounding. In particular, does not collapse .

Last week, David Chodounsky let me know that Bohuslav Balcar showed him the following "folklore" result.

OptimalityIfis not a P-point, then collapses the continuum.

This result is mentioned in Jech's Multiple Forcing book, but without proof and I have never seen one published. (Which, to tell the truth, is the reason I thought it was wrong but more about that later).

#### a proof

- If
is not a P-point, then there exists a partition such that every intersects infinitely many in an infinite set. - In particular, no
and, without loss, all are infinite. - In the ground model
, let's enumerate each as (bijectively). **First observation**The generichas for every . - Fix
. - Since
, we can decide any condition arbitrarily on . - In other words, there's a dense set of conditions
with . - But any condition in this dense set forces what we want, i.e.,
-- which is a set in the ground model.

- Fix
- So let
be a generic over . **Second observation**In, we can define a map , mapping to with . - Check that this is possible because this intersection is a ground model set, hence appears in the enumerations we fixed earlier.

**Third Observation**is cofinal. - Given any
, we want to find such that . - For a density argument, fix any condition
. - Since
is a small set, we can find such that

- Therefore, we can find
such that for some . - Extend
to all of such that . - Then
-- as desired.

- Given any

### An honest mistake

David and I thought we had a proof that Grigorieff forcing with a stable ordered union ultrafilter is proper and

*Comments*.

**François**, 2011/10/16

That's a neat argument! I always find situations like this where you only show thata little strange. Is there any hope to get a nice surjection from onto ?

(There is a small typo where you haveinstead of or .) **Peter**, 2011/10/16

Well, Andreas thought of one when I talked to him about this but writing this post I felt it didn't work (I'll ask again).

He wanted to enumerateinstead of but I didn't see how this helps, i.e., I don't see how can be extended to be equal mod fin to .

(Thanks, I'll fix the typo.)**François**, 2011/10/17

Right, that trick would work if you knew thatis finite for some , but I don't see why that would be the case... **Peter**, 2011/10/20

Ah! I forgot to update this -- Francois, Andreas fixed it. It's not hard, really: instead of a regular enumeration just pick a map so that the power set of every infinite subset ofhas full range. Proceed as in the proof and after finding , choose the subset with number . **François**, 2011/10/21

Full range?**Peter**, 2011/10/29

Surjective range. Inductively build a mapin such a way that for every infinite , we get that the restriction of our map to is still surjective.

Then we can use the above argument: as before we find anwhere our condition has left out an infinite set. Now pick from the power set of that infinite set a suitable candidate for -- done.

Does that make more sense?