(This is the fifth lecture in Stevo Todorcevic’s Forcing class, held in the fall of 2012. You can find the fourth lecture here. Quotes by Stevo are in dark blue; some are deep, some are funny, some are paraphrased so use your judgement. As always I appreciate any type of feedback, including reporting typos, in the comments below.)
- Mention something about Cohen and Random Forcing.
- We recall some facts about .
- Establish some lemmas about .
- Give most of the proof of the functor on ccc posets.
Cohen and Random Forcing
Fact. Cohen forcing gives you an with measure 0, such that for each uncountable .
See the Lorentz Lemma (?) (in additive combinatorics).
Question. Is there a chain condition that distinguishes Random * Cohen from Cohen * Random? (They are both ccc.)
Back to Chang’s Conjecture Stuff
Recall the theorem we are aiming for:
Functor Theorem. Suppose that . Let be a powerfully ccc poset of size . Then there is a (powerfully) ccc poset of size such that if contains a centred subset of size , then is -centred.
Fact. MM implies .
“My guess is that PFA doesn’t say anything here.”
“ is the combinatorial version of .”
Facts about :
- Chang’s Conjecture implies
- If Chang’s conjecture fails, there is a ccc poset which forces . (i.e. forces )
- Under MA(), Chang’s Conjecture is equivalent to
“I don’t like becuase the well-ordering of the reals uses too strong of parameters: a stationary subset of . This is like using AC to prove AC. You want the weakest form of choice.”
Fact. Chang’s Conjecture implies .
The First of Two Preparatory lemmas
Recall (from wikipedia) that is a Jónsson Cardinal if for every function there is a set of order type such that for each , restricted to -element subsets of omits at least one value in .
Problem. Can be a Jónsson cardinal?
Lemma 1. the first Jónsson cardinal.
proof. Let be a Jónsson Cardinal and let be given. Let , so but .
Useful Trick here: Use a wellordering of and the corresponding Skolem function to generate an algebra on that and apply the fact fact that is Johnson on a subalgebra generated by with the extra function symbol which bijects and .
Let . Note that is a limit ordinal and that . So we can find of size , and such that for all .
Note that by elementarity of , for all finite , the following is unbounded in :
Choose of size and such that , using the pigeonhole principle and set to be a singleton.
Let . Choose of size and such that .
“It is very important you go on top. That is why I used unboundedness.“
Continuing with this we get two infinite sequences and such that . [QED]
Note. Compare this with .
The Second Preparatory Lemma
“Why do you have a chance to make interesting ccc posets on ?”
Lemma 2. Suppose is not constant on any product of two infinite sets. Let be an uncountable family of pairwise disjoint subsets of . Then for all there is an in such that for .
Question. Can you get infinitely many such witnesses? I.e. An infinite witness family? Yes, if you force.
proof of Lemma 2.
Without loss of generality, . Assume that for some , all have size . So every can be written in an increasing enumeration as .
Suppose (for the sake of contradiction) that for some there is an and a such that .
Pick two uniform ultrafilters on and on .
Then there exists , and such that there are -many and -many such that .
“It is important that the ultrafilters are on different index sets. It would not work on the same set,” the pigeonhole principle.
“Now we are in business.”
So this means that there is an such that for all we have . Continue as before to get two infinite sets such that , which is a contradiction. [QED]
Compare this with the previous proof. There we already had the finite intersection property. We didn’t need ultrafilters.
Here we give most of the proof of the functor theorem.
proof of Functor Theorem. By lemma 1, fix (“here is a little business with Skolem functions. This is a good exercise.“) such that for all with .
Fix not constant on any rectangle. Fix a 1-1 sequence of reals , (which can be done because ).
Let be defined by
For of some length , and define
“Allowing infinite demands is usually trouble for ccc!“
Claim (To be proved next time): This poset is (powerfully) ccc.
So for such .
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