(This is the eighth lecture in Stevo Todorcevic’s Forcing class, held in the fall of 2012. You can find the seventh lecture here. Quotes by Stevo are in dark blue; some are deep, some are funny, some are paraphrased so use your judgement. I didn’t take these notes, so there are few Stevo quotes. Thanks to Dana Bartasova for letting me reproduce her notes here. As always I appreciate any type of feedback, including reporting typos, in the comments below.)
- Show Chang’s Conjecture implies .
- Show the failure of CC is equivalent to a nice colouring existing. This is fairly technical.
- Introduce definable posets and “Definable CH”.
Lemma: CC implies
proof. Given a function we need to find infinite sets such that is constant.
By CC, find an elementary submodel such that , is countable, but is uncountable. Find an uncountable and such that for all .
By elementarity of , for all finite the set is unbounded in .
We are done by the previous agrument: Construct , such that for all . [QED]
The CC Theorem
Theorem. CC failing is equivalent to the following:
There is an such that:
- For any uncountable family of pairwise disjoint finite subsets of and there is an uncountable such that
Note 1: For all , is countable-to-1.
Otherwise, we get an uncountable and such that for all . By (b) for all .
Note 2: If is a square sequence and , then has these properties. Consequently, implies CC. (Use to construct a ccc poset that adds a kurepa tree, but that would be killed by CC.)
Proof of Theorem.
 Fix as above. We need to force a function which is not constant on the product of any two infinite subsets of .
Let be the collection of all finite maps such that
For , set if and
Claim: is ccc (even property K!)
proof. Let be uncountable. We may assume that
- forms a Delta system with root
- forms a Delta system on with root , increasing
- The are isomorphic as finite models (+ identity on ??)
By Lemma 3(c) find an uncountable such that
Claim: is a 2-linked subset of .
Take . Define to extend and to be 1-1 on new pairs, avoiding odd values.
Check (*) on r.
Let be given in such that . We need to show that .
We may assume . So we may assume , and .
By properties (a), (b) of we know that:
implies (A contradiction. This line might be superfluous.)
Check that . (We check only ).
Choose in and such that . We need to show that .
This is automatic if one of the pairs is new. So WLOG, . But then and .
Now we look at (*) for q, (i.e. to see that .)
Otherwise, since .
By the isomorphism condition we get that , where is the image of relative to the isomorphism between and .
So is an uncountable subset of on which is constant. (A contradiction with note 1).
Forcing with gives such that for all .
- 0 if
Note is not constant on any product of infinite sets.
Recall: CH is equivalent to “Every compact ccc space of weight ” has a Luzin set.
Note: CH implies CH
CH says “Every ccc poset of size has an uncountable collection of centred subsets of such that for all dense open the set is countable.”
Note: If we replace by filters, then this is equivalent to CH.
, where iff such that for all .
is definable if and there exists in such that and for all .
Exercise: Every -centred poset in definable.
All posets we have constructed so far (except for the one in CH implies ) are definable.
“Definable CH” is the statement “CH for definable posets”.
Note: Definable CH implies that all sets of reals are -Borel.
Theorem: Definable CH is not equivalent to CH. (Large cardinals, )
Question: Is CH the same as CH?
4 thoughts on “Stevo’s Forcing Class Fall 2012 – Class 8”
Here are some quotes I recorded:
“What about Namba in ? I like this better. We will analyze Namba forcing on , using . (We really only need a weaker hypothesis, such as either or .)”
“This proof has several interesting useful things in it.” (referring, I think, to the theorem that CC is equivalent to the statement “Every ccc poset forces “)
“What is the combinatorial argument we always use?”
“You can’t control infinite sets. Infinite sets show up all over the place.”
“Whenever you construct a partial ordering of size , you have to ask: ‘What is the influence of Chang’s Conjecture on it?’ ”
(Regarding the Claim that is ccc:) “Actually, I’m going to show it has Property K. It probably also has precalibre , but I haven’t tried to show that. It would be a good exercise to show that it has precalibre .”
“-system on is very tricky, not like on . On , -systems are beautiful: root, tail, tail. On , you have: root, tail, tail, pieces of root, tail, tail, etc.”
“If you look at the analogue between set theory and geometry (and there are many such analogues), we have: has dimension 3, has dimension 4, etc. We cannot visualize 4 dimensions, so it is harder to understand .”
“So far I haven’t seen an interesting ccc partial ordering on . Maybe there will be one.”
That \alepg should have been , of course, but I can’t fix it. Can you?
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