(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?