(This is the fourth lecture in Stevo Todorcevic’s Forcing class, held in the fall of 2012. You can find the third 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.)
- We look at the real reason that regular open algebras don’t collapse to .
- Show there is a -linked, non -centred poset of size .
- Introduce the cardinal , related to Chang’s Conjecture.
- Introduce the idea of a functor on ccc posets.
When talking about how Cohen Forcing does not collapse it is better to think about how the regular open algebra does not collapse . This we can tell from some facts about the distributivity of the algebra:
Fact. In , and there are (for ) such that
Recall the analogous distributive law with replaced by .
Weak Distributive Law. and there are (for ) such that
Fact. If is ccc, then has the weakly--distributive.
Examples of ccc and weakly--distributive algebras.
Example. Try where .
“Weakly Distributive and -centred do not go together.”
Question. Is there a -centred, weakly distributive algebra? (Note that measure algebras do not work.)
For a Measure Algebra, strictly positive, is countably additive or, it is finitely additive and weakly distributive.
Question. Are there nonseparable (non trivial) Maharam algebras? (You need to “make submeasure far away from measure” and “find -finite-cc, not -bounded-cc“.)
Back to the “Suslin Family”.
Theorem 2. There is a -linked, non -centred poset of size .
“Smaller families are more interesting.”
Compare this with the following adjustments:
Theorem 1. There is a -linked, non -centred poset of size
Theorem 0. There is a non -linked poset of size with no 2-linked subset of size .
The proof is similar to Theorem 1, which we did in Class 2.
proof. We may assume that , by Theorem 1.
Let have the finite intersection property, that is non-extendible. That is, there is no such that for all .
We build (“if possible, or in other words we simply try“), , a tower such that
- , for all .
“Of course it is going to be a recursive construction. This tells you you have hope. Otherwise, there is no future.”
Given , let .
“Everybody intersects everybody.”
Suppose is a limit .
Case 1, where .
Pick a sequence increasing to . Define . And for , define by .
Since we are assuming that , fix such that for all . Let . (Remember we were trying to define, recursively, .)
Then for , and is infinite for all .
Case 2, where .
“We force , but we are not allowed to force in the obvious (-centred) way.”
Let iff , and . The ordering is coordinate-wise inclusion.
(See Class 2 notes, for this notation, and intuition.)
Claim: is ccc (and in fact is -k-linked for all )
“If the claim is true, if is not -centred we are done. So assume is -centred.“
Let be such that is centred in , for all . Let . We may assume that each is cofinal in , so that .
“The next move is a dangerous move in general. Because we are asking for infinitely many requirements.“
Claim: These are infinite.
Since is cofinal in , , for all . They are infinite since is infinite and .
“Now what do we know?“
If , such that for some and , , then is infinite.
“Now is going to help us. You do the usual thing. What is the usual thing?“
For , define such that …
. “Now dominate“. Let be such that for all .
“This does not give us enough chances to succeed.”
minimal such that for every if has some points above then it has some point below .
Then, as before, .
“We have that and form a gap, but we know that kills -gaps.“
So we can continue on and either get that , (so we have the last theorem), or we can continue to what we want. [QED]
Theorem. The following are equivalent:
- MA() (i.e. );
- Every first-countable ccc compact space is separable.
Some Preliminaries about the Cardinal
A useful cardinal for us is , where the arrow notation means there are such that is constant.
Theorem (). iff Chang’s Conjecture.
Theorem *. If Chang’s Conjecture fails, there is a ccc poset which forces .
Exercise. Chang’s Conjecture implies .
[Chang’s Conjecture]. For every structure of the form there is an elementary substructure of such that and .
Problem. Is it true (in ZFC) that ? “Any would be interesting.”
Now Amoebas and Functors.
We are going to produce a functor that goes from a ccc poset to another ccc poset . The important thing is that we will be going “from local to global control“.
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.
“We are getting away from meeting dense sets.”
Sketch. implies that for every algebra there is a subalgebra such that but .
This means that we have a map such that (this is global information), for all of cardinality , (where we think of allowing to be any “thin” set.)