(This is the sixth lecture in Stevo Todorcevic’s Forcing class, held in the fall of 2012. You can find the fifth 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 talk about some caliber properties.
- Establish that the ccc functor really goes to ccc spaces.
- Use this to create an interesting ccc poset.
- Talk about some combinatorial versions of CH.
- Mention a few consequences of CH.
Definition. has caliber if for every -sized family can be refined to a -sized such that all families of size have a lower bound in .
- has pre-caliber iff has caliber .
- has property iff has caliber .
- has property K(naster) iff has caliber .
Topological Definition (Šanin): A compact space is -caliber iff the poset of non-empty open subsets of have pre-caliber .
This definition was introduced because separability is not preserved by products, but caliber properties are preserved.
Exercise: Show that Cohen Random has caliber while Cohen * Random does not.
The ccc Functor is ccc
Here is a summary of the notation from Class 5, where we defined this functor.
We are starting with a powerfully ccc poset of cardinality and we are building its specializing poset and
- such that for all ;
- Fix a 1-1 sequence of reals ;
- is not constant on any product of two infinite sets;
- For of some length , define ;
- iff , is centred in .
Claim: is a (powerfully) ccc poset.
proof. Let be a given uncountable family forming a -system with some root .
We may assume (by refining) that for some and all that
- , for all ;
- , for all
We may also assume that for some for all , and
- , for all ;
- , for is a constant mapping, and in fact that we get , for all , (listing ).
“Now we do something unusual.”
Using the fact that is ccc for , we may assume that for each that is a centred family in .
“This exists in a forcing extension by , where becomes -centred. This looks like I’m cheating, going to a forcing extension. But I’m only looking for 2 compatible elements. Use absoluteness of the colouring’s property.“
Find and two uncountable (“this is a separable metric space thing“) such that we have .
“Here the key comes.”
By the property of given in a previous lemma, there are such that for and “because of the unbounded property of .”
“What does it mean? You have to check the defining property of .“
Take , where , and for some particular .
There are three cases:
“The important thing is what these numbers represent.” – Ivan Khatchatourian
Case 1 :
Let , which means , and for all .
“Okay, so let us see… Why you cannot start in one and go to another?“
[Picture argument to be filled in later]
Then or .
Case 2 :
“Again, the picture tells you much more than you can get by chasing definitions.”
Take , and , for all where .
There are no such that and . So or .
“The main thing is now comes case 3.”
Case 3 :
“Suppose not, so you have two possibilities:”
and , and for all .
If , then .
“The projections” “are 1-1” for .
“The place below where they split is below , a contradiction.” [QED]
“It is a slippery argument but … *Shrug*.”
“Showing that is powerfully ccc is the same argument.“
- Every compact ccc scpace has caliber ;
- Every ccc poset has precaliber .
“The proof of this is like the previous proof comparing and .“
Note that this functor also specializes trees:
Let be a centred subset of size . Then and and and
Question: ATFE? (Are The Following Equivalent)
- Every ccc poset has property K.
“This involves checking something for every pair. As the previous proof shows.“
Now we go to the CH
“Log of weight is density.“
Note that it is comparable to replace weight with -weight here.
Question: Is ? In combinatorial language, how many centred sets do you need to cover a poset?
Some Combinatorial CHs
CH: , i.e. every ccc poset of size is -centred.
CH: Every ccc poset of size is -n-linked.
CH: Every ccc poset of size is -2-linked.
Open Question: Are they different?
“We will see that most consequences of CH are actually consequences of these forms of CH.”
Theorem: CH implies .
“These give you another way to create ccc posets. Here ccc is a quite weak condition.“
Theorem*: If every ccc poset has property , then .
“In this course we pretend .“
Theorem**: There is a productively ccc poset without property .
“So plugging in CH gives , i.e. , so .“
Corollary: CH implies .
(Downward) Lemma: Suppose is a well-ordered separable metric space. “We know well orderedness usually has nothing to do with metric spaces.” Let be a (closed) set mapping which respects , i.e. . Let be the collection of finite -free sets ordered by reverse inclusion. THEN, is productively ccc.
Recall that is ccc iff “ is ccc”.
proof. Let be a given uncountable family of finite -free subsets of . We need to find such that is -free.
“You see, being free is a 2-dimensional property.“
As usual, by forming a -system and removing the root, we may assume that elements of are pairwise disjoint and of the same size .
“Showing they’re free on a tail is enough.“
Think of as a subset of . Let be a countable dense subset of .
Since is a separable metric space, we may assume that for some fixed sequence of pariwise disjoint open subsets of we have that
- ; and
- for all , and we have .
We only need to check that and are compatible, by separability. That is, we don’t need to check against , for .
“Now we use the well-ordering.”
“There is always this ‘moreover’ business.“
Moreover, we may assume that for all we have .
Find such that for all and we have . (For this, use uncountability.)
Take such that .
“Closed sets look downwards, they don’t hit ‘s.“
So there is such that . So is -free. “ cannot hit because it is above. cannot hit because I picked it out. was just for non-emptyness.” [QED]
Question: Do we need the well-ordering? It is unnatural for metric spaces.
Remark: Changing this lemma to arbitrary posets is equivalent to OCA.
“Now we come to this example there.”
Example: Let be a -well-ordered, -unbounded family of cardinality .
Define by .
So the poset of finite -free sets is ccc.
Recall from oscillation theory: does not have a 2-linked subset of size .
[Assume .] Apply the functor to the poset to get a poset of size with no 3-linked subset of size .
“We are losing 1-dimesnsion.“
Aside: “CH seems to be quite strong.”
Some consequences of CH.
- Every can be decomposed into continuous sub-functions.
CH implies the density of Lebesgue measure algebra is . This gives you Luzin sets, Sierpinksi functions, etc..