(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.)

**Summary**

- 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.

## Caliber Properties

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 .

**Examples**:

- 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 .”

**Claim**: .

“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** []:

**Claim**:

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** []:

**Claim**: .

“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** []:

**Claim**: .

“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.“

**Corollary**. TFAE:

- MA;
- 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)

- MA;
- 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.“

**Definition**: .

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 .

Then

“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..

Hey Mike,

I think that the notion of caliber was introduced by Sanin (in fact, the S should have a hacek on it), not Shannon. Thanks for posting these notes, it’s very helpful!

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Fixed! Thanks for finding that.

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