Stevo’s Forcing Class Fall 2012 – Class 6

(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.)

Caliber Properties

Definition. \mathbb{P} has caliber (\theta, \lambda, \kappa) if for every \theta -sized family F \subseteq \mathbb{P} can be refined to a \lambda -sized G \subseteq F such that all families of H \subseteq G size \kappa have a lower bound in \mathbb{P} .


  • \mathbb{P} has pre-caliber \theta iff \mathbb{P} has caliber (\theta, \theta, <\omega) .
  • \mathbb{P} has property K_\theta iff \mathbb{P} has caliber (\theta, \theta, 2) .
  • \mathbb{P} has property K(naster) iff \mathbb{P} has caliber (\omega_1, \omega_1, 2) .

Topological Definition (Šanin): A compact space K is \theta -caliber iff the poset of non-empty open subsets of K have pre-caliber \theta .

This definition was introduced because separability is not preserved by products, but caliber properties are preserved.

Exercise: Show that Cohen \times Random has caliber (\omega_1, \omega, \omega) 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 \mathbb{P} of cardinality <\theta_2 and we are building its specializing poset \mathcal{S}(\PP) and

  • d: [\kappa]^{<\omega} \rightarrow \mathbb{P} such that d[X]^{<\omega} = \mathbb{P} for all X \in [\kappa]^\kappa ;
  • Fix a 1-1 sequence of reals \{r_\xi : \xi < \kappa\} \subseteq \{0,1\}^\omega ;
  • c: \kappa \times \kappa \rightarrow \omega is not constant on any product of two infinite sets;
  • c^* (\alpha_0, ..., \alpha_{k-1}) := \max \{c (\alpha_i, \alpha_j) : i \neq j < k\} ;
  • \mathcal{S}(\PP) \subseteq [\kappa]^{<\omega} ;
  • For F \in [\kappa]^{<\omega}, \vec s \in (2^{<\omega})^{<\omega} of some length k , define \PP_F := \{d(\alpha_0, ..., \alpha_{k-1}) : \forall i < k, \alpha_0 < ... < \alpha_{k-1} \in F, r_{\alpha_i} \upharpoonright c^* (\alpha_0, ..., \alpha_{k-1}) = s_i \} ;
  • F \in \mathcal{S}(\PP) iff \forall \vec s \in (2^{<\omega})^{<\omega} , \PP_F is centred in \mathbb{P} .

Claim: \mathcal{S}(\PP) is a (powerfully) ccc poset.

proof. Let \mathcal{F} \subseteq \mathcal{S}(\PP) be a given uncountable family forming a \Delta -system with some root D .

We may assume (by refining) that for some m < \omega and all f \in \mathcal{F} that

  • \Delta(r_\xi, r_\eta) < m , for all \xi \neq \eta ;
  • c(\xi, \eta) < m , for all \xi \neq \eta

We may also assume that for some n, \vert F \vert = n < \omega for all F \in \mathcal{F} , and

  • r_\xi \upharpoonright m \neq r_\eta \upharpoonright m , for all \xi \neq \eta ;
  • F \mapsto \{r_\xi \upharpoonright m : \xi \in F\} , for F \in \mathcal{F} is a constant mapping, and in fact that \forall F,G \in \mathcal{F} we get r_{F(i)} \upharpoonright m = r_{G(i)} \upharpoonright m , for all i < n , (listing F = \{F(0), ..., F(n-1)\} ).

Now we do something unusual.”

Using the fact that \mathbb{P}^k is ccc for k = \vert (2^{\leq m})^{\leq n}\vert , we may assume that for each \vec s \in 2^{\leq m})^{\leq n} that \bigcup \{\bigcup\PP_F (\vec s) : F \in \mathcal{F}\} is a centred family in \mathbb{P} .

This exists in a forcing extension by \mathbb{P} , where \mathbb{P} becomes \sigma -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 \overline{m} > m and two uncountable (“this is a separable metric space thing“) \mathcal{F}_0, \mathcal{F}_1 \subseteq \mathcal{F} such that \forall F \in \mathcal{F}_0, \forall G \in \mathcal{F}_1, \forall i < n we have \Delta(r_{F(i)}, r_{G(i)}) < \overline{m} .

Here the key comes.”

By the property of c given in a previous lemma, there are F \in \mathcal{F}_0, G \in \mathcal{F}_1 such that c(\alpha, \beta) > \overline{m} for \alpha \in F \setminus D and \beta \in G \setminus D because of the unbounded property of c .

Claim: F \cup G \in \mathcal{S}(\PP) .

What does it mean? You have to check the defining property of \mathcal{S}(\PP) .

Take \vec s \in (2^{<\omega})^{<\omega} , where \vec s = (s_0, ..., s_k) , and \vert s_i \vert = \vert s_j \vert = l, \forall i,j for some particular l .

There are three cases:

  1. l < m ;
  2. m \leq l < \overline{m} ;
  3. \overline{m} \leq l .

The important thing is what these numbers represent.” – Ivan Khatchatourian

Case 1 [l < m ]:

Claim: \PP_{F\cup G} (\vec s) = \PP_{F} (\vec s) \cup \PP_{G} (\vec s)

Let d (\alpha_0, ..., \alpha_k) \in \PP_{F\cup G} (\vec s) , which means \{\alpha_0, ..., \alpha_k\} \subseteq F\cup G , and r_{\alpha_i}\upharpoonright l = s_i for all i \leq k .

Okay, so let us see… Why you cannot start in one and go to another?

[Picture argument to be filled in later]

Then \{\alpha_0, ..., \alpha_k\} \subseteq F or \{\alpha_0, ..., \alpha_k\} \subseteq G .

Case 2 [m \leq l < \overline{m} ]:

Claim: \PP_{F \cup G} = \emptyset .

Again, the picture tells you much more than you can get by chasing definitions.”

Take d (\alpha_0, ..., \alpha_k) \in \PP_{F\cup G} (\vec s) , and r_{\alpha_i} \upharpoonright l = s_i , for all i \leq k where l = c^* (\alpha_0, ..., \alpha_k) .

There are no i,j such that \alpha_i \in F \setminus D and \alpha_i \in G \setminus D . So \{\alpha_i\} \subseteq F or \{\alpha_i\} \subseteq G .

The main thing is now comes case 3.”

Case 3 [\overline{m} \leq l ]:

Claim: \vert \PP_F (\vec s) \vert \leq 1 .

Suppose not, so you have two possibilities:”

\{\alpha_0, ..., \alpha_k\}, \{\alpha_0^\prime, ..., \alpha_k^\prime\} \subseteq F \cup G and c^* (\alpha_0, ..., \alpha_k) = l = c^* (\alpha_0^\prime, ..., \alpha_k^\prime) , and r_{\alpha_i}\upharpoonright l = s_i = r_{{\alpha_i}^\prime} \upharpoonright l for all i \leq k .

If \alpha_i \neq \alpha_i^\prime , then \alpha_i \in F \setminus D, \alpha_i^\prime \in G \setminus D .

The projections” r_\alpha \mapsto r_\alpha \upharpoonright m “are 1-1” for \alpha \in F .

“The place below where they split is below \overline{m} , a contradiction.” [QED]

“It is a slippery argument but … *Shrug*.”

“Showing that \mathcal{S}(\PP) is powerfully ccc is the same argument.

Corollary. TFAE:

  • MA(\aleph_1) ;
  • Every compact ccc scpace has caliber \aleph_1 ;
  • Every ccc poset has precaliber \aleph_1 .

The proof of this is like the previous proof comparing \mathfrak{m}_{SH} and \mathfrak{m} .

Note that this functor also specializes trees:

Let X \subseteq \mathcal{S}(\PP) be a centred subset of size \kappa . Then \Gamma := \bigcup X \in [\kappa]^\kappa and [\Gamma]^{<\omega} \in \mathcal{S}(\PP) and d [X]^{<\omega} = \mathbb{P} and
\displaystyle  \bigcup_{\vec s \in (2^{<\omega})^{<\omega}} \PP_X (\vec s) = \mathbb{P}

Question: ATFE? (Are The Following Equivalent)

  • MA(\aleph_1) ;
  • 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: \cc_d := \sup \{ d(K) : K \textrm{ is compact, ccc, weight }\leq \mathfrak{c}\} .

Note that it is comparable to replace weight with \pi -weight here.

Question: Is \cc_d = \mathfrak{c} ? In combinatorial language, how many centred sets do you need to cover a poset?

Some Combinatorial CHs

CH^\omega : \cc_d = \aleph_1 , i.e. every ccc poset of size \leq \mathfrak{c} is \aleph_1 -centred.
CH^n : Every ccc poset of size \leq \mathfrak{c} is \aleph_1 -n-linked.
CH^2 : Every ccc poset of size \leq \mathfrak{c} is \aleph_1 -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^3 implies \theta_2 = \aleph_2 .

These give you another way to create ccc posets. Here ccc is a quite weak condition.

Theorem*: If every ccc poset has property K_{\bb^+} , then \bb \leq \theta_2 .

In this course we pretend \mathfrak{c} = \theta_2 .

Theorem**: There is a productively ccc poset \mathbb{P} without property K_{\bb^+} .

So plugging in CH^3 gives \bb \leq \aleph_1 , i.e. \bb = \aleph_1 , so \bb^+ = \aleph_2 .

Corollary: CH^2 implies \bb = \aleph_1 .

(Downward) Lemma: Suppose (X, \leq_W) is a well-ordered separable metric space. “We know well orderedness usually has nothing to do with metric spaces.” Let F: X \rightarrow \textrm{Exp}(X) be a (closed) set mapping which respects \leq_W , i.e. F(X) \subseteq \{ y : y\leq x\} . Let \mathbb{P} = \{ A \in [X]^{<\omega}: \forall x \neq y \in A, y \notin F(x)\} be the collection of finite F -free sets ordered by reverse inclusion. THEN, \mathbb{P} is productively ccc.

Recall that \mathbb{P} \times \dot \mathbb{Q} is ccc iff \vdash_\mathbb{P}  \mathbb{Q} is ccc”.

proof. Let A \subseteq \mathbb{P} be a given uncountable family of finite F -free subsets of X . We need to find a \neq b such that a \cup b is F -free.

You see, being free is a 2-dimensional property.

As usual, by forming a \Delta -system and removing the root, we may assume that elements of A are pairwise disjoint and of the same size n .

Showing they’re free on a tail is enough.

Think of A as a subset of X^n . Let A_0 \subseteq A be a countable dense subset of A .

Since X is a separable metric space, we may assume that for some fixed sequence V_0, ..., V_n of pariwise disjoint open subsets of X we have that

  • A \subseteq V_0 \times V_1 \times ... \times V_{n-1} ; and
  • for all a \in A , and i < n we have F(a(i)) \cap V_j = \emptyset .

We only need to check that a(i) and b(i) are compatible, by separability. That is, we don’t need to check a(i) against b(j) , for i \neq j .

Now we use the well-ordering.

There is always this ‘moreover’ business.

Moreover, we may assume that for all i<n we have \textrm{otp}\{a(i): a \in A\} = \omega_1 .

Find c \in A such that for all i<n and a \in A_0 we have a(i) <_W c(i) . (For this, use uncountability.)

Take b \in A such that b(i) >_W c(i), \forall i < n .

\displaystyle  b \in (X \setminus F(c_0)) \times (X \setminus F(c_1)) \times ... \times (X \setminus F(c_{n-1})) =: \vec U

Closed sets look downwards, they don’t hit b(i) ‘s.

So there is a \in A_0 such that a \in \vec U . So a \cup c is F -free. “a cannot hit c because it is above. c cannot hit a because I picked it out. b 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 X \subseteq \omega^{\uparrow \omega} be a <^* -well-ordered, <^* -unbounded family of cardinality \bb .

Define F: X \rightarrow\textrm{Exp}(X) by F(x) := \{y \in X : \forall n, y(n) \leq x(n)\} \subseteq \{y \in X : y \leq^* x\} .

So the poset \mathbb{P} of finite F -free sets is ccc.

Recall from oscillation theory: \{\{x\}: x \in X\} \subseteq \PP_X does not have a 2-linked subset of size \bb .

[Assume \bb^+ < \theta_2 .] Apply the functor \mathbb{P} \mapsto \mathcal{S}(\PP) to the poset \PP_X to get a poset \mathcal{S}(\PP) of size \bb^+ with no 3-linked subset of size \bb^+ .

We are losing 1-dimesnsion.

Aside: “CH^3 seems to be quite strong.

Some consequences of CH^3 .

  • \textrm{cf}(\cc) = \aleph_1 ;
  • Every f: \mathbb{R} \rightarrow \mathbb{R} can be decomposed into \aleph_1 continuous sub-functions.

CH^\omega implies the density of Lebesgue measure algebra is \aleph_1 . This gives you Luzin sets, Sierpinksi functions, etc..

3 thoughts on “Stevo’s Forcing Class Fall 2012 – Class 6”

  1. 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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