Stevo’s Forcing Class Fall 2012 – Class 6 – Mike Pawliuk

(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 $\mathbb{P}$ .

Examples:

$\mathbb{P}$ 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:

$l < m$ ;
$m \leq l < \overline{m}$ ;
$\overline{m} \leq l$ .

“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 $b(i) >_W c(i), \forall i < n$ .

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

“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”

Pingback: Stevo’s Forcing Class Fall 2012 – Class 7
Chris Eagle says:

October 11, 2012 at 9:26 pm

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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1. Micheal Pawliuk says:
  
  October 17, 2012 at 4:32 pm
  
  Fixed! Thanks for finding that.
  
  LikeLike

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Caliber Properties

The ccc Functor is ccc

Some Combinatorial CHs

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