# 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 $\mathbb{P}$.

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:

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 we have $\textrm{otp}\{a(i): a \in A\} = \omega_1$.

Find $c \in A$ such that for all $i 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”

1. Chris Eagle says:

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:

Fixed! Thanks for finding that.

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