Stevo’s Forcing Class Fall 2012 – Class 8 – Mike Pawliuk

(This is the eighth lecture in Stevo Todorcevic’s Forcing class, held in the fall of 2012. You can find the seventh lecture here. Quotes by Stevo are in dark blue; some are deep, some are funny, some are paraphrased so use your judgement. I didn’t take these notes, so there are few Stevo quotes. Thanks to Dana Bartasova for letting me reproduce her notes here. As always I appreciate any type of feedback, including reporting typos, in the comments below.)

Warmup Lemma

Lemma: CC implies $\theta_2 = \aleph_2$

proof. Given a function $f: \omega_2 \times \omega_2 \rightarrow \omega$ we need to find infinite sets $A,B \subseteq \omega_2$ such that $f \upharpoonright [A \times B]$ is constant.

By CC, find an elementary submodel $M \prec H_{\aleph_3}$ such that $f \in M$ , $\alpha := M \cap \omega_1$ is countable, but $M \cap \omega_2$ is uncountable. Find an uncountable $x \subseteq M \cap \omega_2$ and $n < \omega$ such that $f(\alpha, \beta) = n$ for all $\beta \in X$ .

By elementarity of $M$ , for all finite $F \subseteq X$ the set $\{\xi \neq \alpha : \forall \beta \in F, f(\alpha,\beta = n)\}$ is unbounded in $\alpha$ .

We are done by the previous agrument: Construct $\{\alpha_i : i < \omega\} \subseteq \alpha$ , $\{\beta_i : i < \omega\} \subseteq X$ such that $f(\alpha_i, \beta_j) = n$ for all $i,j \in \omega$ . [QED]

The CC Theorem

Theorem. CC failing is equivalent to the following:

There is an $e:[\omega_2]^2 \rightarrow \omega_1$ such that:

$e(\alpha, \gamma) \leq \max_{\alpha \leq \beta \leq \gamma < \omega_2} \{e(\alpha, \beta), e(\beta, \gamma)\}$ ,
$e(\alpha, \beta) \leq \max_{\alpha \leq \beta \leq \gamma < \omega_2} \{e(\alpha, \gamma), e(\beta, \gamma)\}$ ,
For any uncountable family $\mathcal{A}$ of pairwise disjoint finite subsets of $\omega_2$ and $\nu < \omega_1$ there is an uncountable $\mathcal{B} \subseteq \mathcal{A}$ such that $(\forall a \neq b \in \mathcal{B})(\forall \alpha \in a)(\forall \beta \in b)[e(\alpha, \beta) > \nu]$

Note 1: For all $\beta$ , $e(\dot, \beta): \beta \rightarrow \omega_1$ is countable-to-1.

Otherwise, we get an uncountable $\mathcal{A} \subseteq \beta$ and $\nu < \omega_1$ such that $e(\alpha, \beta) \leq \nu$ for all $\alpha \in \mathcal{A}$ . By (b) $e(\alpha, \alpha^\prime) \leq \nu$ for all $\alpha, \alpha^\prime \in \mathcal{A}$ .

Note 2: If $(C_\alpha : \alpha < \omega_2)$ is a square sequence $(\square_{\omega_1})$ and $\rho = \rho(C_\alpha : \alpha <\omega_2)$ , then $\rho$ has these properties. Consequently, $\square_{\omega_1}$ implies $\neg$ CC. (Use $\square_{\omega_1}$ to construct a ccc poset that adds a kurepa tree, but that would be killed by CC.)

Proof of Theorem.

[ $\neg 1 \Rightarrow \neg 2$ ] Fix $e$ as above. We need to force a function $g: \omega_2 \times \omega_2 \rightarrow \omega$ which is not constant on the product of any two infinite subsets of $\omega_2$ .

Let $\mathbb{P}$ be the collection of all finite maps $p : [\mathcal{D}_p]^2 \rightarrow \omega$ such that
$\displaystyle \textbf{(*)}\forall \xi<\alpha<\beta \in \mathcal{D}_p, e(\xi, \alpha) > e(\alpha, \beta) \Rightarrow p(\xi, \alpha) \neq p(\xi, \beta)$

For $p,q \in \mathbb{P}$ , set $p \leq q$ if $p \supseteq q$ and
$\displaystyle \textbf{(**)} \forall \alpha < \beta \in \mathcal{D}_q, \forall \xi \in \mathcal{D}_p \setminus \mathcal{D}_q, p(\xi, \alpha) \neq p(\xi, \beta)$

Claim: $\mathbb{P}$ is ccc (even property K!)

proof. Let $A \subseteq \mathbb{P}$ be uncountable. We may assume that

$\{\mathcal{D}_p : p \in A\}$ forms a Delta system with root $D$
$F_p := e^{\prime\prime} [\mathcal{D}_p]^2$ forms a Delta system on $\omega_1$ with root $F$ , increasing
The $p$ are isomorphic as finite models (+ identity on $\mathcal{D}$ ??)

By Lemma 3(c) find an uncountable $B \subseteq A$ such that
$\displaystyle \textbf{(***)} \forall p \neq q \in N \forall \alpha \in \mathcal{D}_p \setminus D \forall \beta \in \mathcal{D}_q \setminus D e(\alpha, \beta) > \max (F)$

Claim: $B$ is a 2-linked subset of $\mathbb{P}$ .

Take $p \neq q \in B$ . Define $r : [\mathcal{D}_p \cup \mathcal{D}_q]^2 \rightarrow \omega$ to extend $p,q$ and to be 1-1 on new pairs, avoiding odd values.

Check (*) on r.

Let $\xi < \alpha <\beta$ be given in $\mathcal{D}_r$ such that $e(\xi, \alpha) > e(\alpha, \beta)$ . We need to show that $r(\xi, \alpha) \neq r(\xi, \beta)$ .

We may assume $\{\xi, \alpha\}, \{\xi, \beta\} \in [\mathcal{D}_p]^2 \cup [\mathcal{D}_q]^2$ . So we may assume $\xi \in D$ , $\alpha \in \mathcal{D}_p \setminus \mathcal{D}_q$ and $\beta \in \mathcal{D}_q \setminus \mathcal{D}_p$ .

By properties (a), (b) of $e$ we know that:
$e(\xi, \alpha) > e(\alpha, \beta)$ implies $e(\xi, \alpha) = e(\xi, \beta)$
implies $e(\xi, \alpha) = e(\xi, \beta) \in F$ (A contradiction. This line might be superfluous.)
$e (\xi, \beta) \leq \max \{e(\xi, \alpha), e(\alpha, \beta)\} = e(\xi, \alpha)$
$e (\xi, \alpha) \leq \max \{e(\xi, \beta), e(\alpha, \beta)\} = e(\xi, \beta)$

Check that $r \leq p,q$ . (We check only $r \leq p$ ).

Choose $\alpha <\beta$ in $\mathcal{D}_p$ and $q \in \mathcal{D}_q \setminus \mathcal{D}_p$ such that $\xi < \alpha <\beta$ . We need to show that $r(\xi, \alpha) = r(\xi, \beta)$ .

This is automatic if one of the pairs is new. So WLOG, $\{\xi, \alpha\}, \{\xi, \beta\} \in [\mathcal{D}_p]^2 \cup [\mathcal{D}_q]^2$ . But then $r(\xi, \alpha) = q (\xi, \alpha)$ and $r(\xi, \beta) = q (\xi, \beta)$ .

Now we look at (*) for q, (i.e. to see that $e(\xi, \alpha) > e(\alpha, \beta)$ .)

Otherwise, $e(\xi, \alpha) \in F$ since $\alpha, \beta \in D$ .

By the isomorphism condition we get that $\forall s \in B, e(\xi(s), \alpha) \in F$ , where $\xi(s)$ is the image of $\xi$ relative to the isomorphism between $s$ and $q$ .

So $X = \{ \xi(s) : s \in B\}$ is an uncountable subset of $\alpha$ on which $e( \dot, \alpha)$ is constant. (A contradiction with note 1).

Forcing with $\mathbb{P}$ gives $f: [\omega_2]^2 \rightarrow \omega$ such that $\{\xi < \alpha : f(\xi, \alpha) = f(\xi, \beta)\}$ for all $\alpha, \beta$ .

Define $g: \omega_2 \times \omega_2 \rightarrow \omega$ by $g(\alpha, \beta) =$

$2 f(\alpha, \beta) + 1$ if $\alpha <\beta$
0 if $\alpha = \beta$
$2 f(\alpha, \beta) + 2$ if $\alpha > \beta$

Note $g$ is not constant on any product of infinite sets.

Definable CH

Recall: CH $^2$ is equivalent to “Every compact ccc space of weight $\leq \mathfrak{c}$ ” has a Luzin set.

Note: CH $^2$ implies CH $^{<\omega}$

CH $^{<\omega}$ says “Every ccc poset $\mathbb{P}$ of size $\leq \mathfrak{c}$ has an uncountable collection $\mathcal{F}$ of centred subsets of $\mathbb{P}$ such that for all dense open $D \subseteq \mathbb{P}$ the set $\{F \in \mathcal{F} : F \cap D \neq \emptyset\}$ is countable.”

Note: If we replace $\mathcal{F}$ by filters, then this is equivalent to CH.

$\mathbb{P} := (P, \leq_\mathbb{P}, \textrm{Comp}_\mathbb{P}^n)_{n<\omega}$ , where $\textrm{Comp}_\mathbb{P}^n (p_0, ..., p_{n-1})$ iff $\exists q \in \mathbb{P}$ such that $q \leq p_i$ for all $i < n$ .

$\mathbb{P}$ is definable if $\mathbb{P} \subseteq \mathbb{R}$ and there exists $\leq, \textrm{Comp}^n (n < \omega)$ in $L(\R)$ such that $\leq_\mathbb{P} = <\upharpoonright P$ and $\textrm{Comp}_\mathbb{P}^n = \textrm{Comp}^n \upharpoonright P$ for all $n < \omega$ .

Exercise: Every $\sigma$ -centred poset in definable.

All posets we have constructed so far (except for the one in CH $^3$ implies $\theta_2 = \aleph_2$ ) are definable.

“Definable CH” is the statement “CH $^2$ for definable posets”.

Note: Definable CH $^2$ implies that all sets of reals are $\aleph_1$ -Borel.

Theorem: Definable CH $^2$ is not equivalent to CH. (Large cardinals, $\mathfrak{c} = \aleph_{\omega_1}$ )

Question: Is CH $^2$ the same as CH?

4 thoughts on “Stevo’s Forcing Class Fall 2012 – Class 8”

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Ari Brodsky says:

October 18, 2012 at 12:55 am

Here are some quotes I recorded:

“What about Namba in $\omega_1$ ? I like this better. We will analyze Namba forcing on $\omega_1$ , using $\mathfrak m > \omega_1$ . (We really only need a weaker hypothesis, such as either $\mathfrak p$ or $\mathfrak b$ .)”

“This proof has several interesting useful things in it.” (referring, I think, to the theorem that CC is equivalent to the statement “Every ccc poset forces $\theta_2 = \omega_2$ “)

“What is the combinatorial argument we always use?”

“You can’t control infinite sets. Infinite sets show up all over the place.”

“Whenever you construct a partial ordering of size $\aleph_2$ , you have to ask: ‘What is the influence of Chang’s Conjecture on it?’ ”

(Regarding the Claim that $P$ is ccc:) “Actually, I’m going to show it has Property K. It probably also has precalibre $\aleph_1$ , but I haven’t tried to show that. It would be a good exercise to show that it has precalibre $\aleph_1$ .”

“ $\Delta$ -system on $\omega_2$ is very tricky, not like on $\aleph_1$ . On $\omega_1$ , $\Delta$ -systems are beautiful: root, tail, tail. On $\omega_2$ , you have: root, tail, tail, pieces of root, tail, tail, etc.”

“If you look at the analogue between set theory and geometry (and there are many such analogues), we have: $\omega_2$ has dimension 3, $\omega_3$ has dimension 4, etc. We cannot visualize 4 dimensions, so it is harder to understand $\omega_3$ .”

“So far I haven’t seen an interesting ccc partial ordering on $\omega_3$ . Maybe there will be one.”

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1. Ari Brodsky says:
  
  October 19, 2012 at 12:40 am
  
  That \alepg should have been $\aleph$ , of course, but I can’t fix it. Can you?
  
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  1. Micheal Pawliuk says:
    
    October 21, 2012 at 7:53 pm
    
    Fixed.
    
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