# Stevo’s Forcing Class Fall 2012 – Class 8

(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:

1. $e(\alpha, \gamma) \leq \max_{\alpha \leq \beta \leq \gamma < \omega_2} \{e(\alpha, \beta), e(\beta, \gamma)\}$,
2. $e(\alpha, \beta) \leq \max_{\alpha \leq \beta \leq \gamma < \omega_2} \{e(\alpha, \gamma), e(\beta, \gamma)\}$,
3. 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”

1. Ari Brodsky says:

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:

That \alepg should have been $\aleph$, of course, but I can’t fix it. Can you?

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1. Micheal Pawliuk says:

Fixed.

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