# Stevo’s Forcing Class Fall 2012 – Class 5

(This is the fifth lecture in Stevo Todorcevic’s Forcing class, held in the fall of 2012. You can find the fourth 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.)

## Cohen and Random Forcing

Fact. Cohen forcing gives you an $M \subseteq \mathbb{R}$ with measure 0, such that $M + X = \mathbb{R}$ for each uncountable $X \subseteq \mathbb{R}^V$.

See the Lorentz Lemma (?) (in additive combinatorics).

Question. Is there a chain condition that distinguishes Random * Cohen from Cohen * Random? (They are both ccc.)

## Back to Chang’s Conjecture Stuff

Recall the theorem we are aiming for:

Functor Theorem. Suppose that $\kappa < \theta_2$. Let $\mathbb{P}$ be a powerfully ccc poset of size $\leq \kappa$. Then there is a (powerfully) ccc poset $\mathcal{S}(\PP)$ of size $\kappa$ such that if $\mathcal{S}(\PP)$ contains a centred subset of size $\kappa$, then $\mathbb{P}$ is $\sigma$-centred.

Fact. MM implies $\theta_2 = \aleph_2$.

My guess is that PFA doesn’t say anything here.”

$\theta_2$ is the combinatorial version of $\mathfrak{c}^+$.”

Recall: $\theta_2 := \min \{\theta : {{\theta}\choose{\theta}} \rightarrow {{\omega}\choose{\omega}}^{1,1}_\omega\}$.

Facts about $\theta_2$:

1. $\theta_2 \geq \aleph_2$
2. Chang’s Conjecture implies $\theta_2 = \aleph_2$
3. If Chang’s conjecture fails, there is a ccc poset which forces ${{\omega_2}\choose{\omega_2}} \not\rightarrow {{\omega}\choose{\omega}}^{1,1}_\omega$. (i.e. forces $\theta_2 > \aleph_2$)
4. Under MA($\aleph_2$), Chang’s Conjecture is equivalent to $\theta_2 = \aleph_2$

I don’t like $\mathfrak{c} = \aleph_2$ becuase the well-ordering of the reals uses too strong of parameters: a stationary subset of $\omega_2$. This is like using AC to prove AC. You want the weakest form of choice.

Fact. Chang’s Conjecture implies $\neg \square$.

## The First of Two Preparatory lemmas

Recall (from wikipedia) that $\kappa$ is a Jónsson Cardinal if for every function $f : [\kappa]^{<\omega}$ there is a set $H$ of order type $\kappa$ such that for each $n$,$f$ restricted to $n$-element subsets of $H$ omits at least one value in $\kappa$.

Problem. Can $\aleph_\omega$ be a Jónsson cardinal?

Lemma 1. $\theta_2 \leq$ the first Jónsson cardinal.

proof. Let $\kappa$ be a Jónsson Cardinal and let $c : \kappa \times \kappa \rightarrow \omega$ be given. Let $M \prec H(\theta)$, so $c \in M, \vert M \cap \kappa \vert = \kappa$ but $M \cap \kappa \neq \kappa$.

Useful Trick here: Use a wellordering of $\kappa^+$ and the corresponding Skolem function to generate an algebra on that and apply the fact fact that $\kappa$ is Johnson on a subalgebra $\mathfrak{A}$ generated by $\{c\} \cup \kappa$ with the extra function symbol which bijects $\kappa$ and $\vert \mathfrak{A} \vert$.

Let $\alpha = \min (\kappa \setminus M)$. Note that $\alpha$ is a limit ordinal and that $\alpha^+ \leq \kappa$. So we can find $X \subseteq (M \cap \kappa) \setminus \alpha$ of size $\alpha^+$, and $\overline{m} < \omega$ such that $c (\alpha, \beta) < \overline{m}$ for all $\beta \in X$.

Note that by elementarity of $M$, for all finite $F \subseteq X$, the following is unbounded in $\alpha$:
$\displaystyle A_F := \{ \xi < \alpha : c (\alpha, \beta) = \overline{m}, \forall \beta \in F\}$

Choose $X_0 \subseteq X$ of size $\alpha^+$ and $\alpha_0 < \alpha$ such that $c[\{\alpha_0\} \times X_0] = \{\overline{m}\}$, using the pigeonhole principle and $F$ set to be a singleton.

Let $\beta_0 = \min X_0$. Choose $X_1 \subseteq X_0$ of size $\alpha^+$ and $\alpha_0 < \alpha_1 < \alpha$ such that $c[\{\alpha_0\} \times (\{\beta_0\}\cup X_1)] = \{\overline{m}\}$.

It is very important you go on top. That is why I used unboundedness.

Continuing with this we get two infinite sequences $A = \{\alpha_i : i < \omega\}$ and $B = \{\beta_i : i < \omega\}$ such that $c [A \times B] = \{\overline{m}\}$. [QED]

Note. Compare this with $\square$.

## The Second Preparatory Lemma

Why do you have a chance to make interesting ccc posets on $\aleph_2$?

Lemma 2. Suppose $c : \kappa \times \kappa \rightarrow \omega$ is not constant on any product of two infinite sets. Let $\mathcal{F}$ be an uncountable family of pairwise disjoint subsets of $\kappa$. Then for all $m < \omega$ there is an $F \neq G$ in $\mathcal{F}$ such that $c (\alpha, \beta) > m$ for $\alpha \in F, \beta \in G$.

Question. Can you get infinitely many such witnesses? I.e. An infinite witness family? Yes, if you force.

proof of Lemma 2.

Without loss of generality, $\mathcal{F} = \{ F_\xi : \xi < \omega_1\}$. Assume that for some $n$, all $F_\xi$ have size $n$. So every $F \in \mathcal{F}$ can be written in an increasing enumeration as $F = \{F(0), F(1), ..., F(n-1)\}$.

Suppose (for the sake of contradiction) that for some $m < \omega, \forall F \neq G \in \mathcal{F}$ there is an $\alpha \in F$ and a $\beta \in G$ such that $c(\alpha, \beta) \leq m$.

Pick two uniform ultrafilters $\mathcal{U}$ on $\omega_1$ and $\mathcal{V}$ on $\omega$.

Then there exists $i,j < n$, and $\overline{m} \leq m$ such that there are $\mathcal{U}$-many $\beta$ and $\mathcal{V}$-many $k$ such that $c (F_k (i), F_\beta(j)) = \overline{m}$.

It is important that the ultrafilters are on different index sets. It would not work on the same set,” the pigeonhole principle.

So this means that there is an $X \in \mathcal{U}$ such that for all $\beta \in X$ we have $\{k : c (F_k (i), F_\beta(j)) = \overline{m}\} \in \mathcal{V}$. Continue as before to get two infinite sets $A,B \subseteq \kappa$ such that $c [A \times B] = \overline{m}$, which is a contradiction. [QED]

Compare this with the previous proof. There we already had the finite intersection property. We didn’t need ultrafilters.

## Functor Theorem

Here we give most of the proof of the functor theorem.

proof of Functor Theorem. By lemma 1, fix $d: [\kappa]^{<\omega} \rightarrow \mathbb{P}$ (“here is a little business with Skolem functions. This is a good exercise.“) such that $d[X]^{<\omega} = \mathbb{P}$ for all $X \subseteq \kappa$ with $\vert X \vert = \kappa$.

Fix $c : \kappa \times \kappa \rightarrow \omega$ not constant on any rectangle. Fix a 1-1 sequence of reals $\{r_\xi : \xi < \kappa\} \subseteq \{0,1\}^\omega$, (which can be done because $\kappa < \theta_2 \leq \mathfrak{c}^+$).

Let $c^* : [\kappa]^{<\omega} \rightarrow \omega$ be defined by
$\displaystyle c^* (\alpha_0, ..., \alpha_{k-1}) := \max \{c (\alpha_i, \alpha_j) : i \neq j < k\}$

For $\vec s \in (2^{<\omega})^{<\omega}$ of some length $k$, and $F \in [\kappa]^{<\omega}$ define
$\displaystyle \PP_F := \{d(\alpha_0, ..., \alpha_{k-1}) : \forall i < k, \alpha_0 < ... < \alpha_{k-1} \in F, r_{\alpha_i} \upharpoonright c^* (\alpha_0, ..., \alpha_{k-1}) = s_i \}$

Thus $\mathcal{S}(\PP) := \{ F \in [\kappa]^{<\omega} : \forall \vec s \in (2^{<\omega})^{<\omega}, \PP_F (\vec s) \textrm{ is centred in } \mathbb{P} \}$

Allowing infinite demands is usually trouble for ccc!

Claim (To be proved next time): This poset is (powerfully) ccc.

So $\mathbb{P} = \bigcup \{\PP_X (\vec s) : \vec s \in (2^{<\omega})^{<\omega}\}$ for such $X$.