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.

Now we are in business.”

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 .

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