# Stevo’s Forcing Class Fall 2012 – Class 4

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

## Measure Algebras

When talking about how Cohen Forcing $\mathcal{C}$ does not collapse $\omega_1$ it is better to think about how the regular open algebra $\B = \textrm{ro}(\mathcal{C})$ does not collapse $\omega_1$. This we can tell from some facts about the distributivity of the algebra:

Fact. In $\B$, $\forall \{a_{n_\xi} : \xi < \omega_1, n < \omega\} \subseteq \mathcal{B}^+$ and $\forall b \in \mathcal{B}^+$ there are $\alpha_n < \omega_1$ (for $n<\omega$) such that
$\displaystyle (b \cap \bigwedge_{n<\omega} \bigvee_{\xi < \alpha_n} a_{n_\xi} )> 0$
.

Recall the analogous distributive law with $\omega_1$ replaced by $\omega$.

Weak Distributive Law. $\forall \{a_{n_k} : k < \omega, n < \omega\} \subseteq \mathcal{B}^+$ and $\forall b \in \mathcal{B}^+$ there are $l_k < \omega$ (for $k<\omega$) such that
$\displaystyle (b \cap \bigwedge_{n<\omega} \bigvee_{k \leq l_k} a_{n_k} ) > 0$
.
Fact. If $\B$ is ccc, then $\B$ has the weakly-$(\omega_1, \omega_1)$-distributive.

#### Examples of ccc and weakly-$(\omega, \omega)$-distributive algebras.

Example. Try $\B = \textrm{ro}(\PP)$ where $\mathbb{P} = \{K \subseteq [0,1] : K \textrm{ is compact and has positive measure}\}$.

Weakly Distributive and $\sigma$-centred do not go together.”

Question. Is there a $\sigma$-centred, weakly distributive algebra? (Note that measure algebras do not work.)

For a Measure Algebra, $\mu: \mathcal{B} \rightarrow [0,1]$ strictly positive, $d_\mu (a,b) = \mu (a \Delta b)$ is countably additive or, it is finitely additive and weakly distributive.

Question. Are there nonseparable (non trivial) Maharam algebras? (You need to “make submeasure far away from measure” and “find $\sigma$-finite-cc, not $\sigma$-bounded-cc“.)

## Back to the “Suslin Family”.

Theorem 2. There is a $\sigma$-linked, non $\sigma$-centred poset of size $\mathfrak{p}$.

Smaller families are more interesting.”

Compare this with the following adjustments:

Theorem 1. There is a $\sigma$-linked, non $\sigma$-centred poset of size $\mathfrak{t}$

Theorem 0. There is a non $\sigma$-linked poset of size $\bb$ with no 2-linked subset of size $\bb$.

The proof is similar to Theorem 1, which we did in Class 2.

proof. We may assume that $\mathfrak{p} < \mathfrak{b}$, by Theorem 1.

Let $\{a_\xi : \xi < \mathfrak{p}\} \subseteq [\omega]^\omega$ have the finite intersection property, that is non-extendible. That is, there is no $b \in [\omega]^\omega$ such that $b \subseteq^* a_\xi$ for all $\xi < \mathfrak{p}$.

We build (“if possible, or in other words we simply try“), $C$, a tower $\{b_\alpha : \alpha <\pp\}$ such that

• $b_\alpha \neq b_\beta$ for $\alpha \neq \beta$
• $b_{\xi + 1} \subseteq a_\xi$
• $\vert b_\alpha \cap a_\xi \vert = \aleph_0$, for all $\alpha, \xi < \mathfrak{p}$.

Of course it is going to be a recursive construction. This tells you you have hope. Otherwise, there is no future.”

Given $b_\xi$, let $b_\xi = b_\xi \cap a_\xi$.

Everybody intersects everybody.”

Suppose $\xi$ is a limit $<\pp$.

Case 1, where $\cof(\xi) = \omega$.

Pick a sequence $\xi_n$ increasing to $\xi$. Define $c_n := \bigcap_{i \leq n} b_{\xi_i}$. And for $\alpha < \mathfrak{p}$, define $f_\alpha \in \omega^\omega$ by $f_\alpha (n) = \min (a_\alpha \cap c_n) \setminus (f_\alpha (n-1) + 1)$.

Since we are assuming that $\mathfrak{p} < \mathfrak{b}$, fix $g \in \omega^\omega$ such that $f_\alpha <^* g$ for all $\alpha < \mathfrak{p}$. Let $b_\xi := \bigcup_{n < \omega} c_n \cap g(n)$. (Remember we were trying to define, recursively, $b_\xi$.)

Then $b_\xi \subseteq^* b_\eta$ for $\eta < \xi$, and $b_\xi \cap a_\alpha$ is infinite for all $\alpha < \mathfrak{p}$.

Case 2, where $\cof(\xi) > \omega$.

We force $b_\xi$, but we are not allowed to force in the obvious ($\sigma$-centred) way.”

Let $ \in \mathbb{P}$ iff $F \in [\xi]^{<\omega}$, $G \in [\pp]^{<\omega}$ and $\forall k < \omega, \forall \alpha \in G, \vert b_F \cap a_\alpha \cap k \vert \geq \vert \Delta_F \cap k \vert$. The ordering is coordinate-wise inclusion.

(See Class 2 notes, for this notation, and intuition.)

Claim: $\mathbb{P}$ is ccc (and in fact is $\sigma$-k-linked for all $k$)

If the claim is true, if $\mathbb{P}$ is not $\sigma$-centred we are done. So assume $\mathbb{P}$ is $\sigma$-centred.

Let $\mathbb{P} = \bigcup_{n<\omega} \PP_n$ be such that $\PP_n$ is centred in $\mathbb{P}$, for all $n$. Let $A_n := \bigcup\{ F : \exists G, \in \PP_n\}$. We may assume that each $A_n$ is cofinal in $\xi$, so that $\xi = \bigcup A_n$.

The next move is a dangerous move in general. Because we are asking for infinitely many requirements.

Let $c_n := \bigcap \{ b_\eta : \eta \in A_n\}$.

Claim: These $c_n$ are infinite.

Since $A_n$ is cofinal in $\xi$, $c_n \subseteq^* b_\eta$, for all $\eta < \xi$. They are infinite since $\Delta_{A_n}$ is infinite and $\vert b_F \cap a_\alpha \cap k \vert \geq \vert \Delta_F \cap k \vert$.

Now what do we know?

If $\alpha \in G$, such that for some $n$ and $F$, $ \in \PP_n$, then $a_\alpha \cap c_n$ is infinite.

Now $\bb$ is going to help us. You do the usual thing. What is the usual thing?

For $\alpha < \mathfrak{p}$, define $f_\alpha \in \omega^\omega$ such that …

$f_\alpha (n) =$ minimal $k > f_\alpha (n-1)$ such that for every $m \leq n$ if $a_\alpha \cap c_n$ has some points above $f_\alpha (n-1)$ then it has some point below $k$.

Then, as before, $b_\xi := \bigcup_{n < \omega} c_n \cap g(n)$.

We have that $\{b_\eta : \eta < \xi\}$ and $\{c_n : n<\omega\}$ form a gap, but we know that $\bb$ kills $\omega, *$-gaps.

So we can continue on and either get that $\mathfrak{p} < \mathfrak{t}$, (so we have the last theorem), or we can continue to what we want. [QED]

Theorem. The following are equivalent:

1. MA($\aleph_1$) (i.e. $\mathfrak{m} > \aleph_1$);
2. Every first-countable ccc compact space is separable.

## Some Preliminaries about the Cardinal $\theta_2$

A useful cardinal for us is $\theta_2 := \min \{\theta : {{\theta}\choose{\theta}} \rightarrow {{\omega}\choose{\omega}}^{1,1}_\omega\}$, where the arrow notation means $\forall c: \theta \times \theta \rightarrow \omega$ there are $A,B \in [\theta]^{\aleph_0}$ such that $c \upharpoonright A \times B$ is constant.

Exercise. $\theta_2 \geq \aleph_2$

Theorem ($\mathfrak{m} > \omega_2$). $\theta_2 = \aleph_2$ iff Chang’s Conjecture.

Theorem *. If Chang’s Conjecture fails, there is a ccc poset which forces ${{\omega_2}\choose{\omega_2}} \not\rightarrow {{\omega}\choose{\omega}}^{1,1}_\omega$.

Exercise. Chang’s Conjecture implies $\theta_2 = \aleph_2$.

[Chang’s Conjecture]. For every structure of the form $\mathfrak{A} = (\omega_2, \omega_1, )$ there is an elementary substructure $\mathfrak{B} = (B, B\cap \omega_1, )$ of $\mathfrak{A}$ such that $\vert B \vert = \aleph_1$ and $\vert b \cap \omega_1 \vert \leq \aleph_0$.

Problem. Is it true (in ZFC) that $\theta_2 \leq \aleph_3$? “Any $\aleph_\alpha$ would be interesting.”

## Now Amoebas and Functors.

We are going to produce a functor that goes from a ccc poset $\mathbb{P}$ to another ccc poset $\mathcal{S}(\PP)$. The important thing is that we will be going “from local to global control“.

Theorem. Suppose that ${{\kappa}\choose{\kappa}} \not\rightarrow {{\omega}\choose{\omega}}^{1,1}_\omega$ 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 centred.

We are getting away from meeting dense sets.”

Sketch. ${{\kappa}\choose{\kappa}} \not\rightarrow {{\omega}\choose{\omega}}^{1,1}_\omega$ implies that for every algebra $\mathfrak{A} = (\kappa, (f_n)_{n<\omega})$ there is a subalgebra $\mathfrak{B} = (B, (f_n \upharpoonright B^{k_n})_{n < \omega})$ such that $\vert B \vert = \kappa$ but $B \neq \kappa$.

This means that we have a map $f: [\kappa]^{<\omega} \rightarrow \mathbb{P}$ such that $f''[\Gamma]^{<\omega} = \mathbb{P}$ (this is global information), for all $\Gamma \subseteq \kappa$ of cardinality $\kappa$, (where we think of allowing $\Gamma$ to be any “thin” set.)

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

1. Ari Brodsky says:

Here are some quotes from Stevo’s fourth class, on Sept. 21:

“You can do lots of forcing without mentioning forcing. The problem is, where do you get the intuition? Where do you get the ideas?”

“Does anyone see a proof of what I just said? You could also have a disproof.”

“All these little things to help us, we assume.” (re: the arbitrarily chosen family of size $\mathfrak p$ is closed under finite intersections)

“It’s like tennis: Moving from one side to the other, then running from the other side.”

“We can be relaxed when stating the theorems, because we’ll figure it out when we do the proofs. It’s the proofs we remember, not the theorems.”

“I hate writing ‘in ZFC’. ”

$\theta_d$ is related to the $d$-dimensional version of the Halpern-Lauchli theorem.”

Sorry I didn’t write down any content from these lectures; just the quotes!

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