Stevo’s Forcing Class Fall 2012 – Class 17

(This is the seventeenth (and final) lecture in Stevo Todorcevic’s Forcing class, held in the fall of 2012. You can find the sixteenth 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.)

[TITLE]

I want to go a little over this proof.

Theorem 1 (Carlson-Laver, 1989). Assuming PFA, $\vDash_\textrm{Sacks} \textrm{MA} (\aleph_1)$.

Recall the notation:

• $\mathcal{S} := \{p \subseteq 2^{<\omega}: p \textrm{ is a perfect tree }\}$;
• $\mathcal{A} := \{(p,n) : p \in \mathcal{S}, n<\omega\}$, the Amoeba;
• $\dot{\PP}$ is an $\mathcal{S}$-name for a ccc poset, identified with $\{$ all $\mathcal{S}$-names for elements of $\dot{\PP}\}$;

In $V^\mathcal{A}$ we order $\dot{\PP}$ as follows: $\tau \leq_\mathcal{A} \sigma$ iff $\exists (p,n)$ in the generic filter such that $p \vDash \tau \leq_{\dot{\PP}} \sigma$.
Aside: We can assume $\dot{\PP} = (X^\smile, \leq_\mathbb{P}), X \subseteq 2^\omega, \vert X \vert \leq \aleph_1$ and $\tau : 2^\omega \rightarrow 2^\omega$ is continuous such that $\textrm{int}_G (\tau) = \tau(\textrm{sacks real})$.

I would like to show this is a ccc poset.”

Claim 2: [In $V^\mathcal{A}$] $(\dot{Q}, \leq_\mathcal{A})$ is a ccc poset.
We actualy need claim 0. (Remember claim 1 was about a caliber property.)

Claim 0: For every sequence $\{\tau_\alpha : \alpha < \omega_1\}$ of elements of $\dot{Q}$ and $p \in \mathcal{S}$ there is an uncountable $\Gamma \subseteq \omega_1$ and a family $\{p_\alpha : \alpha \in \Gamma\}$ of pairwise compactible elements of $\mathcal{S}$ below $p$ such that:
$\displaystyle p_\alpha \wedge p_\beta \not \vDash \tau_\alpha \perp \tau_\beta$

(“Double negation!“)

proof of Claim 0.

Apply PFA to $\mathcal{S} \star \dot{\PP}$ which will force uncountably many generics into ‘the set'(???). [End of proof]

proof of Claim 2.

Let $\{\tau_\alpha : \alpha < \omega_1\}$ of elements of $\dot{Q}$. By lemma 1, some uncountable subset of $\{\tau_\alpha : \alpha < \omega_1\}$ belongs to $V$ so we may work in $V$.

So we need to produce $(p,n) \in \mathcal{A}$ (below any given $(q,m) \in \mathcal{A}$) and $\alpha \neq \beta$ such that $(p,n) \vDash_\mathcal{A} \tau_\alpha \not \perp_\mathcal{A} \tau_\beta$.

In fact, $n=m$. Set:

• $\{s_1, ..., s_k\} = q \cap 2^m$; and
• $q^{s_i} = q \cap \{x \in 2^{< \omega} : x \not \perp s_i\}$. (“Perfect tree, clopen set.”)

Applying lemma 0, we get an uncountable $\Gamma_1 \subseteq \omega_1$ and $\{p_\alpha^1 : \alpha \in \Gamma_1\} \subseteq \mathcal{S} \upharpoonright q^{s_1}$ such hat
$\displaystyle \forall \alpha, \beta \in \Gamma_1, \textrm{ we have } p_\alpha^1 \wedge p_\beta^1 \not \vDash \tau_\alpha \perp \tau_\beta$

Apply lemma 0 again to find an uncountable set $\Gamma_2 \subseteq \Gamma_1$ and $\{p_\alpha^2 : \alpha \in \Gamma_2\} \subseteq \mathcal{S} \upharpoonright q^{s_2}$ such that
$\displaystyle \forall \alpha, \beta \in \Gamma_1, \textrm{ we have } p_\alpha^2 \wedge p_\beta^2 \not \vDash \tau_\alpha \perp \tau_\beta$

Now if $\alpha \neq \beta \in \Gamma$, then for each [… SEE DANA]

Last time I mentioned these names: Souslin, Sacks, Solovay (random).

I decided not to talk about this now as I will be giving a lecture in the Czech Republic in February.

[Here is] exposition about two results, actually 3 (one which unifies them) which are actually quite complete.

Theorem 2 (1985, Laver). $\textrm{MA} (\aleph_1)$ implies $\vDash_\mathcal{R} \textrm{SH}$, for any measure algebra $\mathcal{R}$.

Assuming $\diamondsuit$ isn’t interesting as measure algebras preserve Souslin trees. You’d like to use CH, but the next result says there is no hope.

Ultrapowers of Measure Algebra $(\mathcal{R}, \mu)$

Let $I$ be an index set, $\mathcal{U}$ a non-principal ultrafilter on $I$, and consider $(\mathcal{R}^I /\mathcal{U}, \mu_\mathcal{U})$ be the ultrapower where $\mu_\mathcal{U} ([f]_\mathcal{U}) = \lim_{i \rightarrow \mathcal{U}} \mu(f(i))$.

Theorem (Todorcevic, 1995). If the character (i.e. density of the metric space) of $\mathcal{R}$ is bigger than $I, \theta$ then for every sequence $([f_\xi]: \xi < \theta) \subseteq \mathcal{R}^I /\mathcal{U}$, there is a sequence $(c_\xi: \xi < \theta) \subseteq \mathcal{R}$ such that:

• (“The $c_\xi$‘s reflect.“) $\mu (c_\xi) = \lim_{i \rightarrow \mathcal{U}} \mu(f_\xi (i))$ for all $\xi < \theta$;
• (“More important:”) $\forall K \in \mathcal{U}, \forall \Gamma \in [\theta]^{<\omega}, \bigwedge_{\xi \in \Gamma} c_\xi \leq \bigvee_{i \in K} \bigwedge_{\xi \in \Gamma} f_\xi (i)$

You should dare to do something. It is a Brave New World.

Proof of Theorem 2

Proof of Theorem 2. “Once you do the proof, you will see you do not want to check if PID implies $\vDash_\mathcal{R}$ PID.

Let $(\omega_1, \leq_T)$ be an $\mathcal{R}$-name for an Aronszajn tree nicely put on $\omega_1$. I.e. $[\delta, \delta+\omega)$ is its $\delta^\textrm{th}$ level for $\delta$ countable limit.

This idea comes from the original source for PID.

$\dot{I}_T = \{X \subseteq \omega_1 : X \textrm{ countable}, X \textrm{ strongly unbounded}\} = \{\textrm{pred}_T (\alpha) : \alpha < \omega_1\}$, where $X$ is strongly unbounded if every infinite subset of $X$ is unbounded.

$\vDash_\mathcal{R} \dot{I}_T$ is a P-ideal.

This of course is wht PID implies SH, but not here, because we just have names.

Let $\mathcal{I} = \{A \subseteq \omega_1 : \llbracket A^\smile \in \dot{I}_T \rrbracket_\mathcal{R} = 1\}$.

Claim
: $\mathcal{I}$ is a P-ideal.

I strongly suggest you do this with other posets and see if it is P-ideal. e.g. Sacks. Might have something to do with weakly-distributive or $\omega^\omega$-bounding.”

$\{A_n : n<\omega\} \subseteq \mathcal{I}$. “What are the pieces to chop off?

$\bigcup_{n < \omega} A_n \subseteq \delta$ for $\delta$ countable limit.

Let $\dot{A}_{n,k}$ be the $\mathcal{R}$-name for $\{\alpha \in A_n : \alpha <_T \delta + k\}$.

For $r,m < \omega$, choose a _finite_ set $F_{n,m} \subseteq A_n$ such that
$\displaystyle \mu(\llbracket \bigcup_{k \leq n} A_{n,k} \subseteq F_{n,m}^\smile \rrbracket > 1 - \frac{1}{2^m})$

(“i.e. is very close to 1“)

[PICTURE 1]

Let
$\displaystyle A := \bigcup_{m < \omega} (A_m \setminus \bigcup_{n \leq m} F_{n,m})$

Note $A_m \subseteq^* A, \forall m < \omega$. We need to show this is in $\mathcal{I}$.

For $m < \omega$, $A^m := A \cap \bigcup_{n \geq m} A_n$.

[PICTURE 2]

$\displaystyle \mu(\llbracket \textrm{pred}(\delta + h?) \cap A^m \neq \emptyset \rrbracket < 2^{-m+1})$

(“You don’t need to worry about the $A_n$‘s.“)

By PID, we have that either:

1. There is an uncountable $B \subseteq \omega_1$ such that $[B]^{\aleph_0} \subseteq \mathcal{I}$. (So $\llbracket B^\smile$ contains an uncountable antichain $\rrbracket = 1$.)
2. There is an uncountable $B \subseteq \omega_1$ such that $B \perp \mathcal{I}$.

($\star$) “Here is where we need this measure theory lemma about ultrapowers.

What are the functions?

Pick a uniform ultrafilter $\mathcal{U}$ on $\omega_1$ such that $B \in \mathcal{U}$. For $\alpha \in B$ let $f_\alpha : \omega_1 \rightarrow \mathcal{R}$ be defined by $f_\alpha (\beta) = \llbracket \alpha <_T \beta \rrbracket$.

Claim
. $\{\alpha \in B : \lim_{\beta \rightarrow \mathcal{U}} \mu[f_\alpha (B) \neq 0]\}$ is countable.
proof. Otherwise, let $B_0 = \{\alpha \in B : \mu_\mathcal{U}([f_\alpha])>0\}$.

Apply the ultrapower theorem. We will get some $c_\alpha \in \mathcal{R}$ for ($\alpha \in B_0$) that somehow reflects.
$\displaystyle \forall K \in \mathcal{U}, \forall \Gamma \in [B_0]^{<\omega}, \bigwedge_{\xi \in \Gamma} c_\xi \leq \bigvee_{\beta \in K} \bigwedge_{\alpha \in \Gamma} f_\alpha (\beta)$

(“Here the $<\omega$ is really just 2. Just pairs.“)

If you decode what I wrote, that’s what it says.”

Find $c \in \mathcal{R}$ such that $c \vDash \{\alpha \in B_0 : c_\alpha \in \textrm{ generic object}\} =: B_1$ is countable.

So $c \vDash \dot{B_1}$ is a branch of $T$.

This [$B_2$] is a good name, a fantastic name! I can’t believe I picked such a good name. It is a branch.”

The generic must pick up some $\beta \in K$ for which $\alpha$ [SOMETHING]. So [SOMETHING ELSE]. [End of Proof]

At ($\star$) I force to add to the density, so the theorem applies. If I broke the Suslin tree, great!

[SECTION TITLE]

Let us see that MA($\aleph_1$) is enough to push through SH.”

People are interested in some Martin principles.”

Statement of SM$_\theta$. For every *S*et-*M*apping $f: \theta \rightarrow [\theta]^{\leq \aleph_0}$ either:

1. $\theta = \bigcup_{n < \omega} X_n$, where each $X_n$ is $f$-Free.
2. There is an uncountable $B \subseteq \theta$ such that for every finite $\Gamma \subseteq B$, $\{\beta \in B : \Gamma \subseteq f(\beta)\}$ is uncountable. (“Very much not free.”)

Compare this with the Free-Set Lemma:

Free-Set Lemma. [STATEMENT]

Exercise. SM$_\theta$ implies that trees on $\theta$ with no uncountable chains are special.

(Use pred mapping, $f(\beta) = \{\alpha : \alpha <_T \beta\}$. Use $T=(\theta, <_T)$, so you kill Suslin trees.)

Proposition. SM$_{\omega_1}$ implies that if a compact space $K$ contains an L-subspace,then $K$ contains an uncountable free sequence.

Recall. An L-space is a regular, hereditarily Lindelof, non-separable space. (These exist in ZFC.)

Proof of Proposition.

Assume $K \subseteq \omega_1$ is an L-space. This means (by possibly passing to a subsequence) that
$\displaystyle \overline{\{\alpha : \alpha < \beta\}}^K \cap \{\alpha : \alpha \geq \beta\} = \emptyset$

By regularity, pick $U_\beta \ni \beta$ open in $K$ such that $\overline{U_\beta}^K \cap \overline{\{\alpha : \alpha < \beta\}}^K$. Define $f: \omega_1 \rightarrow [\omega_1]^{\leq \aleph_0}$, by $f(\beta) = \{\alpha : \beta \in U_\alpha\}$.

Free set is a discrete space.” (??)

For $B$ in the second alternative we get $\forall \gamma \in B$, we have that the collection of $\overline{\{\alpha: \alpha < \beta\}}^K$ and $\overline{U_\beta}^K$ (for $\beta \in B \cap \gamma$) has the finite intersection property. [End of proof]

Now We Try to Prove the Theorem.

proof. Fix an $\mathcal{R}$-name $\dot{f}$ for a set-mapping as in SM$_\theta$. For $\beta \in \theta$, $\Delta(\beta) := \{\alpha : \llbracket \alpha \in \dot{f}(\beta) \rrbracket > 0\}$ is countable, by measure algebra.

I want to apply MA$_\theta$, so I need a poset. The idea seems to be to force [the first alternative], and if it isn’t ccc, then [the second alternative].

I will force one free set, but there is the usual trick to get many.

Let $\mathcal{R}_\beta$ be the subalgebra generated by $\{\llbracket \alpha \in \dot{f}(\beta) \rrbracket : \alpha \in \Delta(\beta)\}$, which is separable as a metric space.

Poset. Let $\mathbb{P}$ be the collection of all finite partial maps $p: D_p \rightarrow \mathcal{R}$ such that:

1. $p(\alpha) \in \mathcal{R}_\alpha, \mu(p(\alpha)) > 1/2$ for $\alpha \in D_p$.
2. $p(\alpha) \wedge p(\beta) \leq \llbracket \alpha \notin \dot{f}(\beta) \rrbracket$ for $\alpha \neq \beta \in D_p$.

[1] is a tricky move. For measure algebras making the requirement rarely gives ccc. But I’m hopeful because I’veput it in the separable algebra $\mathcal{R}_\alpha$.

Ordering. $p \leq q$ iff

• $D_p \supseteq D_q$; and
• $p(\alpha) \leq q(\alpha)$ for $\alpha \in D_q$.

If it is (powerfully) ccc, you are forcing alternative 1.

So we focus on alternative 2.

Claim. If $\mathbb{P}$ is not ccc, then there is an $\mathcal{R}$-name $\dot{B}$ for an uncountable subset of $\theta$ satisfying the second alternative of SM$_\theta$.

First a $\Delta$-system, etc.

Let $(p_\alpha : \alpha < \omega_1) \subseteq \mathbb{P}$ be a sequence of pairwise incompatible elements in $\mathbb{P}$.

Now we have to be careful.”

Fix $\xi \in D$, $p_\alpha (\xi)$ for $\xi < \omega_1$, sitting in $\mathcal{R}_\xi$ a separable [SOMETHING SOMETHING].

So [SOMETHING], we can ignore the root.

Now the u.p. thing.

Suppose $p_\alpha (\xi) > 1/2 \neq \epsilon ?$, for all $\xi \in D_p$, $\xi \in \Gamma$. ?

[SOMETHING].

Here are some references:

• Singapore Lecture Notes. [SEE DANA]
• Combinatorial Dichotomies of Set Theory. (BSL 2011).