Stevo’s Forcing Class Fall 2012 – Class 16

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

Preserving (Consequences of) Forcing Axioms

We have two branches of forcing axioms, those that give CH type results, and those that give PFA type results.

 MA($\aleph_1$) $\diamondsuit$ PFA CH MM $\mathfrak{b} = \omega_1$

Example: Katetov Problem. Is the following an equivalence?

A compact space $K$ is metrizable iff $K^2$ satisfies the separation axiom $T_5$.

(i.e. $\forall \overline{A} \cap B = \emptyset = A \cap \overline{B}, \exists \textrm{ open } U \supseteq A, V \supseteq B$ such that $U \cap V = \emptyset$.)

Forcing with (1) Sacks Forcing or (2) Souslin Tree Forcing tend to preserve some consequences of forcing axioms.”

For the Katetov Problem, all 6 of the forcing axioms in the above table will give counterexamples.

You want to force somebody to force. It is second order.

Theorem 2 (Todorcevic, 1989, published in dichotomies in Set Theory) $\vDash_\textrm{Sacks}$ OCA fails.

(“You show no new reals are added. It is oscillation.“)

Forcing with a Souslin tree preserves a lot of things.”

Theorem 4. Assume PFA(S), then $\vDash_\textrm{Souslin}$ SH (and PID and $\mathfrak{p} = \omega_1$, so MA$_{\sigma\textrm{-centred}}$ fails.)

Definition. Fix $S \subseteq 2^{<\omega_1}$ a downward closed coherent Souslin Tree. PFA(S) is the forcing axiom for $\aleph_1$-dense sets for the class of all proper posets that preserve $S$.

Question: Why Coherent? What fails for arbitrary? It is a bit more complicated because you have to preserve a tree.

Theorem 3 (Carlson-Laver, 1989). Assuming $\mathfrak{d} = \omega_1$, $\vDash_\textrm{Sacks} \diamondsuit$.

Interesting. You might need CH. That is how they stated it.

On to the Show.

Proof of Theorem 1. This uses Sacks Amoemba (Introduced by Solovay in the Random Real setting.)

Define:

• $\mathcal{S} := \{p \subseteq 2^{<\omega}: p \textrm{ is a perfect tree }\}$.
• $\mathcal{A} := \{(p,n) : p \in \mathcal{S}, n<\omega\}$.

Ordering
. Say $(p,n) \leq (q,m)$ iff $n \geq m, p \subseteq q, p \cap 2^m = q \cap 2^m$.

strongly suggest you go through this proof. There are many wrong proofs in the literature.”

Fact 1. $\mathcal{A}$ is proper.

Fact 2. (Assuming PFA.) Every uncountable set of ordinals in $V^\mathcal{A}$ contains an uncountable subset in $V$, i.e. for every uncountable $X \subseteq \mathcal{A}$ there is an uncountable $Y \subseteq X$ such that $\bigwedge Y > 0$. (“The Poset has caliber $\aleph_1$.“)

proof of Fact 2
. Let $\{\dot{\alpha}_\xi : \xi < \omega_1\}$ be an uncountable colection of $\mathcal{A}$-names for ordinals. Let $\mathcal{D}_\xi := \{(p,n)\in \mathcal{A}: \exists \alpha \textrm{ such that } (p,n) \vDash \dot{\alpha}_\xi = \alpha\}$. Note that each $\mathcal{D}_\xi$ is dense open (for $\xi < \omega_1$).

The corresponding fact about measure algebras is assuming MA($\aleph_1$), then if there is an uncountable set in some random extension, then there is an uncountable subset in the ground model.

Ah! We need more dense sets. (To ensure perfect.)

So let $\mathcal{E}_k := \{(p,k) \in \mathcal{A} : n > k^\textrm{th} \textrm{ splitting level of }p\}$.

Let $\mathcal{G} \subseteq \mathcal{A}$ be a filter that intersects all $\mathcal{D}_\xi, \mathcal{E}_k$. Let $r := r_\mathcal{G} := \bigcap\{p : \exists n (p,n) \in \mathcal{G}\}$.

Then $r$ is perfect.

Now what does this $r$ force?

Note. For every $\xi < \omega_1$ there is $n_\xi < \omega$ such that $(r, n_\xi)$ decides $\dot{\alpha}_\xi$. (Look at $(p,n) \in \mathcal{G} \cap \mathcal{D}_\xi$, $r \leq p$ and $(p,n)$ decides $\alpha_\xi$. So $(r,n)$ decides $\alpha_\xi$.)

Now what is uncountable set?

Fix an $\overline{n}$ so $\Gamma_\overline{n} := \{\xi : n_\xi = \overline{n}\}$ is uncountable.
$\displaystyle (r, \overline{n}) \vDash \{\dot{\alpha}_\xi : \xi \in \Gamma_\overline{n}^\smile \} \in V$

Figure out the rest.” [End of Fact 2]

Now, back to the proof… preserving MA($\aleph_1$).

Let $\dot{\PP}$ be an $\mathcal{S}$-name for a ccc poset and $\dot{\mathcal{D}}_\xi$ for $\xi < \omega_1$ be $\mathcal{S}$-names for $\aleph_1$-many dense open subsets of $\dot{\PP}$. We need to find an $\mathcal{S}$-name $\dot{\mathcal{G}}$ for a filter of $\dot{\PP}$ which is forced by a coordinate in $\mathcal{S}$ to intersect all these dense sets.

We use Amoeba to help us.

Identify $\dot{\PP} \equiv \{$ all $\mathcal{S}$-names for elements of $\dot{\PP}\}$.

We go to $V^\mathcal{A}$ and order $\dot{\PP}$ as follows: $\tau \leq_\mathcal{A} \sigma$ iff $\exists (p,n)$ in the generic filter such that $p \vDash \tau \leq \sigma$.

(Shocking) Lemma: (In $V^\mathcal{A}$). The poset $(\dot{\PP}, \leq_\mathcal{A})$ is ccc.

(“You are requiring $p$ to be in the generic; you have fewer options.“)

By Fact 2, we can go to $V$ and assume to have uncountable $\{\tau_\xi : \xi < \omega_1\} \subseteq \dot{\PP}$.

There is a price for this:”
$\displaystyle (p,n) \vDash \{\tau_\xi : \xi < \omega_1^\smile\} \subseteq \dot{\PP}$

Need to find $\xi \neq \eta$ and $(q,n) \leq (p,n)$ such that $q \vDash$$\tau_\xi$ and $\tau_\eta$ are compatible in $(\dot{\PP}, \leq_\mathcal{A})$

List $p \cap 2^n$ as $s_1, ..., s_k$. Find an uncountable $\Gamma_1 \subseteq \omega_1$ and $q_{s_1} \subseteq p_s$, (where $q_s := q \cap \{x \in 2^{<\omega} : x \supseteq s \textrm{ or }s \subseteq x\}$)

By using the maximal principle, $\forall \xi, \eta \in \Gamma_1$ we have $q_{s_1} \vDash$$\tau_\xi$ and $\tau_\eta$ are compatible in $(\dot{\PP}, \leq_\mathcal{A})$” (by using $\tau_i$)

Be careful. There is a little bit of funny going on here.

Take $\Gamma_n \subseteq \Gamma_{n-1}$ so that $q_{s_n} \vDash$$\tau_\xi$ and $\tau_\eta$ are compatible in $(\dot{\PP}, \leq_\mathcal{A})$

Now take $q := \bigcup_{i\leq k} q_s$. So for $\xi \neq \eta \in \Gamma_k$ we have $(q, n) \vDash$$\tau := \bigvee_{i \leq k} \tau_i$ is a common extension of $\tau_\xi$ and $\tau_\eta$”.

3 thoughts on “Stevo’s Forcing Class Fall 2012 – Class 16”

1. Ari Brodsky says:

How do I get the password to see Class 17?

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2. Micheal, any chance I could get a copy of the password? These are nice notes, and I would like to take a look at the last one, if possible.

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1. Micheal Pawliuk says:

Oops! I had forgotten about that last set of notes. I still need to fix it up, but I will unblock it. Beware, it still has some rough patches.

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