# Namba Forcing with Arnie Miller

(These are the voyages are the starship enterprise notes from Arnie Miller’s guest lecture in Alan Dow’s Forcing course, held on October 11, 2012. The lecture was a self-contained exploration of strange new worlds Namba forcing. (Alright, I will stop doing that. For those who don’t get the reference.) Arnie Miller’s quotes are in Grey. Yes, the forcing checks are ugly. I have yet to figure that one out.)

## Motivation

It starts with a theorem in Prikry’s Thesis.

Theorem (Prikry). We can have $M \sse M[G]$ such that

• $(\textrm{Cards})^M = (\textrm{Cards})^{M[G]}$; and
• $\exists \kappa \in M$ such that $M \models$ “$\omega < \kappa^{\textrm{reg cardinal}}$” but $M[G] \models \textrm{cof}(\kappa) = \omega$.

Of course here, $\kappa$ would have to be a limit cardinal.

## Changing the Cofinality of a Successor

Example. (~1962) Levy collapse of $\omega_2$ to $\omega_1$.

Let $\PP := (\omega_2)^{<\omega_1} = \bigcup_{\alpha < \omega_1} \omega_2^\alpha$ where $\omega_2^\alpha := \{p \vert p: \alpha \rightarrow \omega_2\}$. The ordering is $p \leq q$ iff $p \supseteq q$. The generic function here is $g : \omega_1 \twoheadrightarrow \omega_2^M$, which collapses the cofinality of $\omega_2$.

This poset is countably closed, so $\omega_1$ is not collapsed. So $M[G] \models “\textrm{cof}(\omega_2^M) = \omega_1$”, because it can’t be $\omega$.

## Namba is Something In-Between

The majority of this lecture will be spent proving the second part of this theorem.

Theorem (Namba, 1971). If $M \models “\PP \textrm{ is Namba}”$ and $G$ is $\PP$-generic, then

1. $M[G] \models “\omega_1 = \omega_1^M$ and $\textrm{cof}(\omega_2^M) = \omega$”
2. (Assuming $M \models$ CH) $(\omega_1^\omega)^M = (\omega_1^\omega)^{M[G]}$ or equivalently, $(\mathcal{P}(\omega))^M = (\mathcal{P}(\omega))^{M[G]}$.

It doesn’t add reals, but it isn’t countably closed. (Recall that for separative posets these are equivalent. See XX.XX.X in Newnen.)

I really like it as a bizarre way to collapse $\omega_2$.”

I’m really into proofs, so I think I’m done with my introduction.”

## Defining Namba Forcing

First we define what a Namba Tree is, and the Namba poset will be the collection of all Namba trees, together with a non-obvious ordering.

Notation. For a poset $\PP \ni p, s$, we write $p_s := \{t \sse p : t \sse s \textrm{ or } s \sse t\}$.
$A(p) := \{t \in p : \textrm{root}(p) \sse t\}$

Definition. $p \sse \omega_2^{<\omega}$ is a Namba Tree if

1. $p \neq \emptyset$
2. $p$ is a tree, (i.e. $\forall s \sse t \in p$ we have $s \in p$)
3. $\exists s \in p$ such that $s = \textrm{root}(p)$ (i.e. $p_s = p$)
4. After the root there is $\omega_2$-splitting. I.e. We have $\forall t \in A(p), \vert \{ \alpha < \omega_2 : t^\frown<\alpha> \in p \} \vert = \omega_2$

It is very bushy and wide as soon as you pass the root.”

Ordering: $p \leq q$ iff $\textrm{root}(q) \subseteq \textrm{root}(p)$ and there is still the immediate $\omega_2$ splitting.

What does the generic do? If $G$ is $\PP$-generic, then let $g := \bigcup\{\textrm{root}(p) : p \in G\}$.

Exercises.

1. $g : \omega \rightarrow \omega_2^M$ is cofinal;
2. $\{g\} = \bigcap_{p \in G} [p]$ where $[p] := \{x \in \omega_2^\omega : \forall n, x\upharpoonright n \in p\}$.

The root is the working part, the splitting is the side condition.”

You need to fix the root. If you move it countably many times then you are in trouble.”

To control this we point out a stronger extension property.

Definition. $p \leq_0 q$ iff $p \leq q$ and $\textrm{root}(p) = \textrm{root}(q)$. (I.e. we only change the side condition.)

Lemma 1. If $p \Vdash \tau < \omega_1^\smile$, then $\exists q \leq_0 p$ and $\exists \alpha < \omega_1$ such that $q \Vdash \tau = \alpha^\smile$.

Of course this is totally false for $\omega_2$

Here is the combinatorics: $\omega_2$ is not $\omega_1$.”

#### proof of Lemma 1.

First a useful notion. Say $s \in A(p)$ is good iff $\exists q \leq_0 p_s$ and $\exists \alpha < \omega_1$ such that $q \Vdash \tau = \alpha^\smile$. If $s$ is not good, then it is bad.

Claim. If $s$ is bad, then $\vert\{\alpha <\omega_2: s^\frown<\alpha> \in p \textrm{ is good}\}\vert \leq \omega_1$.

This says there are at most $\omega_1$ many good successors. The reason is the tree technology.”

Suppose $\vert G \vert = \omega_2$, (where $G$ stands for good), and $$\forall \alpha \in G, \exists q^\alpha \leq_0 p_{s^\frown \alpha}, \exists \beta_\alpha < \omega_1 \textrm{ such that }q \Vdash \tau = \beta_\alpha^\smile$$

This is where the $\omega_1 / \omega_2$ contrast comes in.”

Now $\exists G_0 \sse G, \exists \beta, \forall \alpha \in G_0, \beta_\alpha = \beta$ and $\vert G_0 \vert = \omega_2$.

Do the obvious thing.

Let $q := \bigcup_{\alpha \in G_0} q^\alpha \leq_0 p$ and note $q \Vdash \tau = \beta$ “because it is impossible to force $\tau$ to be anything else.” But this contradicts “$s$ is bad.” [End of Claim]

Now look at the things that are hereditary. Start with the root, take all the bad ones beneath it, throw everything else out, then keep taking bad.”

Define $q := \{s \in p : s \sse\textrm{root}(p) \textrm{ or } s \in A(p), s$ is bad, and every $t$ such that root$(p) \sse t \sse s \textrm{ is bad}\}$. Now $q \leq_0 p$ and $\forall s \in A(q)$ we have $p_s$ is bad.

Also, $p \Vdash \tau < \omega_1^\smile$, $\exists r \leq q, \exists \alpha < \omega_1, r \Vdash \tau = \alpha^\smile$ and $r \leq_0 q_s \leq p_s$ with $s = \textrm{root}(r)$. This implies $s$ is good, a contradiction. [QED]

We want to soup this up.”

(Souped Up) Lemma 2. If $p \Vdash \tau : \omega \rightarrow \omega_1$, then there is a $q \leq_0 p$, and $\pi : q \rightarrow \omega_1^{<\omega}$ such that

• $\forall s \in q$, with $\vert s \vert = n$, $q_s \Vdash \tau \upharpoonright n = \pi(s)^\smile$, $\vert \pi(s) \vert = n$; and
• If $t \supseteq s$, then $q_t \leq q_s$ implies that $\pi(t) \geq \pi(s)$.

So this is a continuous map.”

#### soup proof.

We get this by a fusion argument. You can work locally and fuse everything together.”

For ease of the pictures, the root will be assumed to be empty.

Assume $\textrm{root}(p) = \emptyset$. Construct $p \geq_0 p_0 \geq_0 p_1 \geq_0 …$ of stronger conditions such that $\omega_2^n \cap p_n = \omega_2^n \cap p_{n+1}$ and so $\bigcap_{n < \omega} p_n =: q \leq_0 p$.

Note that $q$ is still a Namba tree as previous levels are preserved.

Take $q^s \leq_0 (p_n)_s$ with $s \in \omega_2^n \cap p_n$.

Observe $q^s \Vdash \tau \upharpoonright n = \pi(s)$ and $p_{n+1} = \bigcup_{s \in \omega_2^n \cap p_n} q^s$

A fusion type argument. It turns something not countably closed into something close.” [QED]

## Now Topology

We will be concerned with $\omega_2^\omega$, and its topology. The basic clopen sets are $[s] := \{x \in \omega_2^\omega : s \sse x\}$ where $s \in [\omega_2]^{<\omega}$.

Then $[p] := \{x \in \omega_2^\omega : \forall n, x \upharpoonright n \in p\}$ is a closed set which is “a super fat, (stupid) Namba tree.

For $\pi : q \rightarrow \omega_1^{<\omega}$ with $\vert s \vert = \vert \pi(s) \vert$ and $s \sse t$ implies $\pi(s) \sse \pi(t)$ then $$\tilde {\pi} : [q] \rightarrow \omega_1^\omega \textrm{ is a (Lipschitz) continuous map}$$

Let’s take the CH example. Find a subcondition of $q$ that is forced to be equal to $\tau$.”

Recall: For a filter $\mathcal{F}$ on $X$,

• $\mathcal{I}_{\textrm{Dual}} := \{X\setminus A : A \in \mathcal{F}\}$
• $\mathcal{F}^+ := \{ A : A \notin \mathcal{I}\} = \{A : \forall F \in \mathcal{F}, F \cap A \neq \emptyset\}$

Examples.

1. For $\mathcal{F}$ the measure 1 sets on $[0,1]$, $\mathcal{I}_{\textrm{Dual}}$ is the ideal of measure 0 sets and $\mathcal{F}^+$ is the positive measure sets.
2. For $\mathcal{F}$ the cofinite sets on $X$, $\mathcal{I}_{\textrm{Dual}}$ is the ideal of finite sets and $\mathcal{F}^+$ is the infinite sets.
3. For $\mathcal{F}$ the club filter on $\omega_1$, $\mathcal{I}_{\textrm{Dual}}$ is the ideal of non-stationary sets and $\mathcal{F}^+$ is the stationary sets.
4. For $\mathcal{I}$ is the ideal of meager real subsets, $\mathcal{F}$ is the comeager sets,  $\mathcal{F}^+$ is the non-meager sets.

What’s this got to do with Namba?

Definition. $T$ is a Fat Namba Tree if

1. $T$ is a Namba tree; and
2. The splitting is not just $\omega_2$-splitting, but misses only $\omega_1$ many nodes.

Proposition. If $T_\alpha$, for $\alpha <\omega_1$ are Fat Namba Trees, with the same roots, then $T = \bigcap_{\alpha< \omega_1} T_\alpha$ is a Fat Namba Tree.

This tells us that $\mathcal{F}_s$, the filter generated by $[T]$ (with $T$ Fat Namba having root $s$) is closed under $\omega_1$ intersections.

Also, $\mathcal{F}_s = \{x \sse \omega_2^\omega : \exists T \textrm{ Fat Namba with root }s, [T] \sse X\}$.

For $p$ a Namba Tree with root$(p) = s$, then $[p] \in \mathcal{F}_s^+$, because if $p \cap T = q$, then $q$ is Namba (because $T$ is fat) and $q \leq_0 p$. This is also true for any $s \in A(p)$.

The Namba closed set meets every Fat Namba tree.”

How do things in $\mathcal{F}_s^+$ relate to each other?

Lemma 3. Suppose $\mathcal{A} \in \mathcal{F}_s^+$ with $\mathcal{A} \sse \omega_2^\omega$. THEN for $Q := \{\alpha : \mathcal{A} \in \mathcal{F}_{s^\frown <\alpha>}^+\}$ we have $\vert Q\vert = \omega_2$.

What does the filter technology tell me about extensions?

proof. “This proof starts in a novel way…” Suppose not. Let $\vert Q \vert \leq \omega_1$ so for all $\alpha \in Q^c$ we have $\mathcal{A} \in \mathcal{I}_{s^\frown <\alpha>}$. So there is a Fat Namba tree $T_\alpha$ with root $s$, so $[T_\alpha] \cap \mathcal{A} = \emptyset$.

Note that a fat union of Namba trees is Fat.

[QED]

Lemma 4. Suppose that $\mathcal{A} \sse [p]$ where $\mathcal{A} \in \mathcal{F}_{s_0}^+, s_0 = \textrm{root}(p)$ and $\mathcal{A}$ closed. “I’m going to combine two lemmas.” THEN $\exists q \leq_0 p, [q] \sse \mathcal{A}$.

proof. WLOG, $s_0 = \emptyset$.

Look at $q := \{ s \in p : \mathcal{A} \in \mathcal{F}_t^+, \forall t \sse s\}$.

Now $\mathcal{A} \in \mathcal{F}_t^+$ implies $t \in p$. Also, $q \leq_0 p$.

For all $x \in [q], \forall n, \mathcal{A} \in \mathcal{F}_{x \upharpoonright n}^+$ (it is large). So $\mathcal{A} \cap [x \upharpoonright n] \neq \emptyset$.

In the world of topology, $x \in \mathcal{A}$ because $\mathcal{A}$ is closed. [QED]

## Wrapping Up

Take a look again at the (souped up) Lemma 2.

Under CH,

• $\{x_\alpha : \alpha < \omega_1\} = \omega_1^\omega$;
• $\mathcal{A}_\alpha := \tilde {\pi}^{-1}\{x_\alpha\}$ are closed sets;
• $[p] = \bigcup_{\alpha<\omega_1} \mathcal{A}_\alpha \in \mathcal{F}_s^+$ where $s = \textrm{root}(p)$;
• $\exists \alpha_0$ such that $\mathcal{A}_{\alpha_0} \in \mathcal{F}_s^+$

So by the lemma, $\exists q_0 \leq_0 p$ such that $[q] \sse \mathcal{A}_{\alpha_0}$.

Claim: $q \Vdash \tau = x_{\alpha_0}$

Recall: $\tilde{\pi}([q]) = \{x_{\alpha_0}\}$. So you can’t force it to disagree.

For all $p,\tau$ if $p \Vdash \tau \in \omega_1^\omega$, then $\exists x \in \omega_1^\omega, \exists q \leq p$ such that $q \Vdash \tau = x^\smile$.

So if $M \models$ CH, then $(\omega_1^\omega)^M = (\omega_1^\omega)^{M[G]}$.

Exercise. “Without CH, $\omega_1$ is still not collapsed. There would be too many $\mathcal{A}_\alpha$. You will get an $\omega_2$ sequence of reals.

Theorem (Bukovsky) If $M \models ” 2^\omega = \omega_1$ and $2^{\omega_1} = \omega_2$”, then $M[G] \models “\textrm{cof}(\omega_3^M) = \omega_1″$.

1. Posted October 19, 2012 at 4:56 pm | Permalink

Something is a bit strange in the changing cofinality of a successor part. Are you sure it doesn’t just collapse $\omega_2$? Did you mean change of cofinality as an ordinal, or as a cardinal?

• Micheal Pawliuk
Posted October 21, 2012 at 7:52 pm | Permalink

Yes, the levy collapse must collapse $\omega_2$ and not just its cofinality. In this context we only care about the cofinality, so that was all that was mentioned.

• Posted October 21, 2012 at 8:23 pm | Permalink

I see. I was confused because I was thinking about singular successors very recently… (without choice, obviously)