UCI Summer School, part 7: Sacks forcing (Brent Cody)

This lecture will introduce some basic properties of Sacks forcing for uncountable inaccessible cardinals, and examine an Easton support iteration of such forcing.

The Sacks forcing on $\omega$ adds a real of minimal constructibility degree, and crucially satisfies a fusion property. Although this was reviewed in the summer school, I'm going to omit the discussion for this post.

Instead we will start with Sacks forcing on uncountable cardinals, which traces back to Kanamori (1980), where using $\diamond_\kappa$ it was shown that long products and iterations of $\mathrm{Sacks}(\kappa)$ preserve $\kappa^+$.

Definition: We say $p\subseteq 2^{<\kappa}$ is a perfect $\kappa$-tree if:

1. If $s\in p$ and $t\subseteq s$ then $t\in p$.
2. If $\langle s_\alpha:\alpha<\eta\rangle$ is a sequence of nodes in $p$, then $s=\bigcup_{\alpha<\eta} s_\alpha\in p$.
3. For every $s\in p$ there is $t\supset s$ with $t\frown 0, t\frown 1\in p$.
4. Let $\mathrm{Split}(p)=\{s\in p: s\frown 0, s\frown 1\in p\}$. Then for some unique club $C(p)\subseteq \kappa$, we have $$\mathrm{Split}(p)=\{s\in p: \mathrm{length}(s)\in C(p)\}.$$

$\mathrm{Sacks}(\kappa)$ is the poset of perfect $\kappa$-trees ordered by inclusion. We think of the generic subset of $\kappa$ added by $\mathrm{Sacks}(\kappa)$ as the intersection of the trees in the generic filter.

The only surprising thing in the generalization is (4): splitting happens for every node on certain levels, which form a club in $\kappa$.

Exercise: $\mathrm{Sacks}(\kappa)$ is $<\kappa$-closed.

Assume $\kappa>\omega$ is inaccessible. Then $\mathrm{Sacks}(\kappa)$ is $\kappa^{++}$-c.c. We will really only consider this case.

Definition: $\mathrm{Split}_\alpha(p)$ is the set of all nodes $s\in p$ with $\mathrm{length}(s)=\beta_\alpha$, where $\langle \beta_\alpha:\alpha<\kappa\rangle$ is an enumeration of $C(p)$, i.e., the level of $p$ at the $\alpha$th member of $C(p)$.

For $p,q\in \mathrm{Sacks}(\kappa)$, write $p\le_\beta q$ iff $p\le q$ and $\mathrm{Split}_\alpha(p)=\mathrm{Split}_\alpha(q)$ for all $\alpha<\beta$.

A descending sequence $\langle p_\alpha:\alpha<\kappa\rangle$ in $\mathrm{Sacks}(\kappa)$ is a fusion sequence if for all $\alpha<\kappa$, $p_\alpha\le_\alpha p_\alpha$.

Lemma (fusion lemma): If $\langle p_\alpha:\alpha<\kappa\rangle$ is a fusion sequence, then $p=\bigcap_{\alpha<\kappa} p_\alpha$ is a lower bound in $\mathrm{Sacks}(\kappa)$.

Proof: exercise. Hint: show that any node $p$ in the intersection is in a cofinal branch of the intersection.

This important lemma affords us a kind of $\kappa^+$ closure, with the catch that we require more of our decreasing sequence. We can see this in action in the next lemma.

Lemma: $\mathrm{Sacks}(\kappa)$ preserves $\kappa^+$.

Proof: If $\dot{f}$ is the name of a function $\kappa\rightarrow \kappa^+$, then we will find $q\le p$ with $q\Vdash \mathrm{ran}(\dot{f})$ bounded. 

Let $p_0=p$. Given $p_\alpha$, for each $s\in \mathrm{Split}_\alpha(p_\alpha)$, let $\bar{r}^s_\alpha\le (p_\alpha)_s$ be such that $\bar{r}^s \Vdash \dot{f}(\alpha)=\eta^s_\alpha$. Here the $(p)_s$ means the subtree of $p$ of nodes compatible with $s$. 

Note $\bigcup\{\bar{r}^s_\alpha:s\in \mathrm{Split}_\alpha(p_\alpha)\}$ might not be a condition by the requirement on splitting levels. Let $C=\bigcap \{C(\bar{r}^s_\alpha:s\in \mathrm{Split}_\alpha(p_\alpha)\}$ and thin each $\bar{r}^s_\alpha$ to some $r^s_\alpha\le \bar{r}^s_\alpha$ with $C(r^s_\alpha)=C$.

At limits $\gamma<\kappa$, let $p_\gamma=\bigcap_{\alpha<\gamma} p_\alpha$ by the fusion lemma. This defines a fusion sequence where the limit forces that the range of $f$ is bounded.

Exercise: Suppose ${}^\kappa M\subseteq M$, for an inner model $M$. Suppose $\mathrm{Sacks}(\kappa)\in M\subseteq V$. If $G$ is $V$-generic for $\mathrm{Sacks}(\kappa)$ then ${}^\kappa M[G]\subseteq M[G]$ in $V[G]$.

Note: This holds for $\kappa^+$-c.c. forcing, but $\mathrm{Sacks}(\kappa)$ is not $\kappa^+$-c.c.

Now we will see what happens when we iterate these Sacks forcings with Easton support below, and at, a measurable cardinal $\kappa$. Think of this like a Sacks forcing version of the Kunen-Paris iteration, where we use the nice fusion property to replace the $\gamma^+$ closure of the factors there.

Theorem (Friedman-Thompson 2008): Assume GCH holds. Suppose $\kappa$ is measurable and let $\mathbb{P}$ be the length $\kappa+1$ Easton support iteration with $\mathbb{Q}_\gamma=\mathrm{Sacks}(\gamma)$ (computed in $V^{\mathbb{P}_\gamma}$) for $\gamma\le \kappa$ inaccessible, and $\mathbb{Q}_\gamma$ is trivial forcing otherwise. Then if $G\ast H$ is $V$-generic for $\mathbb{P}= \mathbb{P}_\kappa\ast \dot{\mathbb{Q}}_\kappa$, then every normal ultrapower lifts to $V[G\ast H]$ (and in a particularly interesting way!)

Proof: Let $j:V\rightarrow M$ be a normal ultrapower by $U\in V$. Then $j(\mathbb{P}_\kappa=\mathbb{P}_\kappa\ast \dot{\mathbb{Q}}_\kappa\ast \dot{\mathbb{P}}_{\kappa+1,j(\kappa)}$. We get the actual $\dot{\mathbb{Q}}_\kappa$ factor at the $\kappa$ step by using the $\kappa$ closure of the ultrapower.

Using this closure further, and the last exercise, ${}^\kappa M[G\ast H]\subseteq M[G\ast H]$ in $V[G\ast H]$, so $M[G\ast H] \vDash \dot{\mathbb{P}}_{\kappa+1,j(\kappa)}\textrm{ is }\le \kappa-\textrm{closed.}$ So there are $\kappa^+$ maximal antichains of $\mathbb{P}_{\kappa,j(\kappa)}$ in $M[G][H]$. We can now build as usual a generic $G_{\kappa+1,j(\kappa)}\in V[G\ast H]$ for $\mathbb{P}_{\kappa+1,j(\kappa)}$ over $M[G\ast H]$. Lift to $j:V[G]\rightarrow M[j(G)]$. 

Now we have to lift $j$ through $\mathbb{Q}_\kappa=\mathrm{Sacks}(\kappa)$. Using the Silver method, $j``H$ has size $\kappa^+$, but the target model $M[j(G)]$ does not have this much closure.

The crucial point is to just take $t:=\bigcap j``H$. We claim that $t$ is a "tuning fork": by this we mean that $t$ consists of a single branch up to the level $\kappa$, at which point it splits into two branch which are cofinal (and that's everything in $t$).

1. The function $f:\kappa\rightarrow 2$ determined by $H$ is in $t$, and this is everything in $t$ below $\kappa$.
2. Every condition in $j``H$ splits at $\kappa$ since for each $p\in H$, $p$ splits at club many levels below $\kappa$, and therefore $j(p)$ splits at level $\kappa$. Therefore, $f\frown 0,f\frown 1\in t$.
3. Since $H$ is a filter, $t$ is cofinal in $j(\kappa)$.
4. We will argue that $t$ does not split anywhere else. Given a club $C\subseteq \kappa$, $D_C:=\{p\in \mathrm{Sacks}(\kappa): C(p)\subseteq C\}$ is dense. So there must be $p_C\in H$ so that $C(p_C)\subseteq C$. Now we have:

Claim: $X=\bigcap \{j(C):C\subseteq \kappa \textrm{ club in } V[G]\}=\{\kappa\}$.

Proof of Claim: Clearly $\kappa\in X$. For the other inclusion, suppose $\alpha\in X$, $\alpha>\kappa$. Then choose $f:\kappa\rightarrow \kappa$, $f\in V[G]$ so that $j(f)(\kappa)=\alpha$. Then let $C_f=\{\nu<\kappa: f``\nu\subseteq \nu\}$ is club, but $\alpha\not\in j(C_f)$ since $\alpha$ is not a closure point of $j(f)$ ($\kappa<\alpha$ maps to $\alpha$). This proves the claim.

Let $t_0, t_1$ be the leftmost and rightmost branches through $t$, respectively. Let $K_0=\{p\in j(\mathbb{Q}_\kappa):t_0\subseteq p\}$. Clearly $j``H\subseteq K_0$.

It remains to show that $K_0$ is $M[j(G)]$-generic for $j(\mathbb{Q}_\kappa)$. Let $D$ be a dense open subset of $j(\mathbb{Q}_\kappa$ in $M[j(G)]$. Then there is a sequence $\vec{D}=\langle D_\alpha:\alpha<\kappa\rangle \in V[G]$ such that $j(\vec{D})_\kappa=D$, where each $D_\alpha$ is a dense open subset of $\mathbb{Q}_\kappa$. 

Claim: Every condition $p\in \mathrm{Sacks}(\kappa)$ can be extended to $q_\infty \le p$ so that for every $\alpha<\kappa$ there is $\beta<\kappa$ so that for any node $s\in \mathrm{Split}_\beta(q_\infty)$, the condition $(q_\infty)_s$ meets $D_\alpha$. 

Proof of Claim: exercise, a fusion argument.

Let $q_\infty\in H$ be as in the claim, using genericity of $H$. By elementarity, $j(q_\infty)$ has the property that at some splitting level of $j(q_\infty)$, say $\beta<j(\kappa)$, any node $s\in \mathrm{Split}_\beta(j(q_\infty))$ is such that $(j(q_\infty))_s$ meets $D$. Now we can just take $s$ to be $t_0\upharpoonright \delta_\beta$, where $\delta_\beta$ is the $\beta$th splitting level of $j(q_\infty)$.

Therefore $K_0$ is generic as claimed, and it is in $V[G\ast H]$, so $j$ lifts to

$$j:V[G\ast H]\rightarrow M[j(G)\ast j(H)].$$ 

The rearrangement inequality is everywhere

Recently, I've been talking with John Susice about some elementary Olympiad-style problems. I told him that in high school I was taught that virtually every inequality problem that appeared in this setting follows from the Rearrangement Inequality. This states that if $x_1\le \cdots\le x_n$ and $y_1\le \cdots\le y_n$ are real numbers, then the expression
$$x_1 y_{\sigma(1)}+\cdots+x_n y_{\sigma(n)}$$
for $\sigma$ a permutation on $[n]$ is maximized when $\sigma$ is the identity permutation and minimized when $\sigma$ is the reversing permutation. To me, this neatly isolates a useful and general principle that seems kind of obvious in hindsight.

The proof of the $n=2$ case of the inequality, which is all I'll need below, is just to expand $(x_2-x_1)(y_2-y_1)\ge 0$. This case also implies the AM-GM inequality (taking $x_1=y_1=\sqrt{a}$ and  $x_2=y_2=\sqrt{b}$).

Now sometime later, John told me an interesting problem about factoring numbers. He had a solution using a trick similar to Euler's product, but for me this was a chance to test my thesis about the rearrangement inequality:

Problem: Prove that every natural number $n$ has at least as many factors congruent to 1 (mod 4) as factors congruent to 3 (mod 4).

Solution: By induction. Clearly this holds for 1 and for primes. Now suppose $n$ is composite, so write $n=pq$ where $p,q<n$. For any integer $k$, let $r_k$ denote the number of factors it has which are congruent to 1 (mod 4) and $s_k$ the number of factors congruent to 3 (mod 4).

The product of two 1 (mod 4) numbers is still 1 (mod 4), and the product of two 3 (mod 4) numbers is 3 (mod 4). On the other hand, the product of a 1 (mod 4) and a 3 (mod 4) is a 3 (mod 4), and even factors can never be 1 or 3 (mod 4). This proves that $r_n=r_pr_q+s_ps_q$, which must be greater than or equal to $s_n=r_ps_q+r_qs_p$ by the rearrangement inequality!

UCI Summer School, part 6: the number of normal measures (Brent Cody)

Sorry for the delay, loyal readers. Here is the beginning of Brent's part of the summer school.

Assume that is consistent that there is a measurable cardinal.

Question: How many normal measures can a measurable cardinal carry?

Let $\mathrm{NM}(\kappa)$ denote the set of normal measures on $\kappa$. Trivially, $\#\mathrm{NM}(\kappa)\le 2^{2^\kappa}$.

One interesting case happens in a canonical inner model.

Theorem (Kunen '71): In $L[U]$, there is exactly one normal measure (on $\kappa$, the unique measurable cardinal).

We can also realize the other extreme:

Theorem (Kunen-Paris '71): There is a forcing extension in which $\#\mathrm{NM}(\kappa)=2^{2^\kappa}$.

We will prove this one later today. In the middle, we have

Theorem (Mitchell '74): It is consistent relative to a measurable $\kappa$ of order $\delta$ that $\#\mathrm{NM}(\kappa)=2^{2^\kappa}$.

And for one case, we can lower the large cardinal assumption used.

Theorem (Apter-Cummings-Hamkins '07): It is consistent relative to a measurable cardinal that that $\#\mathrm{NM}(\kappa)=\kappa^+$.

Finally, we can do it in all cases.

Theorem (Friedman-Magidor '09): Assume GCH. Suppose $\kappa$ is measurable and let $\mu\le \kappa^{++}$ be a cardinal. Then in a cofinality-preserving forcing extension, $\#\mathrm{NM}(\kappa)=\mu$.

The goal eventually will be to show the proof of this result.

Lemma: Suppose $j:V\rightarrow M$ is the ultrapower by a normal measure on $\kappa$. Let $G$ be $V$-generic for $\mathbb{P}$. Assume that in $V[G]$,
$$j_0:V[G]\rightarrow M[j_0(G)]$$
and
$$j_1:V[G]\rightarrow M[j_1(G)]$$
are elementary embeddings extending $j$. Then the following are equivalent:
1) $j_0=j_1$.
2) $j_0(G)=j_1(G)$.
3) The normal measure $U_0$ derived from $j_0$ is equal to the normal measure $U_1$ derived from $j_1$.

Proof: Exercise.

Exercise (Levy-Solovay): Show that every normal measure extends uniquely to a normal measure in any forcing extension by small forcing.

As promised, we will now prove the Kunen-Paris Theorem.

Theorem (Kunen-Paris '71): There is a forcing extension in which $\#\mathrm{NM}(\kappa)=2^{2^\kappa}$.

Proof: By a preparation forcing if necessary, assume that $2^\kappa=\kappa^+$. Let $\mathbb{P}$ be the length $\kappa+1$ Easton support iteration that forces at cardinal stages $\gamma\le \kappa$ with $\mathbb{Q}_\gamma:=\mathrm{Add}(\gamma^+,1)$ (computed in the extension by $\mathbb{P}_\gamma$), trivial forcing at other stages. This is a standard way of forcing the GCH to hold below $\kappa$.

Let $G\ast H$ be $V$-generic for $\mathbb{P}=\mathbb{P}_\kappa\ast \dot{\mathbb{Q}}_\kappa$. Let $j:V\rightarrow M$ be the ultrapower by a normal measure. Factor $j(\mathbb{P}_\kappa)\simeq \mathbb{P}_\kappa \ast \dot{\mathbb{P}}_{\kappa,j(\kappa)}$.

Since ${}^\kappa M[G]\subseteq M[G]$ in $V[G]$ (a name exercise which uses closure under $\kappa$-sequences of $M$ in $V$), we have that in $M[G]$, $\mathbb{P}_{\kappa,j(\kappa)}$ is $\le \kappa$-closed (it is the composition of increasingly closed posets starting with $\mathbb{Q}_\kappa=\mathrm{Add}(\kappa^+,1)$.

Furthermore, computing in $V[G]$, $\mathbb{P}_{\kappa,j(\kappa)}$ has at most $\kappa^+$ maximal antichains inside $M[G]$.  since $|j(\kappa)|=\kappa^+$ by our cardinal arithmetic assumption, and $\mathbb{P}_{\kappa,j(\kappa)}$ is $j(\kappa)$-c.c. of size $j(\kappa)$ in $M[G]$. By enumerating the maximal antichains of $M[G]$ in order-type $\kappa^+$, we can meet them one by one, with the closure of the poset in $M[G]$, noticing that closure of the model $M[G]$ gives that the proper initial segments of this enumeration are in $M[G]$.

Since $j``G\subseteq G\ast G_{\kappa,j(\kappa)}$, we can lift the embedding to
$$j^+:V[G]\rightarrow M[j(G)].$$
Using the Lemma/Exercise above, different choices of $j(G)$ will give rise to different embeddings which will give different normal measures. By passing to finer antichains, we can assume in the enumeration $\langle A_\alpha:\alpha<\kappa^+\rangle$ that if $i<j$, then $A_j$ (strictly) refines $A_i$. Looking at how we built $G_{\kappa,j(\kappa)}$, we can form the tree of attempts to build the generic, noticing that at each level there are incompatible ways to extend the generic so far to meet the maximal antichain. This gives $2^{\kappa^+}=2^{2^\kappa}$ many different generics, and hence different normal measures. $\Box$

So it's not hard to force many normal measures. It is harder to force so that there are few normal measures.

We will need to use Hamkins's Gap Forcing Theorem which gives a sufficient condition for an ultrapower embedding in a generic extension to be the lift of a ground model embedding.

Definition: A forcing $\mathbb{P}$ admits a closure point at $\delta$ if it factors as $\mathbb{P}\simeq \mathbb{Q}\ast \dot{\mathbb{R}}$ where $\mathbb{Q}$ is nontrivial, $|\mathbb{Q}|\le \delta$, and $\Vdash_{\mathbb{Q}} \mathbb{R} \textrm{ is }<\delta-\textrm{closed}$.

Theorem (Hamkins '01, Gap Forcing Theorem): If $V\subseteq V[G]$ admits a closure point at $\delta$ and $j:V[G]\rightarrow M[j(G)]$ is an ultrapower in $V[G]$ with $\mathrm{crit}(j)>\delta$, then $j\upharpoonright V:V\rightarrow M$ is a definable class in $V$.

Finally, we prove one more of the theorems in the introduction.

Theorem (Apter-Cummings-Hamkins '07): It is consistent relative to a measurable cardinal that that $\#\mathrm{NM}(\kappa)=\kappa^+$.

Idea of the proof: Start with at least $\kappa^+$ normal measures on $\kappa$ (e.g., by Kunen-Paris forcing). Force with $\mathrm{Col}(\kappa^+,2^{2^\kappa})$, and show that no new normal measures are added.

Proof: Again we assume $2^\kappa=\kappa^+$. Start with $\#\mathrm{NM}(\kappa)\ge \kappa^+$. Let $\mathbb{P}=\mathrm{Add}(\omega,1)\ast \dot{\mathrm{Col}}(\kappa^+, 2^{2^\kappa})$. The point of the Cohen forcing is to give  $\mathbb{P}$ a closure point below $\kappa$. Suppose $c\ast G\subseteq \mathbb{P}$ is $V$-generic. Then every normal measure in $V$ generates a normal measure in $V[c]$ by Levy-Solovay, and these remain normal measures in $V[c][G]$ since $\mathrm{Col}(\kappa^+, 2^{2^\kappa})$ is $\le \kappa$-closed in $V[c]$.  So there are at least $\kappa^+$ normal measures in $V[c][G]$.

To show the other inequality, suppose $U$ is a normal measure on $\kappa$ in $V[c][G]$. Let
$$j:V[c][G]\rightarrow M[c][j(G)]$$
be the ultrapower. By the gap forcing theorem, $j\upharpoonright V:V\rightarrow M$ is a definable class in $V$. So we can lift to $j\upharpoonright V[c]:V[c]\rightarrow M[c]$. We can now define $U$ in $V[c]$, as the derived measure from this embedding. $\Box$