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:

- If $s\in p$ and $t\subseteq s$ then $t\in p$.
- If $\langle s_\alpha:\alpha<\eta\rangle$ is a sequence of nodes in $p$, then $s=\bigcup_{\alpha<\eta} s_\alpha\in p$.
- For every $s\in p$ there is $t\supset s$ with $t\frown 0, t\frown 1\in p$.
- 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 $t$).

- The function $f:\kappa\rightarrow 2$ determined by $H$ is in $t$, and this is everything in $t$ below $\kappa$.
- 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, $f\frown 0,f\frown 1\in t$.
- Since $H$ is a filter, $t$ is cofinal in $j(\kappa)$.
- 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

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

## No comments:

## Post a Comment