## Tuesday, September 1, 2015

### UCI Summer School, part 5 (Monroe Eskew)

This is just a placeholder, for now. My notes for this part are quite rough, so it will be a while before I will try to record it here. The next installment of these notes will cover Brent Cody's lectures on some results about the number of normal measures.

### UCI Summer School, part 4: Measure algebras (Monroe Eskew)

Here are some more applications of the ideas we have been considering.

Definition: $\mathcal{B}$ is a measure algebra if it is a complete Boolean algebra equipped with some function $\mu:\mathcal{B}\rightarrow [0,1]$ with $\mu(0)=0, \mu(1)=1, \mu(b)>0$ for $b\neq 0$, and $\mu$ countably additive (i.e., if $\langle b_i:i<\omega\rangle$ is an antichain, then $\mu(\sum b_i)=\sum \mu(b_i)$.

Exercise: All measure algebras are c.c.c.

Example: Let $\kappa$ be a cardinal. We will describe a topology on ${}^\kappa 2$. Fix some $x\in [\kappa]^{<\omega}$ and $s:x\rightarrow 2$ (i.e., $s$ is a finite domain partial function from $\kappa$ to $2$). Then basic open sets are of the form $\mathcal{O}_s=\{r\in {}^\kappa 2: \forall \alpha\in x(r(\alpha)=s(\alpha))\}$. Let $\mathcal{B}_\kappa$ be the $\sigma$-algebra generated by these basic open sets, and define $\mu$ on $\mathcal{B}_\kappa$ by setting $\mu(\mathcal{O}_s=\frac{1}{2^{|s|}}$ (standard theorems from real analysis give that $\mu$ extends uniquely to a countably additive probability measure on $\mathcal{B}_\kappa$.

Define $\mathrm{Null}=\{A\in\mathcal{B}_\kappa: \mu(A)=0\}$. Then  $\mathcal{R}_\kappa:=\mathcal{B}_\kappa/\mathrm{Null}$ is a measure algebra.

Exercise: Prove that $\mathcal{R}_\kappa$ forces $2^\omega\ge \kappa$.

Exercise: If $\mathcal{A}$ is a measure algebra and $\Vdash_A \dot{\mathcal{B}}$ is a measure algebra, then $\mathrm{r.o.}(\mathcal{A}\ast \dot{\mathcal{B}})$ is a measure algebra. (Note: this is not as easy as it may seem at first since for example $\mathcal{A}$ might even add new reals which can be measures of elements of $\mathcal{B}$! We use r.o. for the Boolean completion here since the letter $\mathcal{B}$ is overloaded).

Exercise: If $\mathcal{A}$ is a complete sublagebra of a measure algebra $\mathcal{B}$, then $\mathcal{A}$ is a measure algebra.

Continuing along this line,

Theorem: If $\mathcal{B}$ is a measure algebra, $\mathcal{A}$ a complete subalgebra of  $\mathcal{B}$, and $G\subseteq \mathcal{A}$ is generic over $V$, then in $V[G]$ we have that $\mathcal{B}/G$ is a measure algebra.

Note that in $V[G]$, $G$ is a filter on $\mathcal{B}$, so $\mathcal{B}/G=\{[b]_G:b\in \mathcal{B}\}$. We use $G^*$ for the dual ideal. It's important to distinguish between the orderings of the two Boolean algebras here, and will be good to see how to translate between them using the forcing relation.

Lemma: If $\mathcal{B}$ is complete and $\mathcal{A}$ is a complete subalgebra and $G\subseteq \mathcal{A}$ is generic, then $\mathcal{B}/G$ is complete in $V[G]$.

Proof of Lemma: Suppose $\langle [b_\alpha]_G:\alpha<\kappa\rangle \in P(\mathcal{B}/G)\cap V[G]$. For each $\alpha<\kappa$, let  $X_\alpha:=\{b:1\Vdash_{\mathcal{A}} [b]_{\dot{G}}\le [\dot{b}_\alpha]_{\dot{G}}\}$. Let $c_\alpha=\sum X_\alpha\in V$ (meet taken in $\mathcal{B}$).

We claim that $1\Vdash_{\mathcal{A}} [c_\alpha]_{\dot{G}}=[\dot{b}_\alpha]_{\dot{G}}$ for each $\alpha$--this suffices to prove the lemma. The proof of the claim is straightforward but a little tedious. First we show $1\Vdash_{\mathcal{A}} [c_\alpha]_{\dot{G}}\le[\dot{b}_\alpha]_{\dot{G}}$. If this doesn't hold, then there are $d,p$ so that $p\in \mathcal{A}$, $d\wedge p\neq 0$, and
$$p\Vdash [\check{d}]\le [\check{c}_\alpha] \textrm{ and }[\check{d}]\wedge [\dot{b}_\alpha]=0.$$
The first conjunct implies that $p\wedge d\le c_\alpha$ in $\mathcal{B}$, and since $c_\alpha$ is a lub for $X_\alpha$, there is some $b\in X_\alpha$ so that $p\wedge d\wedge b\neq 0$. So $1\Vdash [p\wedge d\wedge b]\le [b_\alpha]$ by the definition of $b\in X_\alpha$. But the second conjunct gives $p\wedge d\wedge b_\alpha=0$, contradiction.

Now to show $1\Vdash_{\mathcal{A}} [c_\alpha]_{\dot{G}}\ge[\dot{b}_\alpha]_{\dot{G}}$, assume for a contradiction that there are $p,a\in \mathcal{A}$ so that $p\Vdash [\check{a}]\le [\dot{b}_\alpha]$ and $p\Vdash [a\wedge \neg c_\alpha]\neq 0$. Now $p$ forces $[p\wedge a \wedge \neg c_\alpha]\le [\dot{b}_\alpha]$. Trivially, $\neg p$ forces $[p\wedge a \wedge \neg c_\alpha]=0$. So it's just outright forced that $[p\wedge a \wedge \neg c_\alpha]\le [b_\alpha]$ and thus $p\wedge a\wedge \neg c_\alpha \in X_\alpha$, which contradicts $c_\alpha$ is an upper bound for $X_\alpha$, completing the proof of the claim and the lemma. $\Box$

Proof of Theorem: Let $\mu$ be a measure on $\mathcal{B}$, $\mathcal{A}$ a complete subalgebra of $\mathcal{B}$. Define
$$\mu(b\mid a)=\frac{\mu(a \wedge b)}{\mu(a)}.$$
(We say the measure of $b$ conditioned on $a$).

Definition: For $a\in A,b\in B, \epsilon>0$, say $a$ is $\epsilon$-stable for $b$ if for all $x\le a$ in $\mathcal{A}$, $|\mu(b\mid x)-\mu(b\mid a)|<\epsilon$.

Lemma: For all $b\in B$ and for all $\epsilon>0$ the set $\{a\in A:a \textrm{ is }\epsilon-\textrm{stable for} b\}$ is dense in $A$.
Proof: Exercise. An interesting one.

In $V[G]$, let $\nu:\mathcal{B}/G\rightarrow [0,1]$ be given by $\nu([b])=r$ if for every $\epsilon>0$ there is some $a\in G$ so that $a$ is $\epsilon$-stable for $b$ and $|\mu(b\mid a)-r|<\epsilon$. The idea is that $G$ could add new reals, so we can only have approximations to the measure of $[b]$ using ground model reals attached to the members of $\mathcal{A}$.

We can check that this is well-defined: if $[b]_G=[c]_G$, then some $a\in G$ forces $b\Delta c\in G^*$. This means that $a\perp (b\Delta c)$, so $\mu(a\wedge (b\Delta c)=0$. Therefore $\mu(b\mid x)=\mu(c\mid x)$ for all $x\le a$. Suppose $r_0\neq r_1$ both satisfy $\nu(b)=r_i$. Take $\epsilon<|r_1-r_0|$. Let $a_0,a_1\in G$ be $\epsilon/4$-stable for $b$ with $|\mu(b\mid a_i)-r_i|<\epsilon/4$. Now take $a\le a_0, a_1$ in $\mathcal{A}$. By a triangle inequality argument, we have $|r_1-r_0|<\epsilon$, a contradiction.

Exercise: Check that $\nu(b)>0$ for all $b\neq_G 0$.

Exercise: Check that $\nu$ is countably additive. First prove that it is finitely additive.
$\Box$.

Now suppose $P(Z)/I$ is a measure algebra. The duality theorem (ccc case) says that for any $\theta$, $\mathcal{R}_\theta\ast P(Z)/\bar{I}\cong P(Z)/I\ast j(\mathcal{R}_\theta)$,
where $j:V\rightarrow M$ is the generic embedding in $V[G]$, $G$ generic for $P(Z)/I$.

The right hand side of this isomorphism is a measure algebra, since $P(Z)/I$ is a measure algebra by assumption, and $j(\mathcal{R}_\theta)$ is a measure algebra of $M$, a model which is closed under countable sequences (and so has all the countable sequences to witness countable additivity and completeness of the Boolean algebra).

We have a map $e:\mathcal{R}_\theta\rightarrow \mathrm{r.o.}(P(Z)/I\ast j(\mathcal{R}_\theta)$, so $\mathcal{R}_\theta$ is isomorphic to a complete subalgebra of the right hand side. Now if $H\subseteq \mathcal{R}_\theta$ is generic, then $B/e''H$ is a measure algebra. Therefore $P(Z)/\bar{I}$ is also a measure algebra.

A real-valued measurable cardinal is a cardinal $\kappa$ which carries a $\kappa$-additive probability measure on all subsets of $\kappa$ which gives measure 0 to singletons. It is atomless if every set of positive measure has a subset of strictly smaller positive measure.

Corollary: If $\langle \kappa_i:i<\theta\rangle$ is a sequence of measurable cardinals, then if $\gamma=\sup \kappa_i$, $\mathcal{R}_\gamma$ forces all $\kappa_i$ to be atomless real-valued measurable cardinals (RVMs).

We note the fact that if $\kappa$ is atomlessly RVM, then $2^\omega\ge \kappa$, so we can't get class many RVMs.

However, if $\kappa$ is strongly compact, then $\mathcal{R}_\kappa$ forces that for all regular $\lambda\ge \kappa$, there is a countably additive real-valued probably measure $\mu_\lambda$ on $\lambda$ giving measure 0 to all subsets of size $<\lambda$.

## Friday, August 7, 2015

### UCI Summer School, part 3: Applications of Duality Theorem (Monroe Eskew)

We now turn towards applications of the duality theorem. It is recommended that the reader recalls the notation ($I,j,K,J,\hat{H},e,\iota$, etc.) from the previous lecture before proceeding.

The basic idea is that one uses the isomorphism there:
$$\mathcal{B}(\dot{\mathbb{P}\ast P(Z)/J})\equiv \mathcal{B}(P(Z)/I\ast j(\mathbb{P})/\dot{K})$$
to calculate the quotient algebra $P(Z)/J$ as $\mathcal{B}(P(Z)/I\ast j(\mathbb{P})/\dot{K})/e''H$.

As discussed near the end of the last lecture, under certain assumptions, the statement of the duality theorem becomes somewhat simpler. The first examples will fall into this case.

Special Case: If $I$ is $\kappa$-complete and $\mathbb{P}$ is $\kappa$-c.c., then the hypothesis of the duality theorem holds, $K=\{0\}$ and $J=\bar{I}$, the ideal generated by $I$ in $V^\mathbb{P}$.

Exercise: Show that in the above case, if $p\in \mathbb{P}$ and $A\in (P(Z)/I)\cap V$, then
$$\iota(p,\check{A})=(A,j(\dot{p})).$$

A further simplification will be that we will usually start with a measurable cardinal $\kappa$ and take $I$ to be  the dual to the measure on $\kappa$, so $P(\kappa)/I$ is the trivial Boolean algebra.

A measurable cardinal $\kappa$ has a 2-saturated, $\kappa$-complete ideal, namely the dual to the measure on $\kappa$, and under GCH every cardinal $\kappa$ carries a $\kappa^{++}$-saturated, $\kappa$-complete ideal, namely the ideal of bounded subsets. This motivates the following natural questions, which are the main focus of this lecture:

Question: Suppose $\mu\le \kappa^+$ is a regular cardinal. Is it consistent that there is a cardinal $\kappa$ which is not measurable, but still $\kappa$ carries a $\mu$-saturated, $\kappa$-complete ideal? (Here we want the amount of saturation to be exactly $\mu$).

Digression: does the answer change if we require $\kappa$ to be a successor cardinal?

For the case where $\kappa$ is a successor cardinal, $\kappa^+$-saturation is the strongest we can hope to achieve.

Exercise: Prove using the method of generic ultrapowers that if $\kappa$ is a successor cardinal then there is no $\kappa$-complete, $\kappa$-saturated ideal on $\kappa$.

Kunen showed that if $\kappa$ is a successor cardinal, then getting a $\kappa^+$-saturated ideal on $\kappa$ requires large cardinals much stronger than a measurable, although we won't do this argument here (you can find it in this previous Specinar post. We now return to the original question.

Answer to question 1, if $\mu<\kappa$ ($\mu$ regular): We will use the basic technique of computing the quotient algebra $P(\kappa)/J$ in $V[H]$ using the duality theorem. Start with $\kappa$ measurable in the ground model. Let $\theta\ge \kappa$, and consider $\mathrm{Add}(\mu,\theta)$ which adds $\theta$ Cohen subsets of $\mu$. $\mathrm{Add}(\mu,\theta)$ is $\kappa$-c.c., and under the GCH it is even $\mu^+$-c.c. Let $I=U^*$, where $U$ is a $\kappa$-complete normal ultrafilter on $\kappa$ (here the star means taking the dual ideal). Let $j:V\rightarrow M$ be the ultrapower embedding.

The duality theorem gives the isomorphism:
$$\mathrm{Add}(\mu,\theta)\ast P(\kappa)/\bar{I}\cong P(\kappa)/I\ast \mathrm{Add}(\mu,j(\theta))\cong\mathrm{Add}(\mu,j(\theta)),$$
since $P(\kappa)/I$ is trivial.

If $H$ is generic for $\mathrm{Add}(\mu,\theta)$ over $V$, then
$$e''H=\{(1,j(p)):p\in H\}.$$
So
$$P(\kappa)/\bar{I}\cong \mathrm{Add}(\mu,j(\theta))/e''H\cong \mathrm{Add}(\mu,j(\theta)).$$
In $V[H]$, $P(\kappa)/\bar{I}\cong \mathcal{B}(\mathrm{Add}(\mu,j(\theta))$, so $\bar{I}$ is $(\mu^{<\mu})^+$ saturated. Furthermore, it is easy to check that $\bar{I}$ is $\kappa$-complete, and $2^\mu\ge \kappa$ in $V[H]$, so $\kappa$ is not measurable. This answers question 1 for the case where $\mu<\kappa$. $\Box$

We might ask what large cardinal properties of $\kappa$ are implied by this ideal hypothesis.

Proposition: If $\kappa$ carries a $\kappa$-complete $\mu$-saturated ideal for some $\mu<\kappa$, then:

1. $\kappa$ is weakly Mahlo
2. $\kappa$ has the tree property.
Exercise: Prove (1) of the proposition using generic ultrapowers.

Proof of Proposition (2): Suppose $T$ is a $\kappa$-tree. If $G\subseteq P(\kappa)/I$ is generic, then in $V[G]$, $T$ has a branch $b$ given by taking any member of level $\kappa$ of $j(T)$, where $j$ is the generic ultrapower embedding. Now for each $\alpha<\kappa$, $S_\alpha=\{x\in T_\alpha: \exists p(p\Vdash \check{x}\in \dot{b})\}<\mu$ by the saturation. Now $\bigcap_{\alpha<\kappa} S_\alpha$ is a $\kappa$-tree all of whose levels have size $<\mu<\kappa$. It is well-known (or a good exercise) that such trees have cofinal branches. $\Box$

Definition: An ideal $I$ is nowhere prime if there is no $I$-positive set $A$ so that $I\upharpoonright A$ is prime (i.e., dual to an ultrafilter).

Exercise: Show that if there is a nowhere prime, $\kappa$-complete, $\mu^+$-saturated ideal, where $\mu<\kappa$, then $2^\mu\ge \kappa$.

We continue with Question 1 with other arrangements of $\mu$ and $\kappa$.

Answer to question 1, if $\mu=\kappa^+$: Start with $\kappa$ measurable with $2^\kappa=\kappa^+$, $U$ a normal ultrafilter and $j:V\rightarrow M$ the ultrapower embedding. Let $\langle \mathbb{P}_\alpha,\dot{\mathbb{Q}}_\alpha:\alpha<\kappa\rangle$ be the Easton support iteration where for regular $\alpha$, $\Vdash_{\mathbb{P}_\alpha} \dot{Q}=\mathrm{Add}(\alpha,1)$. It is straightforward to verify that $\mathbb{P}_\kappa$ has the $\kappa$-c.c., so we are in the special case again.

Note that $j(\mathbb{P}_\kappa)=\mathbb{P}_\kappa\ast \mathbb{Add}(\alpha,1)\ast (\mathbb{P}_{\kappa+1,j(\kappa)})^M$ (We use the notation $\mathbb{P}_{\xi,j(\kappa)}$ for $j(\mathbb{P}_\kappa)_{\xi,j(\kappa)}$). The tail part is computed differently in $M$ than in $V$, e.g., the support is on $M$-regular cardinals.

If $G_\kappa\subseteq \mathbb{P}_\kappa$ is generic over $V$, then the special case of the duality theorem says that  in $V[G_\kappa]$, $P(\kappa)/\bar{I}\cong (\mathbb{P}_{\kappa,j(\kappa)})^M$. However, $P(\kappa)/\bar{I}\cong(\mathbb{P}_{\kappa,j(\kappa)})^M$ does not have the $\kappa^+$-c.c. since there are $M$-regular cardinals between $\kappa^+$ and $j(\kappa)$. So $\bar{I}$ is not $\kappa^+$-saturated.

Now we could also satisfy the hypothesis of the duality theorem of adding a $j(\mathbb{P}_\kappa)=P(\kappa)/J\cong \mathrm{Add}(\alpha,1)\ast (\mathbb{P}_{\kappa+1,j(\kappa)})^M$ generic filter $\hat{H}$ over $M$ in a different way. By a standard technique, a $j(\mathbb{P})$-generic over $M$ exists in $V[G_{\kappa+1}]$ (where $G_{\kappa+1}$ is $\mathbb{P}_\kappa\ast \mathrm{Add}(\alpha,1)$-generic). This is because we clearly get a generic for the initial part $\mathbb{P}_\kappa\ast \mathrm{Add}(\alpha,1)$, just $G_{\kappa+1}$ itself. For the tail, $(\mathbb{P}_{\kappa+1,j(\kappa)})^{M[G_{\kappa+1}]}$ is is $j(\kappa)$-c.c. of size $j(\kappa)$ in $M[G_{\kappa+1}]$, so $M[G_{\kappa+1}]$ thinks the poset has at most $j(\kappa)$ maximal antichains. In $V[G_{\kappa+1}]$, $|j(\kappa)|=\kappa^+$, and the poset is $\kappa^+$-closed (in $M[G_{\kappa+1}]$, but also in $V[G_{\kappa+1}]$ by the agreement between these models). So we can construct a generic by hand in $V[G_{\kappa+1}]$ by enumerating all of the maximal antichains in $M[G_{\kappa+1}]$. This completes the construction of $\hat{H}$ in the extension by $\mathrm{Add}(\kappa,1)$.

In this construction, we have that $j(\mathbb{P})/K\cong \mathbb{P}_\kappa\ast \mathrm{Add}(\kappa,1)$, since the Boolean algebra homomorphism $j(\mathbb{P}_\kappa)\rightarrow \mathbb{P}_\kappa\ast \mathrm{Add}(\kappa,1)$ given by $p\mapsto \|p\in \hat{H}\|$.  has kernel exactly $K$ as defined in the last lecture, and the map is surjective since the codomain completely embeds into the domain. So in the duality theorem calculation, we obtain an ideal $J$ so that $P(\kappa)/J\cong \mathrm{Add}(\kappa,1)$. So $J$ is a $\kappa^+$-saturated ideal on $\kappa$. $\Box$

Note that $\kappa$ is inaccessible.

Exercise: Prove that $\kappa$ is weakly compact in $V[G_\kappa]$. (Hint: use the tree property characterization.)

Exercise: Prove that $\kappa$ is not measurable in $V[G_\kappa]$, but it is measurable in $V[G_{\kappa+1}]$.

Remark: By forcing with $(\mathbb{P}_{\kappa,j(\kappa)})^{M[G_{\kappa+1}]}$ instead of just $\mathrm{Add}(\kappa,1)$ to add the  $j(\mathbb{P})$-generic, we can get a nowhere prime $\kappa$ complete $\kappa^+$-saturated ideal on $\kappa$ in $V[G_{\kappa+1}]$.

Answer to question 1, if $\mu=\kappa$: We will find an example so that $\kappa$ is not weakly compact (compare to earlier results for saturation below $\kappa$), and in fact the quotient algebra is isomorphic to a $\kappa$-Suslin tree.

In the exercises, we will describe how to construct, for $\alpha$ regular with $\alpha^{<\alpha}=\alpha$, a forcing $\mathbb{Q}_\alpha$ which adds an $\alpha$-Suslin tree $\dot{T}_\alpha$ so that $\mathcal{B}(\mathbb{Q}_\alpha\ast \dot{T}_\alpha)\cong\mathrm{Add}(\alpha,1)$. This is due to Kunen.

In the construction for $\mu=\kappa^+$ we got a model (there called $V[G_\kappa]$) where there was an inaccessible $\kappa$ and an ideal $J$ on $\kappa$ so that
$$P(\kappa)/J\cong\mathrm{Add}(\kappa,1)\cong \mathbb{Q}_\kappa\ast \dot{T},$$
where $\dot{T}$ is the $\kappa$-Suslin tree added by $\mathbb{Q}_\kappa$.

Now start with this to be our ground model $V$. Let $H\subseteq \mathbb{Q}_\kappa$ be generic. We want to show that in $V[H]$, there is an ideal $J_1$ on $\kappa$ so that $P(\kappa)/J_1\cong T$.

If $G\subseteq \mathrm{Add}(\kappa,1)$ is generic over $V$, then take in $V[G]$ an embedding
$$j:V\rightarrow M$$
which was constructed before. We want to extend the embedding to $V[H]$.

Now $G\in M$ since $M$ is closed under $\kappa$-sequences in $V[G]$. We can extend $j$ to $V[G]$ by constructing a generic $\hat{G}$ for $\mathrm{Add}(j(\kappa))^M$ over $M$ with $\hat{G}\upharpoonright \kappa=G$ (using the standard method; cf the second exercise following previous construction).

In $V$, by duality theorem there are $J_1$ and $K$ so that
$$\mathbb{Q}_\kappa\ast P(\kappa)/J_1\cong P(\kappa)/J\ast j(\mathbb{Q}_\kappa)/K.$$
In this case, $K$ is a maximal ideal since the $j(\mathbb{Q}_\kappa)$-generic over $M$ is already just added by $P(\kappa)/J$. So in $V[H]$, $P(\kappa)/J_1\cong T$.

Now we turn to Kunen's forcing construction. Conditions in Kunen's forcing $\mathbb{Q}$ are normal trees of successor ordinal height $<\kappa$ which are homogeneous: for all $t\in T$ not on the top level, $T_t\cong T$, where $T_t$ is the tree $\{s\in T: t\le_T s\}$ with the order inherited from $T$.

Exercise:

1. Show that Kunen's forcing is $\kappa$-strategically closed. Hint: the strategy will go by choosing a particular branch through each of the small trees chosen in a play of the game so far.
2. Show that $\mathbb{Q}\ast \dot{T}$ has a $\kappa$-closed dense subset, and deduce that $\mathbb{Q}\ast \dot{T}\cong \mathrm{Add}(\kappa)$.
3. Show that $\dot{T}$ is a Suslin tree.

$\Box$

We will do one last application to construct a precipitous ideal on a cardinal $\kappa$ which is not measurable so that its quotient algebra is $\kappa^+$ closed.

Start with $\kappa$ measurable and $2^\kappa>\kappa^+$. We will use the Easton support iteration $\langle \mathbb{P}_\alpha,\dot{\mathbb{Q}}_\alpha:\alpha<\kappa\rangle$, where $\dot{\mathbb{Q}}_\alpha=\dot{\mathrm{Add}(\alpha^+,1)}$ for inaccessible $\alpha<\kappa$ (and is trivial forcing otherwise).

Then $\mathbb{P}_\kappa$ is $\kappa$-c.c., and forces that for all inaccessible $\alpha<\kappa$, $2^\alpha=\alpha^+$ (this is a standard coding trick that was assigned as an exercise in one of Spencer Unger's lectures here). By duality,
$$\mathbb{P}_\kappa\ast P(\kappa)/\bar{I}\equiv j(\mathbb{P}_\kappa).$$
If $G_\kappa\subseteq \mathbb{P}_\kappa$ is generic, then $j(\mathbb{P}_\kappa)/e''G_\kappa\equiv \mathbb{P}_{\kappa,j(\kappa)}$. Since $M[G_\kappa]$ is closed under $\le \kappa$ sequences in $V[G_\kappa]$, this tail is $\kappa^+$-closed forcing over $V[G_\kappa]$.

However, GCH holds at every inaccessible $\alpha<\kappa$ and fails at $\kappa$ in $V[G_\kappa]$. By a reflection argument, $\kappa$ cannot be measurable in $V[G_\kappa]$. $\Box$

Exercise: Show that if $H\subseteq P(\kappa)/\bar{I}$ is generic, then $\kappa$ is measurable in $V[G_\kappa\ast H]$.

## Monday, July 27, 2015

### UCI Summer School part 2: Duality Theorem (Monroe Eskew)

The Duality Theorem gives a general technique for forcing to make an ideal whose quotient algebra has various properties. It appears in Matthew Foreman's "Calculating quotient algebras of generic embeddings." My version of these notes omits a lot of the dots which indicate that certain objects are just names in a forcing extension. This is for aesthetic reasons, and hopefully does not lead to confusion.

Duality Theorem: Suppose $I$ is a precipitous ideal on $Z$ and $\mathbb{P}$ is any partial order. If: there is a further generic extension of the extension by $P(Z)/I$ so that if $j:V\rightarrow M\subseteq M\subseteq V[G]$ is the ultrapower embedding from $G\subseteq P(Z)/I$, there is $H\subseteq \mathbb{P}$ generic over $V$ and $\hat{H}\subseteq j(\mathbb{P})$ generic over $M$ and some extension of $j$ to $\hat{j}:V[H]\rightarrow M[\hat{H}]$.

Then: there is a $\mathbb{P}$-name for an ideal $J$ on $Z$ and a $P(Z)/I$-name for an ideal $K$ on $j(\mathbb{P})$ and a canonical isomorphism $$\iota:\mathcal{B}(\dot{\mathbb{P}\ast P(Z)/J})\equiv \mathcal{B}(P(Z)/I\ast j(\mathbb{P})/\dot{K}).$$

So a very general statement of lifting a generic ultrapower map to a forcing extension gives a useful isomorphism for computing $P(Z)/J$ in the generic extension by $\mathbb{P}$. We remark that in some cases, this will be an equivalence.

Note: In what follows, we tacitly identify all of the posets involved with their Boolean completions. Occasionally for emphasis, this identification will be explicit.

Proof: Assume (1). There is some $A\in I^+$ and some $P(Z)/I$-name for a forcing $\dot{\mathbb{Q}}$ so that
$$A\Vdash_{P(Z)/I}(\Vdash_\dot{\mathbb{Q}} \dot{H_0}\subseteq j(\mathbb{P}) \textrm{ is generic over }M \textrm{ and }H:=j^{-1}[H_0]\subseteq \mathbb{P}\textrm{ is generic over }V).$$

Note that the set $\{p\in\mathbb{P}:\quad \Vdash_{P(A)/I\ast \mathbb{Q}} j(p)\not\in H_0\}$ cannot be dense, since it is the complement of the generic $j^{-1}[H_0]$. So there is $p_0\in \mathbb{P}$ so that for all $p\le p_0$, $\| j(p)\in H_0 \|_{P(A)/I\ast \mathbb{Q}}\neq 0$. We will constrain ourselves to work below this $p_0$ in $\mathbb{P}$ and below $A$ in $P(Z)/I$. For simplicity, assume that $A=Z$ and $p_0=1_\mathbb{P}$.

In $V^{P(Z)/I}$, define
$$K=\{p\in j(\mathbb{P}):\quad \Vdash_{\mathbb{\dot{Q}^G}} p\not\in H_0\}.$$
Let $G\ast h$ be generic for $P(Z)/I\ast j(\mathbb{P})/K$. From the $j(\mathbb{P})/K$-generic $h$, we can define a $j(\mathbb{P})$-generic $\hat{H}=\{p:[p]_K\in h\}$.

Claim: The following properties of $H_0$ are also true of $\hat{H}$:

1. $\Vdash_{P(Z)/I\ast \mathbb{Q}} \hat{H}$ is $j(\mathbb{P})$-generic over $M$.
2. $\Vdash_{P(Z)/I\ast \mathbb{Q}} j^{-1}[\hat{H}]$ is $\mathbb{P}$-generic over $V$.
3. For all $p\in \mathbb{P}$, $\not\Vdash_{P(Z)/I\ast \mathbb{Q}} j(p)\not\in \hat{H}$.
Proof of Claim: For (1), suppose $D\in M$ is open dense in $j(\mathbb{P})$. Then $\{[d]_K:d\in D\textrm{ and }d\not\in K\}$ is dense in $j(\mathbb{P})/K$. Otherwise, there would exist $p\in j(\mathbb{P})/K$ so that $p\wedge d\in K$ for all $d\in D$. But this is impossible because we could then force with $\mathbb{Q}$ over $V[G]$ to get a generic $H_0$ containing $p$ (as $p\not\in K$), and then $H_0\cap D=\emptyset$, contradicting genericity of $H_0$ over $M$.

The remaining parts of the claim can be checked similarly, and are left as an exercise. $\Box$.

Now let $e:\mathbb{P}\rightarrow \mathcal{B}(P(Z)/I\ast \dot{j(\mathbb{P})/K})$ be defined by $e(p)=\|j(p)\in \hat{H}\|$.

By (3) of the claim above, $\mathrm{ker}(e)=0$. Also $e$ preserves Boolean operations simply by the elementarity of $j$. By (2) of the claim, $e$ is a regular embedding (maps maximal antichains pointwise to maximal antichains).

Exercise: $e:\mathbb{P}\rightarrow \mathbb{Q}$ is a regular embedding iff for every $q\in\mathbb{Q}$ there is $p\in \mathbb{P}$ so that for every $p'\le p$, $e(p')$ is compatible with $q$.

Thus, if $H\subseteq \mathbb{P}$ is generic over $V$, we can force with the quotient $\mathcal{B}(P(Z)/I\ast \dot{j(\mathbb{P})/K})/e"H$ over $V[H]$ to obtain a generic $G\ast h$ for $P(Z)/I\ast j(\mathbb{P})/K$. By the definition of $e$, we have $j_G"H\subseteq \hat{H}$, where $\hat{H}$ is defined from $h$ as before. So we can extend the embedding $j_G$ to $\hat{j}:V[H]\rightarrow M[\hat{H}]$.

In $V[H]$ we can finally define $J=\{A\subseteq Z:1\Vdash [\mathrm{id}]\not\in \hat{j}(A)\}$, where the forcing is with the quotient $\mathcal{B}(P(Z)/I\ast \dot{j(\mathbb{P})/K})/e"H$. In $V$, let
$$\iota(p,\dot{A})=e(p)\wedge \|[id]\in\hat{j}(\dot{A})\|.$$

Exercise: $\iota$ is order and incompatibility preserving.

It remains to show that the range of $\iota$ is dense. So take an arbitrary $(B,\dot{q})\in P(Z)/I\ast j(P)/K$. By strengthening this condition, we may assume without loss of generality that there is $f:Z\rightarrow \mathbb{P}$ in $V$ so that $B\Vdash [[f]_M]_K=\dot{q}$.

By regularity of $e$ (using the characterization in the exercise), there is a $p$ so that for all $p'\le p$, $e(p')\wedge (B,\dot{q})\neq 0$. Let $\dot{A}$ be a $\mathbb{P}$-name for a subset of $Z$ such that $p \Vdash \dot{A}=\{z\in B: f(z)\in H\}$ and $\neg p \Vdash \dot{A}\in J^+$.

We check that $(p,\dot{A})$ is actually a condition, which involves checking that $p\Vdash \dot{A}\in J^+$. So take a generic $G\ast \dot{h}$ containing $e(p)\wedge (B,\dot{q})$ (which is nonzero by choice of $p$). Then clearly $B\in G$ and since $[[f]_M]_K=q\in h$, we have $[f]_M=j(f)([\mathrm{id}])\in\hat{H}$. Therefore $[\mathrm{id}]\in \hat{j}(A)$, so this generic $G\ast \dot{h}$ shows that it is not forced by $\mathcal{B}(P(Z)/I\ast \dot{j(\mathbb{P})/K}/e"H$ that $[\mathrm{id}]\not\in \hat{j}(A)\}$.

By definition $\iota:=\iota(p,\dot{A})$ forces $j(p)\in \hat{H}$ and $[\mathrm{id}]\in\hat{j}(\dot{A})$. Since $\hat{j}$ extends $j$ and it's forced that $\dot{A}\subseteq B$, $\iota$ must force $B\in G$. And since $j(p)\Vdash_{j(\mathbb{P})} j(\dot{A})=j(\{z:j(f)(z)\in \hat{H}\})$, $\iota$ must force $\dot{q}=[j(f)(\mathrm{id})]_K\in h$. $\Box$

Remark: Suppose $K$ as in the Duality Theorem is forced to be principal, i.e., there is $m$ so that $$\Vdash K=\{p\in j(\mathbb{P}:p\le \neg m\}.$$
Then the Duality Theorem is easily seen to be an equivalence.

We can compute some nice properties of the ideal $J$ arising from the previous theorem.

Proposition: Using the notation of the previous theorem, $J$ is forced to be precipitous, with the same completeness as $I$. If $I$ is normal, then $J$ is also normal. Also, if $\bar{G}\subseteq P(Z)/J$ is generic over $V[H]$ and $G\ast h=\iota[H\ast \bar{G}]$ and $\hat{j}:V[H]\rightarrow M[\hat{H}]$ are as before, then $V[H]^Z/{\bar{G}}=M[\hat{H}]$ and $\hat{j}$ is the ultrapower embedding.

Finally, we relate $J$ to the ideal in $V[H]$ generated by $I$.

Proposition: Suppose $K$ as in the Duality Theorem is forced to be principal, with $m$ so that $\Vdash K=\{p\in j(\mathbb{P}):p\le \neg m\}.$ Suppose further that there exist $f$ and $A$ so that $A\Vdash \dot{m}=[f]_G$ and $\dot{B}$ is a $\mathbb{P}$-name for $\{z\in A: f(z)\in \dot{H}\}$. Then $\bar{I}\upharpoonright B=J\upharpoonright B$, and $A\setminus B\in J$, where $\bar{I}$ is the ideal in $V[H]$ generated by $I$.

## Friday, July 17, 2015

### UCI Summer School part 1: Basics (Monroe Eskew)

The UCI summer school in set theory just finished, and I'll be posting my notes on this blog. They will not be added in chronological order, and the final product will be very different from the actual presentation, assume that all errors were added by me.

There are some exercises which we solved during problem sessions during the day; I will preserve these in the text and can post solutions as well if there is interest.

We start off with Monroe Eskew's lectures on the duality theorem.

Basic facts: This section is meant to be supplemented by my earlier postings about ideal properties. The basic situation is that $I$ is an ideal on $Z\subseteq P(X)$, $G$ is a generic for the forcing $P(Z)/I$, and $j:V\rightarrow M=V^Z/G$ is the generic ultrapower by $G$.

Definition: An ideal $I$ has the disjointing property if every antichain $A$ in $P(Z)/I$ has a pairwise disjoint system of representatives.

Exercise: If $I$ is a $\kappa$-complete, $\kappa^+$-saturated ideal, then $I$ has the disjointing property.  If $I$ is a normal ideal on $Z\subseteq P(X)$, and $I$ is $|X|^+$-saturated, then $I$ has the disjointing property.

Lemma: Suppose $A\in I^+$ and $I$ has the disjointing property, and $[A]\Vdash \dot{\tau} \in V^Z/\dot{G}$. Then there is $f:Z\rightarrow V$ such that $A\Vdash \tau = [f]_G$.

Proof: Exercise.

Lemma: Suppose $I$ is a countably closed ideal with the disjointing property. Let $\kappa=\mathrm{crit}(j)$. Then $I$ is precipitous, and the generic ultrapower $M$ is closed under $\kappa$-sequences from $V[G]$.
Proof: Precipitousness follows from the combinatorial characterization of precipitousness. Suppose that $\Vdash \langle \tau_\alpha:\alpha<\kappa\rangle \subseteq V^Z/G$.  By the previous lemma, for all $\alpha<\kappa$, there is $f_\alpha:Z\rightarrow V$ in $V$ such that $\Vdash \tau_\alpha=[f_\alpha]_G$ and $k:Z\rightarrow \mathrm{ON}$ in $V$ such that $\Vdash \kappa=[k]_G$. Finally, take $f:Z\rightarrow V$ given by  $f(z)=\langle f_\alpha(z):\alpha<k(z)\rangle$. In $M$, $[f]_G=\langle \tau_\alpha:\alpha<\kappa\rangle$. $\Box$

Definition: $I$ is $(\lambda,\kappa)$-presaturated if for any sequence of antichains $\langle A_\alpha:\alpha<\gamma<\lambda\rangle$, the set $\{X:\forall\alpha<\gamma\, |\{a\in A_\alpha:a\cap x\in I^+\}|<\kappa\}$ is dense.

Exercise: $(\lambda^+, \kappa^+)$-presaturation and $\kappa$-completeness imply that the generic ultrapower $M$ is closed under $\lambda$-sequences from $V[G]$. Also show this for normal and $(\lambda^+,|X|^+)$-presaturated ideals.

We will use the notation $\mathcal{B}(\mathbb{P})$ to denote Boolean completion of a poset. As we have seen in previous posts, this is very handy when dealing with generic ultrapowers.

Exercise: $\mathbb{P}$ and $\mathbb{Q}$ are separative posets. The following are equivalent:

1. $\mathcal{B}(\mathbb{P})=\mathcal{B}(\mathbb{Q})$.
2. There is a $\mathbb{P}$-name $\dot{h}$ and a $\mathbb{Q}$-name $\dot{g}$ so that $$\Vdash_\mathbb{P} \dot{h} \textrm{ is }\mathbb{Q}-\textrm{generic over }V,$$ $$\Vdash_\mathbb{Q} \dot{g} \textrm{ is }\mathbb{P}-\textrm{generic over }V,$$ and$$\Vdash_\mathbb{P} \dot{g}^{\dot{h}^\dot{G}}=\dot{G} \textrm{ and } \Vdash_\mathbb{Q} \dot{h}^{\dot{g}^\dot{H}}=\dot{H}.$$

## Tuesday, June 16, 2015

### Tight stationarity and tree-like scales

My article, "Tight stationarity and tree-like scales," is now available online in the Annals of Pure and Applied Logic.

You can also find the final version here.

I have some slides on this topic from last year's BEST conference available here. (Coincidentally, I'm now at the 2015 BEST conference :)

## Friday, April 17, 2015

### Spencinar: Morley's categoricity theorem, part 1

Unfortunately, I missed posting about some interesting lectures given at the Very Informal Gathering, and the Southern California logic meeting at Caltech. We'll return with this series of posts following Spencer Unger's talks at the first two seminars of the quarter. There may be a larger amount of errors than usual in my transcription because of my limited knowledge of model theory. Basic model theory will be helpful to follow these notes, but I'll do my best to fill in the definitions. The notation follows Marker's text closely.

A general question is:
Given a first order theory $T$, what can we say about the number of non-isomorphic models of $T$ of size $\lambda$, as $\lambda$ varies through the cardinals?

This is addressed by Shelah's classification theory. Historically, the first step was Morley's categoricity theorem (1965) which states:

• If $T$ is a first-order theory in a countable language and $T$ is categorical in some uncountable cardinal, then it is categorical in all uncountable cardinals.
The proof of this theorem, which was the content of the lectures, involves the ideas of stability, indiscernibles, saturated models, and splitting types. It's somewhat different from the one I recall from the basic model theory class I took many years ago.

To begin with, we can define a rank $R$ on sets of formulas (i.e., types) with parameters in some large saturated model $\mathcal{C}$ (the "monster model"). Let $p$ be a set of formulas. Then, working in $\mathcal{C}$,
• $R(p)=-1$ if $p$ is not satisfiable.
• $R(p)\ge 0$ if $p$ is satisfiable.
• $R(p)\ge \alpha+1$ if for any finite $p_0\subseteq p$, there is a formula $\psi$ such that $R(p_0\cup\{\psi\})\ge \alpha$ and $R(p\cup\{\neg \psi\})\ge \alpha$.
The rank $R(p)$ is then defined inductively to be the minimum value so that all of the above conditions are satisfied. Note: this isn't what is known as Morley rank. Andrew Marks noticed that this rank is the Cantor-Bendixson rank on the Stone space of complete types.

Some properties of the rank:
• (Monotonicity) If $p\vdash q$ then $R(p)\le R(q)$.
• (Finite character) For every $p$, there is a finite $p_0\subseteq p$ so that $R(p)=R(p_0)$.
• Proof: If $R(p)=\alpha$, there is a finite $p_0$ witnessing $R(p)\not\ge \alpha+1$, and monotonicity gives the reverse inequality.
• (Invariance) If $f$ is an automorphism of $\mathcal{C}$, then $R(p)=R(f(p))$.
• Proof is by induction on the rank.
Definition: $T$ is $\lambda$-stable if for any $A\subseteq C$ of size $\lambda$, $|S^n(A)|\le \lambda$. Here $S^n(A)$ is the set of complete $n$-types over $A$, that is, maximal satisfiable sets of formulas with parameters over $A$ and free variables $x_1,\ldots, x_n$.

Theorem: Let $T$ be a theory in a countable language. Then $T$ is $\aleph_0$-stable iff $R(\{\bar{x}=\bar{x}\})<\infty$, i.e., has some ordinal value. In fact, if these equivalent conditions hold, then $R(\{\bar{x}=\bar{x}\})<\omega_1$.

Proof: Suppose $R(\{\bar{x}=\bar{x}\})\ge \omega_1$. We will inductively construct a complete binary tree whose nodes are finite sets of formulas $p_\eta$ for $\eta\in 2^{<\omega}$ and whose branches will give $2^{\aleph_0}$ many different complete types, contradicting stability. For the construction, we will maintain that $R(p_\eta)\ge\omega_1$.

Let $p_\emptyset=\{\bar{x}=\bar{x}\}$. Given $p_\eta$, there are formulas $\psi_{\eta,i}(\bar{x},\bar{a}_{\eta,i})$ for each $i<\omega_1$ witnessing that $R(p_\eta)\ge i+1$. Using the fact that the language of $T$  is countable together with $\aleph_0$-stability, there are $\psi$ and $t$ so that for an unbounded set of $i<\omega_1$, $\psi=\psi_{\eta,i}$ and $t=\mathrm{tp}(\bar{a}_{\eta,i})$ (the set of all formulas satisfied by $\bar{a}_{\eta,i}$, over the finitely many parameters in $p_\eta$). Pick $a_\eta$ realizing $t$. Then define
• $p_{\eta\circ 0}=p_\eta\cup\{\psi(\bar{x},\bar{a}_\eta)\}$
• $p_{\eta\circ 1}=p_\eta\cup\{\neg\psi(\bar{x},\bar{a}_\eta)\}$
Using invariance of rank under automorphisms and the fact that in $\mathcal{C}$ for any two tuples realizing the same type there exists an automorphism sending one to the other, the ranks of these sets of formulas are at least $i$ for an unbounded set of $i<\omega_1$, so they must be $\ge\omega_1$, allowing the construction to continue. The branches of the tree we construct can be extended to complete types, and the construction ensures that each of these complete types is distinct (for two different branches, look at the first place of disagreement).

For the converse, assume $R(\bar{x}=\bar{x})<\infty$ and $T$ is not stable. Then there are $A\subseteq \mathcal{C}$, $|A|\le \lambda$ such that there are distinct $\{p_i:i<\lambda^+\}\subseteq S^n(A)$. By assumption, for any $i<\lambda^+$, there is $\alpha_i$ so that $R(p_i)=\alpha_i$. By finite character, there is $q_i\subseteq p_i$ finite with $R(q_i)=R(p_i)=\alpha_i$. By counting, $|\{q_i:i<\lambda^+\}|\le \omega$. Let $i\neq j$ be such that $q_i=q_j$, so $p_i,p_j$ are different completions of $q_i=q_j$. This means there is a formula $\psi$ with $\neg\psi\in p_i$ and $\psi\in p_j$. This gives $R(q_j)\ge R(p_j)+1$, a contradiction to the choice of $q_j$. $\Box$.

## Monday, January 26, 2015

### Spencinar: Forcing clubs in stationary subsets of $P_\kappa(\lambda)$

The first part of this talk follows closely the paper "Forcing closed unbounded sets" by Uri Abraham and Saharon Shelah ([AS]). The second part will follow the paper "Nonsplitting subset of $P_\kappa(\kappa^+)$ by Moti Gitik ([G]).

We have previously seen that given a stationary $S\subseteq \omega_1$, there is a forcing which has the $\omega_2$-c.c. and is $<\omega_1$-distributive (and hence preserves all cardinals) which adds a club $C\subseteq S$, namely the forcing of closed bounded subsets of $S$, ordered by end-extension (Baumgartner--Harrington--Kleinberg).

This generalizes to higher regular cardinals $\kappa$ assuming some cardinal arithmetic (GCH with $\kappa$ the successor of a regular cardinal suffices) and fatness of $S$, which says that for any club $E\subseteq \kappa$, $S\cap E$ contains closed subsets of arbitrarily large order type $<\kappa$ (this is a result of J. Stavi).

The focus of this talk will be to obtain analogues for these theorems for subsets of $P_\kappa(\lambda)=\{x\subseteq \lambda: |x|<\kappa\}$.

This situation is not as clear as for subsets of ordinals, and in fact there are some sets whose stationarity is quite absolute (By stationary and club I mean here in the sense of Jech, see this previous post). For example, Theorem 6 in [AS] gives a stationary subset of $P_{\aleph_1}(\omega_2)$ whose stationarity is preserved by any forcing which preserves $\aleph_2$:

Theorem: Let $W\subseteq V$ be a transitive inner model such that $\omega_2^W=\omega_2^V$, and let $S=(P_{\aleph_1}(\omega_2))^W$. Then $S$ is stationary in $V$.

Proof: The idea of the proof is to fix an arbitrary club $C\subset P_{\aleph_1}(\omega_2)$ and find submodels which are ordinals, or somehow coded by ordinals in $W$. Then these submodels will automatically be in $S$.

By Kueker's theorem, there is a function $F:[\omega_2]^{<\omega}\rightarrow \omega_2$ so that any $x\in P_{\aleph_1}(\omega_2)$ closed under $F$ is in $C$ (this uses the fact that we're working on $\aleph_1$ and $\aleph_2$, but this is a minor technical point). Let $\alpha$ be an ordinal closed under $F$. If $\alpha$ is countable, then we're done; otherwise $\alpha$ has cardinality $\aleph_1$, hence also $W$-cardinality $\aleph_1$, so fix a bijection $h:\omega_1\rightarrow \alpha$ in $W$. Then there is $\xi<\omega_1$ with $h[\xi]$ closed under $F$, and $h[\xi]$ is in $W$ and hence in $S$, so we're done again. $\Box$

In [G], it's shown that any forcing which adds reals to $W$ destroys the club-ness of $S$.

This paper gives some examples of sets whose stationary can be destroyed. We won't go through those results here. Instead, we now turn towards a result in [G] which also gives examples of this kind.

Let $\kappa$ be supercompact in $W$, and let $V$ be obtained from $W$ by Radin forcing. We won't need to know much about Radin forcing, just:
• the continuum function and all cardinalities are preserved,
• there is a club $C\subseteq \kappa$ of $W$-inaccessible cardinals,
• $\kappa$ remains (sufficiently) supercompact in $V$.
From now on, work in $V$. Let $A$ be any subset of the set of all $t\in P_\kappa(\kappa^+)\cap W$ such that $V\vDash |t| \textrm{ is a successor cardinal}"$.

We will find a forcing that adds a club in $S:=P_\kappa(\kappa^+)\setminus A$ which is $<\kappa$-distributive (this is crucial, so that $P_\kappa(\kappa^+)$ itself does not change), and has the $\kappa^+$-c.c.

Define the poset $\mathbb{P}$ to be the collection of subsets of $S$ of size $<\kappa$ which have a maximum element and are closed under increasing unions, ordered by end-extension.

Claim: $\mathbb{P}$ is $<\kappa$-distributive.

Let $\langle D_\beta: \beta<\alpha\rangle$ be a sequence of dense open subsets of $\mathbb{P}$ for some $\alpha<\kappa$.  Then we will prove that $\bigcap_{\beta<\alpha} D_\beta$ is also dense. So let $x\in \mathbb{P}$.

Let $M\prec (H(\theta),\in,\triangleleft,A,\langle D_\beta\rangle,x,\ldots)$ be an elementary substructure of size $\kappa$ so that $M\cap \kappa^+\in \kappa^+$ and $M$ is closed under $<\kappa$-sequences. Let $g\in W$ be a bijection from $\kappa$ onto $M\cap \kappa^+$. Using $<\kappa$-closure of $M$, construct $\langle M_i:i<\kappa\rangle$ a continuous IA chain of elementary submodels of $M$ of size $<\kappa$ so that $\alpha+1\subseteq M_0$ and $gi\subseteq M_i$. There is a club $E\subseteq \kappa$ such that for every $i\in E$, $i\in C$ and $M_i\cap \kappa^+=gi$. Let $\langle i_\beta:\beta<\kappa\rangle$ be the increasing enumeration of $E$, and let $N_\beta=M_{i_\beta}$.

We will inductively construct a decreasing sequence of conditions $\langle x_\beta:\beta\le \alpha\rangle$ such that:

•  $x_0\le x$,
•  $x_{\beta+1}\in D_\beta$,
•  $N_\beta\cap \kappa^+=\max(x_\beta)$,
•  $x_\beta\in N_{\beta+1}$.

Take $x'_0\in N_0$, $x'_0\le x$, and let $x_0=x'_0\cup\{N_0\cap \kappa^+\}$.

Now we construct $x_\beta$, assuming that $x_\gamma$ has been constructed for every $\gamma<\beta$. If $\beta=\gamma+1$ for some $\gamma$, then pick $x'_\beta$ to be the $\vartriangleleft$-least in $N_\beta\cap D_\beta$ extending $x_\gamma$, and define $x_\beta=x'_\beta\cup\{N_\beta\cap \kappa^+\}$. This is a valid condition since $N_\beta\cap \kappa^+$ has $W$-cardinality in $C$, so it can't be a member of $A$.

If $\beta$ is limit, then let $x'_\beta$ be the closure of $\bigcup \{x_{\gamma}:\gamma<\beta\}$ under increasing unions, and $x_\beta=x'_\beta\cup\{N_\beta\cap \kappa^+\}$. By internal approachability, $x_\beta\in N_{\beta+1}$.

It is easy to see that:
Claim: For every $\gamma<\beta$, $x_\beta$ is an end-extension of $x_{\gamma}$.

We now check that $x_\beta\in \mathbb{P}$. If not, then there is $t\in x_\beta\cap A$. Since $t\subseteq N_\beta\cap \kappa^+$, we must have $|t|^W\le i_\beta$. Furthermore, $|t|^W$ is a successor cardinal of $W$ and $i_\beta$ is W-inaccessible, so $|t|^W<i_\beta$.

Since $i_\beta$ is $W$-regular, there is some $\gamma<\beta$ with $t\subseteq gi_\gamma=N_\gamma\cap \kappa^+$. But this is impossible by the claim.

The $\kappa^+$-c.c. follows by a standard $\Delta$-system argument.

Gitik used this forcing as the building block of an iteration to produce a stationary subset $Z$ of $P_\kappa(\kappa^+)$ so that the non-stationary ideal restricted to that stationary set is $\kappa^+$-saturated, which is very interesting in light of results that show that the whole nonstationary ideal is not saturated. Another way to look at this is that it shows the consistency of the failure of a natural analogue of Solovay's splitting theorem for stationary subsets of $P_\kappa(\lambda)$.

Assuming that $\kappa$ is supercompact, the set $Z$ of all $t\in P_\kappa(\kappa^+)\cap W$ such that $W\vDash |t| \textrm{ is a successor cardinal}"$ is stationary. The idea of the iteration is that we will destroy the stationary of the "bad sets" to while maintaining the stationary of $Z$. Maintaining the stationarity of $Z$ is achieved through extending the supercompactness embedding to the final model, but this is not an easy task in this case.

An earlier version of this post had several occurrences of $V$ which should have been $W$.