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 $V[G_\kappa\ast H]$.

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 $I$.

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$$\Vdash_\mathbb{P} \dot{g}^{\dot{h}^\dot{G}}=\dot{G} \textrm{  and  } \Vdash_\mathbb{Q} \dot{h}^{\dot{g}^\dot{H}}=\dot{H}.$$

Learning project #1: a theorem

0th post in this series
1st post (updated with some more facts about IA structures at the end)
(edited 12/10/2014 to improve clarity)

We'll explain the proof of Theorem 27 of FMS:

Theorem 27: Suppose $\kappa$ is a supercompact cardinal and $\mu<\kappa$ is regular. Then in $V^{\rm{Col}(\mu,<\kappa)}$, the nonstationary ideal on $\mu$ is precipitous.

We'll follow the proof in FMS, trying to use Foreman's chapter for some new insights. I think the material is generally covered much better in the Handbook chapter than in the article.

Now we have Lemma 1 for proving ideals are precipitous, we can carry out this gameplan.

Let $G$ be a generic for $\rm{Col}(\mu,<\kappa)$, and let's think in the generic extension $V[G]$. We will show:

 Main Claim: Let $\mathfrak{A}$ be an expansion of $\langle H(\lambda),\in,\Delta\rangle$ (where $\lambda$ is some regular cardinal above, say, $2^{2^{2^\mu}}$). Then for almost all (in the sense of the nonstationary ideal) $N\prec\mathfrak{A}$ with $N\in \rm{IA}$, $|N|<\mu$, and $N\cap \mu$ an ordinal:

For all maximal antichains $\mathcal{A} \in N$ of $P(\mu)/NS$, there is an $A\in \mathcal{A}$ such that 

1. $N\cap \mu\in A$ and
2. letting $N^A$ equal the Skolem hull of $N\cup \{A\}$, $N^A\cap \mu=N\cap \mu$ in $\mathfrak{A}$.

Let's explain the usefulness of these conditions. To apply our Lemma 1, let $\mathcal{A}=\langle A \rangle$ be a tree of maximal antichains for $P(\mu)/\rm{NS}$ with underlying tree $T\subseteq \langle\mu^+\rangle^{<\omega}$. We will construct a branch $f:\omega\rightarrow \mu^+$ so that $\bigcap_n A_{f\upharpoonright n}\neq \emptyset$.

We'll construct the branch $f$ in $\omega$ stages. 

Take $N_0$ containing $\mathcal{A}$ as an element. Let $s_0=\langle\alpha_0\rangle$ be so that $A_{s_0}$ witnesses (1) and (2) with antichain equal to the first level of $\mathcal{A}$. From now on, use the notation $s_i:=\langle \alpha_0,\ldots, \alpha_{i}\rangle$. 

At the $i+1$st stage, let $N_{i+1}:=(N_i)^{\alpha_i}$ (the Skolem hull of $N_i\cup\{\alpha_i\}$ in $\mathcal{L}$). Pick $\alpha_{i+1}$ so that (1) and (2) are satisfied for $N=N_{i+1}$ and $s=\langle \alpha_0,\ldots, \alpha_{i+1}\rangle$ in the maximal antichain consisting of $\langle A_s:s \textrm{ extends }s_i\rangle$ together with some set not containing the single point $N\cap \mu$ (in order to do this construction, we will need to show that the property that $N_i\in\rm{IA}$ is maintained, which is Lemma 2 from last time). 

We have $N_i\cap \mu\in A_{s_i}$ for all $i$, but $N_i\cap\mu$ was always just equal to $N\cap \mu$, so $N\cap \mu$ would be in the intersection of all of the $A_{s_i}$. The key point of absorbing the index of the member of the antichain was so that the construction could continue in a coherent way, ensuring we are actually building a branch of $T$. (Foreman's chapter further develops these ideas, leading to the notion of "catching antichains".)

Let us now prove the main claim. Suppose there is a stationary $S\subseteq \rm{IA}$ of structures for which the conclusion of the main claim fails. The idea of the proof is that the supercompact gives us enough reflection of $S$ to catch the antichain. Let $\mathbb{P}=\rm{Col}(\mu,<\kappa)$ and let $j:V\rightarrow M$ be an $|H(\lambda)|$-supercompact embedding. Since $\mathbb{P}$ is $\kappa$-c.c., we can force to find $H\subseteq j(\mathbb{P})$ which contains $G$ as a subset, so we can extend $j$ to $\hat{j}:V[G]\rightarrow M[H]$ in $V[H]$.  

Subclaim: $S$ reflects to a set of size $\mu$, i.e., there is $Y\subseteq \lambda$ so that $\mu\subseteq Y$ and $|Y|=\mu$ so that $S\cap P(Y)$ is stationary in $P(Y)$.

Proof of Subclaim: By Lemma 3 from last time, $\hat{j}(S)\cap P(\hat{j}``H(\lambda))$ is stationary in $V[H]$. Now because of the collapse, $|\hat{j}``H(\lambda)|=\mu$ and $\mu\subseteq \hat{j}``H(\lambda)$ since $\kappa$ is the critical point. By closure of $M$ (and, it can be seen, of $M[H]$, $\hat{j}``H(\lambda)\in M[H]$. So ($\hat{j}$ of) the statement of the subclaim holds in $M[H]$, so by elementarity the subclaim holds in $V[G]$.

Returning to the proof of the main claim, let $Y$ be given by the subclaim and let $f:\mu\rightarrow Y$ be a bijection. Let $T:=\{\delta<\mu: f``\delta\in S\textrm{ and }\delta=f``\delta\cap\mu\}$. Then it is easy to see that $T$ is stationary, using the fact that $S$ reflects to $Y$. By maximality, there is $A\in\mathcal{A}$ such that $T\cap A$ is stationary. So we can find an IA structure $N'\prec \mathfrak{A}$ containing $A, f$ and with $N'\cap \mu \in T\cap A$. Take $N$ to be $f``(N' \cap\mu)$, so $N\in S$ by definition of $T$. By the second property in the definition of $T$, we also have $N\cap \mu=N'\cap \mu$. But $N'$ contains $A$, so the Skolem Hull of $N\cup\{A\}$ is contained in $N'$ and therefore has the same supremum below $\mu$, and would show that $N$ cannot be in $S$.

This finishes the proof. 

Learning project #1, part 1: preliminaries

0th post in this series (updated with some more motivation):
(other relevant posts for reference: Spencinar #3: precipitous ideals)


Let's begin! Remember that one of our goals is

Theorem: $\rm{Col}(\mu,<\kappa)$ makes the nonstationary ideal restricted to $[\mu]^{<\mu}$ precipitous.

First, let's define notions of stationarity on sets of the form $[\lambda]^{<\kappa}=\{x\subseteq \lambda: |x|<\kappa\}$. We will have $\kappa$ regular.

There are two different notions here. If $X$ is a set and $C\subseteq P(X)$, then $C$ is strongly club if there is a structure $\mathfrak{A}$ on $X$ with countably many function symbols so that $C$ is the set of elementary substructures of $\mathfrak{A}$ (or equivalently, the set of substructures closed under a function $X^{<\omega}\rightarrow X$). $C$ is J-club if it is unbounded in the $\subseteq$ order on $[\lambda]^{<\kappa}$ and closed under increasing unions of size less than $\kappa$. It turns out the two notions are closely connected, since a result of Kueker says that the filter generated by the J-club sets is the filter generated by the strongly club sets together with all sets of the form $\{x\in [\lambda]^{<\kappa}:x\cap \kappa \in \kappa\}$ (this means that the ordinals of $x\cap \kappa$ are an initial segment of all ordinals).

We can correspondingly define weakly stationary (resp. J-stationary) sets as those that intersect every strongly club (resp. J-club) set. For our purposes, strongly club and weakly stationary sets are the right notions, and we will refer to them from now on without the adverbs.

In this post, we will get a simple criterion for precipitousness. Read the Spencinar #1 or #3 posts for basic information about generic ultrapowers.

Our notation will be that $Z$ is a set and $I$ is an ideal on $Z$. We will use $I^+$ to denote the collection of $I$-positive sets, i.e., the set $P(Z)-I$. We will use $I^\wedge$ to mean the dual filter of $I$, i.e., the collection of sets of the form $Z-A$ for some $A\in I$. The equivalence class of $S\subseteq Z$ in $P(Z)/I$ will be denoted by $[S]$.

Definition: A tree of maximal antichains $\mathcal{A}$ for $P(Z)/I$ below $S$ is an underlying tree $T\subseteq \mathrm{ON}^{<\omega}$ labeled with $\langle A_s: s\in T\rangle$ elements of $P(Z)$ such that

1. $A_\emptyset = S$,
2. The levels of the tree form increasingly refined maximal antichains of $P(Z)/I$ restricted to $S$, i.e., for each $s\in T$, the set $\{[A_{s\ast \alpha}]:s\ast\alpha\in T\}$ is a maximal antichain below $A_s$ ($\ast$ denotes concatenation). Further, let us assume that $A_{s\ast \alpha}\subseteq A_s$. Denote the levels by $\mathcal{A}_n$, $n<\omega$. 

Lemma 1: (FMS p. 31, originally Jech--Prikry) $I$ is precipitous iff for every $S\in I^+$ and every tree $\mathcal{A}$ of maximal antichains for $P(Z)/I$ restricted to $S$, there is a branch $f$ of the tree so that $\cap_{n<\omega} A_{f\upharpoonright n}$ is nonempty.

Proof: If $I$ is precipitous, and $\langle A_s: s\in T$ is a tree of maximal antichains below some $S\in P(Z)$, then consider the generic ultrapower (see older posts for references) by $P(Z)/I$ restricted to $[S]$. This is a map
$$j:V\rightarrow M\subseteq V[G]$$
defined in the forcing extension by $G$ which is generic for $P(Z)/I$ restricted to $[S]$. We will think of $G$ as a $V$-ultrafilter extending the dual filter of $I$.

Let's review the construction of this ultrapower. The elements are represented by functions $F: Z\rightarrow V$ in $V$, and we have Los's theorem which says for a formula $\varphi$ in the language of set theory and $F_0,\ldots,F_n$, $M\vDash \varphi([F_0],\ldots,[F_n])$ if and only if $\{z:V\vDash \varphi(F_0(z),\ldots,F_n(z))\}\in G$.

In $M$, let $\mathcal{B}=j(\mathcal{A})$ and $i$ be the element represented in the ultrapower by the identity function on $Z$.

For each $n<\omega$, there must be a unique $A_{s_n}\in \mathcal{A}_n$ so that $i\in j(A_{s_n})$: this $A_s$ is the one in $G$ (such $A_s$ must exist by a basic density argument using the fact that $\mathcal{A}_n$ is a maximal antichain, and it must be unique). The sequence of these $\langle A_{s_n}:n<\omega\rangle$ can be computed in $V[G]$ and is a branch through $\mathcal{B}$. Then there must be a branch in $M$ through the subtree of $\mathcal{B}$ consisting of those nodes which are labeled by a set containing $i$ as an element. (Otherwise, in $M$ there would be a rank function $f$ on this subtree taking ordinal values such that $f(B_s)<f(B_t)$ if $s$ is below $t$, and this would have to be in $V[G]$, so there couldn't be a branch in $V[G]$. Note that this is an important pattern of argument that is very common in many different places in set theory).

Now this branch is a branch through $\mathcal{B}$ with nonempty intersection (containing $i$, for one) so by elementarity there is such a branch through $\mathcal{A}$.

Conversely, assume that $I$ is not precipitous. Then there is $S\subseteq Z$ such that $[S]$ forces that the generic ultrapower is well-founded. By maximality of the forcing language, we can find names $\dot{F}_n$ such that

1. $[S]\Vdash \dot{F}_n:Z\rightarrow V$,
2. $[S]\Vdash \dot{F}_n\in V$,
3. $[S]\Vdash ``M \vDash [\dot{F}_{n+1}]\in [\dot{F}_n]''$. 

We can build a tree of maximal antichains $\mathcal{A}$ below $[S]$ such that

• a condition $A_s$ on the $n$th level decides the value of $\dot{F}_n$, say as $f^s_n$,
• if $s$ is the predecessor of $t$ in the underlying tree, then $f^t_{n+1}(z)\in f^s_n(z)$ for all $z\in A_t$.

If we have a tree of maximal antichains with just the first property, we can throw out sets in $I$ from each node to get the second property.

This tree has no branch with nonempty intersection, otherwise this would give an infinite descending sequence in $V$. $\Box$


Another thing we will need is the notion of an internally approachable (IA) structure. Let $\lambda$ be sufficiently large regular cardinal and consider the structure $\mathfrak{A}=\langle H(\lambda),\epsilon,\Delta,f_i\rangle$ where the $f_i$ are countably many function symbols and $\Delta$ is a fixed well-order of $H(\lambda)$ needed for certain Skolem hull arguments. An elementary substructure $N\prec \mathfrak{A}$ is internally approachable iff it can be written as $N=\bigcup_{j<\delta}N_\delta$ for some $\delta$ such that for all $\beta\in \delta$, $\langle N_\alpha:\alpha<\beta\rangle\in N$ (this definition is slightly weaker than other common definitions of IA structures). Sometimes such an $N$ will be said to be internally approachable of length $\delta$.

Because of the way these structures are built in layers, an IA structure can do some pretty "meta" computations (as we will see). This enables us to do some inductive constructions, where the IA condition gives a kind of strengthened inductive hypothesis.

First we have a basic lemma about IA structures which is useful in many contexts.

Lemma 2: Let $N\prec \langle H(\lambda),\epsilon,\Delta$ be IA of length $\delta$ for $\delta<\lambda$. Suppose $a\in y$ for some $y\in N$. Then the Skolem hull of $N\cup\{a\}$ computed in $H(\lambda)$ is IA of length $\delta$.

Proof: Let $\langle N_\alpha:\alpha<\delta\rangle$ witness approachability. For each $\alpha<\delta$, let $N^*_\alpha$ be the closure of $N_\alpha\cup\{a\}$ under all functions $H(\lambda)\rightarrow H(\lambda)$ in $N_{\alpha+1}$. The $N^*_\alpha$ are elementary substructures of $H(\lambda)$ that witness approachability for the Skolem hull of $N\cup\{a\}$: if $\beta<\delta$, then $\langle N^*_\alpha:\alpha<\beta\rangle$ can be computed from $a$ and the sequence $\langle N_\alpha:\alpha\le \beta\rangle$. $\Box$

Next is the main importance of IA structures in this project. It is very typical of applications of approachability:

Lemma 3: Suppose $S\subseteq \rm{IA}$ be a stationary subset of $[H(\lambda)]^{<\mu}$, for some uncountable regular $\mu<\delta$.  Then $S$ remains stationary in $[H(\lambda)^V]^{<\mu}$ after forcing with $\mathbb{P}=\rm{Col}(\mu,<\kappa)$ for any ordinal $\kappa$.

Proof: Let $\dot{F}$ be a name for a function $H(\lambda)^{<\omega}\rightarrow H(\lambda)$ in the extension. Let $\lambda^*$ be a sufficiently large regular cardinal, and $N\prec \langle H(\lambda^*),\in,\Delta,\mathbb{P},\dot{F},S\rangle$ with $|N<\mu|$ and $N\cap H(\lambda)\in S$.

Then $N\cap H(\lambda)$ is IA of some length $\delta<\mu$, witnessed by $\langle N_\alpha:\alpha<\delta\rangle$. We will define a decreasing sequence $\langle p_\alpha:\alpha<\delta\rangle$ of conditions in $\mathbb{P}$ so that

1. For every $\beta<\delta$, there is $M_\beta\in N$ containing $N_\beta$ and each of the previous $M_\alpha$, $\alpha<\beta$, and so that for each $x\in M_\alpha$, $p_\beta$ forces there is $y\in M_\beta$ with $F(x)=y$.
2. For every $\beta<\delta$, $\langle p_\alpha:\alpha<\beta\rangle\in N$.

We just pick the $\Delta$-least $p_\beta$ so that there is a $M_\beta$ satisfying the above, and then pick the $\Delta$-least such $M_\beta$. These things all exist in $N$ since $N$ is IA, and we can get (2) since this computation can be done inside $N$.

Now use closure of $\mathbb{P}$ to find a $p$ extending all of the $p_\alpha$. This $p$ forces that $N\cap H(\lambda)$ is closed under $F$, so $S$ is stationary.

Learning project #1: Foreman-Magidor-Shelah "Martin's Maximum, saturated ideals, and nonregular ultrafilters I"

The paper "Martin's Maximum, saturated ideals, and nonregular ultrafilters I" by Foreman-Magidor-Shelah seems to be perhaps the most influential set theory paper in recent times (let's use a loose definition of "recent"). In this paper, they introduce the "maximal" forcing axiom known as Martin's Maximum, and then prove some consistency results from large cardinals about the saturation of "natural" ideals. We will focus on the second thing; for example:

1. Forcing with $\rm{Col}(\mu,<\kappa)$ makes the nonstationary ideal restricted to $[\mu]^{<\mu}$ precipitous.

2. Forcing with  $\rm{Col}(\omega_1,<\kappa)$ gives an $\aleph_2$-saturated ideal on $\omega_1$.

3. To make the nonstationary ideal on $\omega_1$ saturated, we must force with something more than the collapse. But this is also possible and can be done with $\kappa$-c.c. semiproper forcing.

All of the conclusions above are also consequences of Martin's Maximum.

Recall the definition and motivation of precipitous ideals from the beginning of a previous post. In that post, we gave the Jech--Magidor--Mitchell--Prikry construction of a precipitous ideal on $\omega_1$. These authors in fact proved that the nonstationary ideal on $\omega_1$ can be precipitous. The method was by an iterated forcing that at each step destroyed the stationarity of sets in the ideal constructed so far.

However, the FMS paper managed to find a different consistency proof of precipitousness of the nonstationary ideal on $\omega_1$. In fact, the proof is general to any other regular cardinals, unlike the JMMP construction which uses something special about $\omega_1$ for destroying stationarity (although generalizations of that construction to other cardinals has been carried out in work of Gitik). The drawback is that the FMS construction uses quite large cardinals, whereas Gitik found the equiconsistency.

I plan to at least cover (1.) and (2.) in a forthcoming series of posts. The plan is to prove (3.) sometime in a future learning project dealing with semiproper forcing, RCS iterations, and the other parts of the FMS paper.

The posts will come out slowly because I want to make sure I fully understand each step. The references will be the FMS paper and Section 8 of Foreman's Handbook chapter.

Spencinar #3: Precipitous ideals (part 1)

This week I'll continue the theme of Spencinar #1 by introducing some more concepts around the idea of generic ultrapowers by ideals. The emphasis will be on using large cardinals to construct ideals on small cardinals. The material comes mostly from Foreman's comprehensive chapter in the Handbook.

Prerequisites: Basic theory of large cardinals, forcing, and ultrapowers, some ideas in the first Spencinar post. The Boolean algebra approach to forcing will be helpful.

Recall the basic situation. If $I$ is an ideal on a set $Z$, then we can force with the poset $P(Z)/I$ as defined in Spencer's talk. Its elements are the mod $I$ equivalence classes of the positive subsets of $P(Z)$, ordered by the subset relation (also taken mod $I$). We denote the class of $S\subseteq Z$ by $[S]_I$. This poset is often called the quotient algebra of $I$. The generic object $G$, obtained by forcing with this poset is (equidefinable with) a $V$-ultrafilter, and so in the extension $V[G]$ we can take the $V$-ultrapower of $V$ by $G$. We denote this ultrapower by $V^Z/G$.

The basic definition, due to Jech and Prikry, is:

Definition: An ideal $I$ on a set $Z$ is precipitous if it is forced by $P(Z)/I$  that:

$V^Z/G$ is well-founded (in V[G]).

There are several combinatorial characterizations of precipitousness for an ideal $I$.

Let's give some nontrivial examples of this. A very basic result is that saturated ideals are precipitous.

Proposition: Suppose $I$ is $\kappa$-complete and $\kappa^+$-saturated. Then $I$ is precipitous.

Proof: Suppose that $I$ is not precipitous, so we can find $[S]_I$ forcing that $V^Z/G$ is ill-founded. Then let $\langle\dot{F}_n:n<\omega\rangle$ be $P(Z)/I$-names for functions $Z\rightarrow V$ in $V$ so that $[S]_I$ forces that for every $n$, $[\dot{F}_{n+1}]_G \in [\dot{F}_n]_G$ in the ultrapower.

The key is that one can replace the  $\langle\dot{F}_n:n<\omega\rangle$ by actual functions in $V$ without changing $S$. Let $n<\omega$. Find a maximal $P(Z)/I$ antichain $\mathcal{A}$ below $[S]_I$ deciding the value of $\dot{F}_n$. By a "disjointifying" argument using saturation and completeness, we can find a system of pairwise disjoint representatives $\langle A_i:i\in \mathcal{A}\rangle$ for the elements of this antichain. Define $f_n(z)$ to be the value of $\dot{F}_n$ forced by $[A_i]_I$ if $z\in A_i$, and 0 otherwise. Then we can see that $[S]_I$ forces $[\dot{F}_n]_G = [f_n]_G$. 

But now (using countable completeness of $I$), in $V$, for almost every $z\in S$, $f_{n+1}(z)\in f_n(z)$ for every $n$, giving an ill-foundedness in $V$. $\Box$

So the generic ultrapowers we considered last time (which didn't exist) were well-founded.

An immediate question is: can there be a precipitous ideal on small cardinals, e.g., $\omega_1$? The generic ultrapower construction and absoluteness of constructibility show that existence of any precipitous ideal implies that it is consistent that there is an embedding $L\rightarrow L$, so existence of a precipitous ideal has consistency strength at least that of the existence of $0^\#$.

We construct a precipitous ideal in a forcing extension starting from a measurable cardinal. The ideal is a fundamental one--start from a large cardinal ideal $J$, force to collapse the large cardinal to a small cardinal, and show that the ideal $I$ generated by $J$ in the extension retains some of the nice properties of $J$.

Actually, we'd like our precipitous idea to be a natural ideal, e.g., the nonstationary ideal on $\omega_1$. This is possible. However, today we will only get precipitousness for an induced (read: unnatural) ideal, that is an ideal which is defined to be the collection of subsets forced not to be in a particular ultrafilter of the form $U(\hat{j},i)=\{X:i\in \hat{j}(X)\}$ derived by a certain generic embedding $\hat{j}$. (We'll see this in action below).

Theorem: Suppose $\kappa$ is a measurable cardinal and let $U$ be a $\kappa$-complete normal ultrafilter on $\kappa$. Let $\mathbb{P}=\mathrm{Col}(\omega, <\kappa)$ be the Levy poset collapsing $\kappa$ to $\omega_1$. Then the ideal $I$ in $V^\mathbb{P}$ generated by the dual of $U$ is precipitous in $V^\mathbb{P}$.

Proof. By the basic theory of elementary embeddings and large cardinals, the ultrapower embedding $j:V\rightarrow M$ given by $U$ can be extended to $j^+:V[G]\rightarrow M[G\ast H]$, where $G$ is generic for $\mathbb{P}$ and $G\ast H$ is $V$-generic for $j(\mathbb{P})$ (here we used that $\mathbb{P}$ completely embeds into $j(\mathbb{P})=Col(\omega, <j(\kappa))^M$, and we'll write $j(\mathbb{P})=\mathbb{P}\ast \mathbb{Q}$). The extension takes the object $\mathbb{P}$-named by $\dot{\tau}$ (evaluated by $G$) to the object $j(\mathbb{P})$-named by $j(\dot{\tau})$ (evaluated by $G\ast H$).

The correct way to think of this is that forcing with $\mathbb{Q}$ over $V$ adds the embedding $j^+$.

First we show that (in $V[G]$) 

Claim: $I$ is equal to the ideal $\{X \subset \kappa : 1\Vdash  \kappa \not\in  j^+(X) \}$. 

Clearly, $I$ is contained in this ideal. Conversely, work in $V$ and suppose $\dot{X}$ is a $\mathbb{V}$ name for a subset of $\kappa$, and $p\in \mathbb{P}$ forces that $1\Vdash_\mathbb{ Q} \kappa \not\in  j^+(X)$. Define $A=\{\alpha:p\Vdash\alpha\not\in \dot{X}\}$. Then $A\in U$ (and so $p$ forces that $\dot{X}\in \dot{I}$, finishing the proof), otherwise we can define $F:(\kappa \setminus A)\rightarrow \mathbb{P}/p$ so that $F(\alpha)$ forces $\alpha\in X$. But then in $M$, $F$ represents a condition below $j(p)=p$ which forces $\kappa$ into $j^+(X)$, contradicting the choice of $p$. This proves the claim.

Now we show the precipitousness. Work in $V[G]$. Define a Boolean algebra embedding $P(\kappa)\rightarrow \mathcal{B}(\mathbb{Q})$ (here the codomain is the Boolean completion of $\mathbb{Q}$) which sends a set $X$ to the truth value $\lVert \kappa\in j^+(X)\rVert$. The use of the Boolean completion is so that this map is defined--in general, many different members of $\mathbb{Q}$ could force $\kappa\in j^+(X)$, but in the Boolean completion there is a greatest such member. Since $I$ is the kernel of this embedding, we see that the embedding factors to $\iota:P(\kappa)/I\rightarrow \mathcal{B}(\mathbb{Q})$. We will show that $\iota$ gives a dense embedding from the quotient algebra of $I$ into $\mathbb{Q}$. 

Claim: The range of $\iota$ is dense in $\mathcal{B}(\mathbb{Q})$. 

Since $\mathbb{Q}$ is densely embedded in $\mathcal{B}(\mathbb{Q})$, it suffices to show that for any $q\in\mathbb{Q}$ there is $X\subseteq \kappa$ with $\iota(X)=q$. Let $q=j(F)(\kappa)$, where $F\in V$ is a function such that for all $\alpha$, $F(\alpha)\in \mathrm{Col}(\omega, <\kappa)$. Define $X=\{\alpha: F(\alpha)\in G\}$. Now for any $H$ which is $V[G]$ generic for $\mathbb{Q}$, we have

 $\iota(X)\in H$ iff $\kappa \in j^+(X)$ (definition of $\iota$)

 iff $j(F)(\kappa)\in j^+(G)$ (definition of $X$)

iff $q\in j^+(G)$ (definition of $F$)

iff $q\in H$

so by separativity, $\iota(X)=q$, proving the claim.

So forcing with the quotient algebra of $I$ is equivalent to forcing with $\mathbb{Q}$; generic filters are equidefinable using the $\iota$ function. So let $H$ be obtained by applying $\iota$ image to an arbitrary $V[G]$-generic filter for $P(Z)/I$.

Now in $V[G\ast H]$, let $U^+=\{X\subseteq \kappa: \kappa\in j^+(X)\}$ be the ultrafilter derived from $j^+$, so we know from general considerations that the ultrapower of $V[G]$ by $U^+$ embeds into $M[G\ast H]$ (the embedding takes the element represented by a function $f $ to $j^+(f)(\kappa)$). But $U^+$ is exactly the filter generated the $\iota$-preimage of $H$, by the definition of $\iota$, so it is the original generic for $P(Z)/I$.