> [!info]
>
We know that if $j:V\to M$ is a non-trivial elementary embedding where $M$ is an inner model of $V$, then $\mathrm{crit}\left(j\right)$ is a measurable cardinal, which is (in $\mathrm{ZFC}$) rather large. So if we want to have embeddings with low critical points, such as $\omega_{1}$ (without waiving AC) we need something else: a *generic embedding* is an elementary embedding $j:V\to M\subseteq V\left[G\right]$ where $G$ is a $V$-generic filter (for some forcing notion).
# Ideals
## Basic definitions
1. For any set $Z$, an *$I\subseteq\mathcal{P}(Z)$ is an ideal on $Z$* if $I$ is closed under finite unions and under subsets.
2. $I$ is *proper* if $Z\notin I$.
3. $I$ is *non-principal* if it contains all singletons from $Z$.
From now on all ideals are proper and non-principal
4. $\breve{I}=\left\{ A\subseteq Z\mid Z\mathord{\smallsetminus}A\in I\right\}$ is the *dual filter* of $I$.
5. $I^{+}=\mathcal{P}\left(Z\right)\mathord{\smallsetminus}I$ are the *$I$-positive* sets.
6. A maximal proper ideal is called *prime*.
Note that $I$ is prime iff $\breve{I}$ is an ultrafilter iff $\breve{I}=I^{+}$.
7. $I$ is *$\kappa$-complete* if it is closed under unions of size
lt;\kappa$.
Equivantly, any partition of $Z$ of size lt;\kappa$ contains an element of $I^{+}$.
8. The *completeness* of an ideal $I$ - $\mathrm{comp}(I)$ is the least $\kappa$ such that $I$ is not $\kappa^{+}$-complete (so $I$ is closed under unions of size lt;\kappa$ but not of size $\kappa$).
9. $\mathcal{A}\subseteq I^{+}$ is called an *antichain* if for every distinct $X,Y\in\mathcal{A}$, $X\cap Y\in I$
10. An ideal $I$ is called *$\lambda$-saturated* if every antichain is of cardinality lt;\lambda$.
11. An ideal $I$ has the *disjointing property* if any antichain has a disjoint set of representitives.
Assume $Z\subseteq\mathcal{P}(X)$ for some set $X$, so $I\subseteq\mathcal{P}(Z)\subseteq\mathcal{P}(\mathcal{P}(X))$ .
12. *Normal* if for every function $f:A\to X$ where $A\in I^{+}$ which is regressive ($\forall a\in A\,f(a)\in a)$ there is $B\in I^{+}$, $B\subseteq A$ such that $f$ is constant on $B$.
13. *Fine* if for every $x\in X$ $\left\{ z\in Z\mid x\in z\right\} \in\breve{I}$.
14. An ideal $I$ on $\omega_1$ is called *presaturated* if for every $I$-positive $T \subset \omega_1$ and every countable set $\mathcal{A}_i(i<\omega)$ of maximal antichains in $\mathcal{P}\left(\omega_1\right) / I$ there exists an $I$-positive $T^{\prime} \subset T$ such that for each $i<\omega,\left\{[S] \in \mathcal{A}_i \mid S \cap T^{\prime} \in I^{+}\right\}$ has cardinality less than or equal to $\aleph_1$.
>[!example]
>The canonical example for a normal fine ideal is the [[Club sets and stationary sets#^ns-ideal|non-stationary ideal]].
## Forcing
> [!info] Definition
> $I^{+}$ is considered a [[forcing]] notion under the pre-order given by $X\leq Y$ ($X$ is stronger than $Y$) iff $X\subseteq Y$.
This is [[forcing equivalent]] to the [[Boolean algebra]] $\mathcal{P}(Z)/I$ - the set of equivalence classes of the relation $X\sim_{I}Y$ iff $X\triangle Y\in I$, ordered by $\left[X\right]_{I}\leq\left[Y\right]_{I}$ iff $X\mathord{\smallsetminus}Y\in I$.
>[!note] Remark
>An ideal is $\lambda$-saturated iff every antichain in this forcing is of cardinality lt;\lambda$ i.e. iff it has $\lambda$-[[Chain conditions|chain condition]].
**Proposition.** Let $G\subseteq I^{+}$ be a generic filter. Then $G$ is a $V$-ultrafilter on $Z$ extending $\breve{I}$, and if $I$ is $\kappa$-complete then $G$ is $\kappa$-complete.
> [!info] Definition - the *Generic embedding*
> Using a generic $G\subseteq I^{+}$, we can define an ultrapower of $V$ in the usual way: for $f,g:Z\to V$ such that $f,g\in V$, define $\begin{aligned}
f\sim_{G}g & \iff\left\{ x\in Z\mid f(x)=g(x)\right\} \in G\\
fE_{G}g & \iff\left\{ x\in Z\mid f(x)\in g(x)\right\} \in G\end{aligned}$ and let $V^{Z}/G=Ult(V,G)=\left\{ \left[f\right]_{G}\mid f:Z\to V,f\in V\right\}$.
We can verify that Łoś's theorem holds: $Ult(V,G)\vDash\varphi([f_{1}]_{G},\dots,[f_{n}]_{G})\iff\left\{ x\in Z\mid V\vDash\varphi\left(f_{1}(x),\dots,f_{n}(x)\right)\right\} \in G$ and that $j(a)=\left[c_{a}\right]_{G}$ where $c_{a}(x)=a$ for all $x\in Z$ gives an elementary embedding $j:V\to Ult(V,G)\subseteq V\left[G\right]$.
> [!note] Remark
> This ultrapower however might not be well-founded, even e.g. when $I$ is a $\kappa$-complete ideal over $\kappa$.
>
> If $Ult(V,G)$ is well-fonded we identify it with its transitive collapse $M$ and write $j:V\to M\cong Ult(V,G)$.
## Precipitousness
>[!info] Definition
>And ideal $I$ is called *precipitous* if for every generic $G\subseteq I^{+}$, $Ult(V,G)$ is well-founded.
>[!example]
>Let $\kappa>\omega$ be regular. Then $I=\left\{ X\subseteq\kappa\mid\left|X\right|<\kappa\right\}$ is a non-precipitous $\kappa$-complete ideal.
>[!info] Definition
> 1. For $S\in I^{+}$, an *$I$-partition of $S$* is a set $W\subseteq I^{+}$ such that for every $X\ne Y$ in $W$, $X\cap Y\in I$. (This is essentially a maximal antichain below $S$ in the forcing notion.)
> 2. For $I$-partitions $W_{1},W_{2}$, denote $W_{1}\leq W_{2}$ ($W_{1}$ *refines* $W_{2}$) if $\forall X\in W_{1}\exists Y\in W_{2}(X\subseteq Y)$.
> 3. A *functional on $S$* is a collection of functions, say *$F$,* such that $W_{F}:=\left\{ \operatorname{Dom}(f)\mid f\in F\right\}$ is an $I$-partition of $S$.
> 4. For functionals $F_{1},F_{2}$, denote $F_{1}<F_{2}$ if
> 1. each $f\in F_{1}\cup F_{2}$ is a function to ordinals
> 2. $W_{F_{1}}\leq W_{F_{2}}$
> 3. if $f\in F_{1}$ and $f'\in F_{2}$ are such that $\operatorname{Dom}(f)\subseteq\operatorname{Dom}(f')$ then $f(z)<f'(z)$ for all $z\in\operatorname{Dom}(f)$
**Lemma.** The following are equivalent for an ideal $I$:
1. $I$ is precipitous.
2. There is **no** $S\in I^{+}$ with a sequence $F_{0}>F_{1}>\dots>F_{n}>\dots$ of functionals.
3. For every $S\in I^{+}$, if $\left\{ W_{n}\mid n<\omega\right\}$ is a descending sequence of $I$-partitions of $S$, there are $X_{0}\supset X_{1}\supset\dots\supset X_{n}\supset\dots$ such that for every $n$ $X_{n}\in W_{n}$ and $\bigcap_{n<\omega}X_{n}\ne\varnothing$.
## Completeness
**Proposition.** If $I$ is $\kappa$-complete then $Ult(V,G)$ is well-founded up to $\kappa$, i.e every $\alpha<\kappa$ is in the well-founded part.
**Proposition.** Every precipitous ideal is countably complete.
>[!note] Notation
> If $I$ an ideal on $Z$ and $S\in I^{+}$ then $I\mathord{\upharpoonright}S$ denotes either the ideal $I\cap\mathcal{P}(S)$ on $S$ or the ideal on $Z$ generated by $I\cup\left\{ Z\mathord{\smallsetminus}S\right\}$. These two ideals give isomorphic forcing notions so the abuse of notation makes sense.
**Proposition.** If $I$ is a precipitous ideal on $Z$, $G\subseteq I^{+}$ generic with $j:V\to M\subseteq V[G]$, then $j$ is not the identity, and $\mathrm{crit}\left(j\right)$ is the largest $\kappa$ such that there is $S\in G$ with $\mathrm{comp}(I\mathord{\upharpoonright}S)=\kappa$.
### An application
**Theorem.** (CH) If there is a precipitous ideal on $\omega_{1}$ then every $\Sigma_{2}^{1}$ set of reals is Lesbegue measurable.
Saturation
----------
An ideal $I$ is called $\lambda$-saturated if every antichain in $\mathcal{P}(Z)/I$ (equivalently -- every $\mathcal{A}\subseteq I^{+}$ such that for every distinct $X,Y\in\mathcal{A}$ $X\cap Y\in I$) is of cardinality lt;\lambda$.
The saturation of $I$ $\mathrm{sat}(I)$ is the least $\lambda$ such that $I$ is $\lambda$-saturated. Note that $\mathrm{sat}(I)\leq\left(2^{\left|Z\right|}\right)^{+}$, and it can be shown to be a regular cardinal.
**Proposition.** Every $\kappa$-complete $\kappa^{+}$-saturated ideal is precipitous.
The usual proof of this proposition shows that what is actually used is the following:
If $I$ is a $\kappa$-complete $\kappa^{+}$-saturated ideal then every antichain in $\mathcal{P}(Z)/I$ has a disjoint set of representitives.
By $\kappa^{+}$-saturation, any antichain is of the form $\left\langle \left[A_{\alpha}\right]\mid\alpha<\gamma\right\rangle$ for some $\gamma\leq\kappa$. By $\kappa$-completeness, for all $\beta<\kappa$ $A_{\beta}^{*}=A_{\beta}\mathord{\smallsetminus}\bigcup_{\alpha<\beta}A_{\alpha}=A_{\beta}\mathord{\smallsetminus}\bigcup_{\alpha<\beta}A_{\alpha}\cap A_{\beta}\in I^{+}$ and $A_{\beta}^{*}\sim_{I}A_{\beta}$.
An ideal $I$ has the *disjointing property* if any antichain has a disjoint set of representitives.
If $I$ is countably complete with the disjointing property then
1. $I$ is precipitous.
2. If $j:V\to M\subseteq V[G]$ is the derived generic embedding with $\mathrm{crit}\left(j\right)=\kappa$ then $\prescript{\kappa}{}{M}\cap V[G]\subseteq M$.
1\. Let's use (3) of lemma [\[lem:equiv-precipitous\]](#lem:equiv-precipitous){reference-type="ref" reference="lem:equiv-precipitous"}. Let $S\in I^{+}$, and $\left\{ W_{n}\mid n<\omega\right\}$ is a descending sequence of $I$-partitions of $S$. These are antichains. By the disjointing property, we can inductively replace each $W_{n}$ by a disjoint $W_{n}^{*}$, keeping the $\geq$ relation. $J=I\mathord{\upharpoonright}S$ is also countably complete, and for every $n$ $\bigcup W_{n}^{*}\in\breve{J}$, so by countable completeness also $\bigcap_{n}\bigcup W_{n}^{*}\in\breve{J}$, so for every $x\in\bigcap_{n}\bigcup W_{n}^{*}$ there is a descending sequence $X_{i}\in W_{n}^{*}\subseteq W_{n}$ with non-empty intersection.
For 2 we first need a general lemma:
If $I$ has the disjointing property, and $\dot{f}$ is an $I^{+}$-name for a function from $Z$ to $V$ that lies in $V$, then there is some $g\in V$ ($g:Z\to V$) such that $\Vdash[\dot{f}]=\left[\check{g}\right]$.
As we noted before, a name for a function $\dot{f}$ corresponds to a functional $F\in V$ (on $Z$). $W_{F}$ is a antichain, so by the disjointing property there is a disjoint $I$-partition of $Z$ $W=\left\{ X'\mid X\in W_{F}\right\}$ such that $X\sim_{I}X'$. So for every $x\in\bigcup W$, define $g(x)=f(x)$ for the unique $f\in F$ such that $x\in\operatorname{Dom}(f)'$, and for $x\in Z\mathord{\smallsetminus}\bigcup W$ $g(x)=0$. Then $\Vdash[\dot{f}]=\left[\check{g}\right]$.
Back to the proof of 2. Let $X=\left\langle X_{\alpha}\mid\alpha<\kappa\right\rangle \in\prescript{\kappa}{}{M}\cap V[G]$. Let $\dot{X}=\left\langle \dot{X}_{\alpha}\mid\alpha<\kappa\right\rangle$ be a name for such an $X$. By the lemma, there are functions $g_{\alpha}\in V$ such that for every $\alpha$, $\Vdash\dot{X}_{\alpha}=\left[\check{g}_{\alpha}\right]$. Let $Y\in G$ such that $Y\Vdash\kappa=\mathrm{crit}\left(j\right)$. So by the lemma there is some function $k\in V$ such that $Y\Vdash\kappa=\left[k\right]$. Let $f(x)=\left\langle g_{\alpha}(x)\mid\alpha<k(x)\right\rangle$, $f:Z\to V$. Then $\left[f\right]$ is a function from $\left[k\right]=\kappa$ such that for all $\alpha$, $\left[f\right]_{\alpha}=\left[g_{\alpha}\right]=X_{\alpha}$, so $\left[f\right]=X\in M$.
Let $\kappa>\omega$ regular and $I$ a $\kappa^{+}$-saturated $\kappa$-complete ideal on $\kappa$. If $\forall\lambda<\kappa$ $2^{\lambda}=\lambda^{+}$ then $2^{\kappa}=\kappa^{+}$.
Let $G\subseteq I^{+}$ be generic, $j:V\to M\subseteq V[G]$ the derived embedding where, by the previous lemmas, $M$ is well founded, $\mathrm{crit}\left(j\right)=\kappa$, and $\prescript{\kappa}{}{M}\cap V[G]\subseteq M$.
By elementarity, $M\vDash\forall\lambda<j(\kappa)\,\,2^{\lambda}=\lambda^{+}$ so in particular $M\vDash\left|\mathcal{P}(\kappa)\right|=\kappa^{+}$. As we already noted, for every $X\subseteq\kappa$ $X=j(X)\cap\kappa\in M$ so $\mathcal{P}^{V}(\kappa)\subseteq\mathcal{P}^{M}(\kappa)$. Since $M$ is an inner model of $V[G]$ we have $\left(\kappa^{+}\right)^{M}\leq\left(\kappa^{+}\right)^{V[G]}$.[^1] So (in $V[G]$) $\left|\mathcal{P}^{V}(\kappa)\right|\leq\left|\mathcal{P}^{M}(\kappa)\right|\leq\left(\kappa^{+}\right)^{M}\leq\left(\kappa^{+}\right)^{V[G]}$ Note that $\kappa^{+}$-saturation means that the forcing is $\kappa^{+}$-cc, so it doesn't collapse cardinals $\geq\kappa^{+}$, so we have (still in $V[G]$) $\left|\mathcal{P}^{V}(\kappa)\right|\leq\left(\kappa^{+}\right)^{V[G]}=\left(\kappa^{+}\right)^{V}$ which was also true in $V$ since $\left|\mathcal{P}^{V}(\kappa)\right|^{V}$ wasn't collapsed.
Normality & fineness
--------------------
As in the case of regular ultrapowers, the generic embeddings are generated by the identity function $\operatorname{Id}:Z\to Z$:
If $I$ is an ideal on $Z$, $G\subseteq I^{+}$ generic and $j:V\to Ult(V,G)$ the derived embedding, then
1. For $A\subseteq Z$, $A\in G$ iff $\left[\operatorname{Id}\right]_{G}Ej(A)$;
2. For $g:Z\to V$ with $g\in V$, $\left[g\right]_{G}=j(g)(\left[\operatorname{Id}\right]_{G})$
This is the same as in regular ultrapowers:
1\. $\left[\operatorname{Id}\right]_{G}Ej(A)$ iff $G\ni\left\{ z\in Z\mid\operatorname{Id}(z)\in c_{A}(z)\right\} =\left\{ z\in Z\mid z\in A\right\} =A$.
2\. By Łoś theorem, $\left[g\right]_{G}=j(g)(\left[\operatorname{Id}\right]_{G})=\left[c_{g}\right]_{G}\left(\left[\operatorname{Id}\right]_{G}\right)$ iff $G\ni\left\{ z\in Z\mid g(z)=c_{g}(z)(\operatorname{Id}(z))\right\} =\left\{ z\in Z\mid g(z)=g(z)\right\} =Z$
Identifying what set does $\left[\operatorname{Id}\right]_{G}$ represents usually requires further assumptions:
Let $Z\subseteq\mathcal{P}(X)$ for some set $X$. An ideal $I$ on $Z$ (so $I\subseteq\mathcal{P}(Z)\subseteq\mathcal{P}(\mathcal{P}(X))$ ) is called
- *Normal* if for every function $f:A\to X$ where $A\in I^{+}$ which is regressive ($\forall a\in A\,f(a)\in a)$ there is $B\in I^{+}$, $B\subseteq A$ such that $f$ is constant on $B$.
- *Fine* if for every $x\in X$ $\left\{ z\in Z\mid x\in z\right\} \in\breve{I}$.
Note that for an ordinal $\kappa$, $\kappa\subseteq\mathcal{P}(\kappa)$ and normality for an ideal on $\kappa$ as defined above is the usual normality.
If $I$ is a normal fine ideal on $Z\subseteq\mathcal{P}(\lambda)$, $G\subseteq I^{+}$ generic and $j:V\to Ult(V,G)$ the generic embedding, then $\left[\operatorname{Id}\right]_{G}=j''\lambda$.
As a corollary, for every $A\subseteq\lambda$, $j(A)\cap j''\lambda$ is represented by the function $f_{A}(z)=A\cap z$, and every $\alpha<\lambda$ is represented by the function $z\mapsto\operatorname{otp}(z\cap\alpha)$.
If $I$ is $\lambda$-complete then the ultrapower is well-founded up to $\lambda$ so $\left[\operatorname{Id}\right]_{G}=\lambda$, and $\left[\operatorname{Id}\right]_{G}<j(\lambda)$ so $\lambda$ is the critical point.
For $\alpha\in\lambda$, $j(\alpha)E\left[\operatorname{Id}\right]$ iff $\left\{ z\in Z\mid\alpha\in z\right\} \in G$, which is true by fineness.
If $I$ is normal then $G$ is a $V$-normal ultrafilter, i.e. normal with respect to functions in $V$, since if $S=\left\{ z\in Z\mid f(z)\in z\right\} \in I^{+}$ then the set of $S'\leq S$ on which $f$ is constant is dense below $S$. So the claim follows as in the usual case: for any $f:Z\to V$ in $V$, $\left[f\right]E\left[\operatorname{Id}\right]$ iff $\left\{ z\in Z\mid f(z)\in z\right\} \in G$ i.e. $f$ is regressive on a set in $G$, so by normality it is constant with some value $\beta$ on a set in $G$, hence $\left[f\right]=j(\beta)$.
For the corollary, $j(A)\cap j''\lambda=\left[c_{A}\right]\cap\left[\operatorname{Id}\right]$, and this is equal $\left[f_{A}\right]$ since $\left\{ z\in Z\mid c_{A}(z)\cap\operatorname{Id}(z)=f_{A}(z)\right\} =Z.$ For $\alpha<\lambda$ note that $\alpha=\operatorname{otp}\left(j(\alpha)\cap j''\lambda\right)=\operatorname{otp}\left(\left[f_{\alpha}\right]\right)$ so $\alpha$ is represented by $\operatorname{otp}\circ f_{\alpha}$.
If $I$ is a normal fine precipitous ideal on $Z\subseteq\mathcal{P}(X)$ where $\left|X\right|=\lambda$ with a gen.emb. $j:V\to M\subseteq V[G]$ then $\mathcal{P}(\lambda)\cap V\subseteq M$. If $I$ has the disjointing property then also $M^{\lambda}\cap V[G]\subseteq M$.
WLoG assume $X=\lambda$. Let $A\in\mathcal{P}(\lambda)\cap V$. If $x\in A\subseteq\lambda$, then $j(x)\in j''\lambda$, and also $j(x)\in j(A)$, so $j(x)\in j''\lambda\cap j(A)$ which is in $M$ by the previous lemma. On the other hand, if $a\in j''\lambda\cap j(A)$, then on one hand $a=j(x)$ for some $x\in V$, but then we must have by elementarity $x\in A$. Note that if $\pi:j''\lambda\to\lambda$ is the transitive collapse, then $\pi^{-1}=j\mathord{\upharpoonright}\lambda$, so the $A=j^{-1}\left[j''\lambda\cap j(A)\right]$ can be computed in $M$.
Now assume $I$ has the disjointing property and let $X=\left\langle X_{\alpha}\mid\alpha<\kappa\right\rangle \in M^{\lambda}\cap V[G]$. As before let $\dot{X}=\left\langle \dot{X}_{\alpha}\mid\alpha<\kappa\right\rangle$ be a name for such an $X$ and $g_{\alpha}\in V$ such that for every $\alpha$, $\Vdash\dot{X}_{\alpha}=\left[\check{g}_{\alpha}\right]$. Let $\mathcal{G}=\left\langle g_{\alpha}\mid\alpha<\lambda\right\rangle$ and denote $j(\mathcal{G})=\left\langle j(g)_{\alpha}\mid\alpha<j(\lambda)\right\rangle$ (if $\alpha=j(\beta)$ then $j(g)_{\alpha}=j(g_{\beta})$ ).
Define $g:Z\to V$ by $g(z)=\left\langle g_{\alpha}(z)\mid\alpha\in z\right\rangle$. Then $\left[g\right]=j(g)(j''\lambda)=\left\langle j(g)_{\beta}(j''\lambda)\mid\beta\in j''\lambda\right\rangle =\left\langle j(g_{\alpha})\mid\alpha\in\lambda\right\rangle \in M$ so also $\left\langle \left[g_{\alpha}\right]\mid\alpha<\lambda\right\rangle =\left\langle j(g_{\alpha})(j''\lambda)\mid\alpha<\lambda\right\rangle \in M.$
The non-stationary ideal
------------------------
The collection of non-stationary sets on a regular uncountable cardinal $\kappa$ is, by Fodor's lemma, a normal ideal. Let's indroduce the generalized notion of stationarity.
Let $X$ be some (infinite) set, $Z=\mathcal{P}(X)$.
1. An *algebra* on $X$ is a structure $\mathfrak{A}=\left\langle X,f_{n}\right\rangle _{n<\omega}$ where $f_{n}:X^{k}\to X$ for some $k$.
2. $C\subseteq\mathcal{P}(X)$ is called *closed unbounded* if there is some algebra $\mathfrak{A}$ such that $C=C_{\mathfrak{A}}=\left\{ z\subseteq X\mid\forall n(f_{n}''z^{k}\subseteq z)\right\}$ i.e the collection of all subsets of $X$ closed under all functions of $\mathfrak{A}$.
3. $S\subseteq\mathcal{P}(X)$ is called *stationary* if for every algebra $\mathfrak{A}$ $S\cap C_{\mathfrak{A}}\ne\varnothing$, i.e for any countable collection of finitary functions on $X$, $S$ contains a set closed under all of them.
4. The sets of the form $C_{\mathfrak{A}}$ form a filter, called the *closed unbounded (club) filter,* and its dual is the *non-stationary ideal* $\mathrm{NS}_{X}$ whose positive sets are the stationary sets.
- Using Skolem functions, the club filter is also generated by club sets which consist of all *elementary* substructures of some structure on $X$. And vice-versa -- a stationary set is one that contains an elementary substructure of any structure on $X$.
- The same notion can be achieved using single functions $F:\left[X\right]^{<\omega}\to X$, where a club is the set of all sets closed under such $F$ and a stationary is a set such that for for each such $F$ has a member closed under $F$ .
- To get the usual notion of stationarity for $X=\kappa$ regular we restrict the club filter to the stationary set $\kappa$: let $C\subseteq\kappa$ be club in the usual sense contains. For $\alpha<\kappa$, let $\gamma(\alpha)>\alpha$ be the minimal limit above $\alpha$, and define $f(\alpha)=\begin{cases}
\alpha+1 & \gamma(\alpha)\in C\\
\gamma(\alpha) & \gamma(\alpha)\notin C
\end{cases}$ and we want to show that $C_{F}\cap\kappa\subseteq C$. Assume $\gamma$ is closed under $f$. Note that $\gamma$ must be limit. If $\gamma\notin C$ then since $C$ is closed, there must be a maximal limit $\delta\in C\cap\gamma$, so $f(\delta)=\gamma$ by contradiction. On the other hand, any $C_{\mathfrak{A}}\cap\kappa$ is closed unbounded using an elementary substructure argument. hence the club filter restricted to $\left\{ z\subseteq\kappa\mid z\cap\kappa\in\kappa\right\} =\kappa$ is the usual club filter.
The non-stationary ideal on $X$ is normal and fine.
Let $S$ be stationary and $f:S\to X$ regressive. We want to claim that there is some $x\in X$ such that $S_{x}=\left\{ z\in S\mid f(z)=x\right\}$ is stationary. Assume otherwise -- each such set is non-stationary. Let $C_{x}$ witness this, i.e $C_{x}$ is a club such that $S_{x}\cap C_{x}=\varnothing$. For simplicity we'll use the definition using one function -- $f_{x}:\left[X\right]^{<\omega}\to X$ defines $C_{x}$. So define $g(x,x_{0},\dots,x_{n-1})=f_{x}(x_{0},\dots,x_{n-1})$, and let $z\in S$ be closed under $g$. So if $x\in z$, then $z$ is closed under $f_{x}$, i.e $z\in C_{x}$. $f$ is regressive so $f(z)\in z$, hence $z\in C_{f(z)}$ but also trivially $z\in S_{f(z)}$, a contradiction.
For any $x\in X$, $\left\{ z\subseteq X\mid x\in z\right\}$ is the set of all $z$ closed under $c_{x}(a)=x$.
The nonstationary ideal is the smallest normal fine ideal:
Let $I$ be a normal fine countably complete ideal on $Z\subseteq\mathcal{P}(X)$. If $A\subseteq Z$ is nonstationary then $A\in I$.
Assume towards contradiction $A\notin I$. Let $\mathfrak{A}=\left\langle X,f_{n}\right\rangle _{n<\omega}$ be an algebra witnessing the nonstationarity of $A$, so each $z\in A$ is not closed under some $f_{n}$. For each $z\in A$ choose some $\left\langle x_{0},\dots,x_{n-1}\right\rangle$ witnessing this for $f_{n}$. By normality and countable completeness, there is some $B\subseteq A$, $B\in I^{+}$ such that for every $z\in B$ the witnesses are the same (using some tricks). So in particular $B\cap\left\{ z\subseteq X\mid f_{n}(x_{0},\dots,x_{n-1})\in z\right\} =\varnothing$, but this contradicts the fineness of $I$.
The stationary tower
====================
The full tower
--------------
Let $\varnothing\ne X\subseteq Y$
1. If $S\subseteq\mathcal{P}\left(Y\right)$ is stationary then $S\mathord{\downarrow}X=\left\{ Z\cap X\mid Z\in S\right\}$ is stationary
2. If $S\subseteq\mathcal{P}\left(X\right)$ is stationary then $S\mathord{\uparrow}Y=\left\{ Z\in\mathcal{P}\left(Y\right)\mid Z\cap X\in S\right\}$ is stationary
1\. Let $\mathfrak{A}$ be an algebra on $X$. Define $\bar{\mathfrak{A}}$ on $Y$ such that for every $n$, $\bar{f}_{n}\mathord{\upharpoonright}X^{k_{n}}=f_{n}$ and some arbitraty value otherwise. So if $Z\in S$ is closed under $\bar{\mathfrak{A}}$ then $Z\cap X$ is closed under $\mathfrak{A}$.
2\. let $\mathfrak{B}$ be an algebra on $Y$. We can assume wlog that its functions are closed under compositions. So define $\bar{\mathfrak{B}}$ on $X$ such that for every $n$ and $\bar{x}\in X^{k}$, if $f_{n}(\bar{x})\in X$ then $\bar{f}_{n}(\bar{x})=f_{n}(\bar{x})$ and otherwise we take some fixed $x_{0}\in X$. Let $Z\in S$ be closed under $\bar{\mathfrak{B}}$. Take $\bar{Z}$ to be the closure of $Z$ in $\mathfrak{B}$, we claim that $\bar{Z}\cap X=Z$. By the assumption on compositions, any $x\in\bar{Z}$ is obtained as $f_{n}(y)$ for some $y\in Z$, but if $f_{n}(y)\in X$ then it is equal $\bar{f}_{n}(y)$ which is in $Z$ by the closure of $Z$.
Note that if $S\subseteq\mathcal{P}(X)$ is stationary, then $\cup S=X$. So we can say that any set $a$ is simply *stationary* if it is stationary in $\mathcal{P}(\cup a)$.
Let $\kappa$ be strongly inaccessible. The (full) stationary tower is $\mathbb{P}_{<\kappa}=\left\{ a\in V_{\kappa}\mid a\text{ is stationary (in \ensuremath{\mathcal{P}(\cup a)}) }\right\}$ with the pre-order $a\geq b$ ($b$ is stronger than $a$) if $\cup a\subseteq\cup b$ and $b$ projects into $a$, i.e for all $Z\in b$, $Z\cap(\cup a)\in a$ (in other words $b\mathord{\downarrow}(\cup a)\subseteq a$). Note that if $X\subseteq Y\subseteq\cup a$ then $(a\mathord{\downarrow}Y)\mathord{\downarrow}X=a\mathord{\downarrow}X$ so this relation is transitive.
Let $G\subseteq P_{<\kappa}$ be a generic filter. For every $X\in V_{\kappa}$, define $\begin{aligned}
U_{X} & =\left\{ b\mathord{\downarrow}X\mid b\in G,X\subseteq\cup b\right\} \\
& =\left\{ a\subseteq\mathcal{P}\left(X\right)\mid a\in G\text{ stationary in }\mathcal{P}(X)\right\} .\end{aligned}$
$U_{X}$ is an ultrafilter on $\mathcal{P}(X)$ that extends the club filter. It is also normal.
It is a filter since $G$ is a filter. It is ultra: let $A\subseteq\mathcal{P}\left(X\right)$, and $a\in\mathbb{P}_{<\kappa}$. Extend $a$ to some $b$ such that $X\subseteq\cup b$. Let $\begin{aligned}
b_{0} & =\left\{ Z\in b\mid Z\cap X\in A\right\} \\
b_{1} & =\left\{ Z\in b\mid Z\cap X\notin A\right\} \end{aligned}$ then one of them is stationary. If $b_{0}$ then $b_{0}\mathord{\downarrow}X\subseteq A$, if $b_{1}$ then $b_{1}\mathord{\downarrow}X\subseteq\mathcal{P}\left(X\right)\mathord{\smallsetminus}A$. So by genericity, one of them is in $U_{X}$.
If $C\subseteq\mathcal{P}\left(X\right)$ is a club, witnessed by $F:\left[X\right]^{<\omega}\to X$, for every $a\in\mathbb{P}_{<\kappa}$, extend it to some $b$ such that $X\subseteq\cup b$, and let $C'$ be the club in $\mathcal{P}\left(\cup b\right)$ defined by $F'\left(\bar{x}\right)=F\left(\bar{x}\cap X\right)$. Note that if $Z$ is closed under $F'$ then $Z\cap X$ is closed under $F$. Then $b\cap C'$ is stationary and is stronger than $C$. Thus by genericity we get that $C\in G$.
Let $a\in\mathbb{P}_{<\kappa}$ and $F:a\to X$ regressive. If $b\leq a$, set $F':b\to X$ by $F'(Z)=F(Z\cap(\cup a))$. Then there is some stationary $b'\leq b$ on which $F'$ is constant. So the set of conditions on which the function defined by $F'(Z)=F(Z\cap(\cup a))$ is constant is dense below $a$. So if $a\in U_{X}$, i.e $a\in G$, there is some $b\in G$ on which $F'$ is constant, and $a\supseteq b\mathord{\downarrow}X\in U_{X}$.
So for every $X$ we can define $j_{X}:(V,\in)\to(M_{X},E_{X})=Ult\left(V,U_{X}\right)$ using functions from $\mathcal{P}\left(X\right)$ to $V$ which are in $V$ (not necessarily well-founded). If $X\subseteq Y$, we want to define $j_{XY}:M_{X}\to M_{Y}$. For $f:\mathcal{P}(X)\to V$, define $f_{Y}:\mathcal{P}(Y)\to V$ by $f_{Y}(Z)=f(Z\cap X)$ and define $j_{XY}\left(\left[f\right]_{U_{X}}\right)=\left[f_{Y}\right]_{U_{Y}}.$ To see it is elementary, note that $S\in U_{X}$ iff $S\mathord{\uparrow}Y\in U_{Y}$. So we get a directed system of embeddings:
$\left\{ M_{X},j_{X},j_{XY}\mid X,Y\in V_{\kappa}\mathord{\smallsetminus}\left\{ \varnothing\right\} ,X\subseteq Y\right\}$ and we can check that they commute -- for $X\subseteq Y\subseteq Z$, $j_{Y}=j_{XY}\circ j_{X}$ and $j_{XZ}=j_{YZ}\circ j_{XY}$. So we can take a directed limit $j:(V,\in)\to\left(M,E\right)$. Members of $\left(M,E\right)$ can be taken as $\left[f\right]_{G}$ such that $f:a\to V$ for some $a\in G$.
[\[prop:the-identity\]]{#prop:the-identity label="prop:the-identity"}For every $X\in V_{\kappa}$, $i^{X}$ -- the identity on $\mathcal{P}\left(X\right)$ -- represents $j[X]=j''X$. In particular, $X\in M$.
To show this, we must show that for every $Y\supseteq X$, the function $i_{Y}^{X}\left(Z\right)=Z\cap X$ (the lift of $i^{X}$ to $Y$) represents $j_{Y}\left[X\right]$, since if $j_{Y}\left[X\right]=\left[i_{Y}^{X}\right]_{U_{Y}}=j_{XY}\left(\left[i^{X}\right]_{U_{X}}\right)$ then in the limit we get $j\left[X\right]=j_{X\infty}\left(\left[i^{X}\right]_{U_{X}}\right)$.
$j_{Y}\left[X\right]=\left[i_{Y}^{X}\right]_{U_{Y}}$.
For $x\in X$, $j_{Y}\left(x\right)E_{Y}\left[i_{Y}^{X}\right]_{U_{Y}}$ iff $\left\{ Z\subseteq Y\mid c_{x}\left(Z\right)\in i_{Y}^{X}\left(Z\right)\right\} =\left\{ Z\subseteq Y\mid x\in Z\cap Y\right\} \in U_{Y}$ which is true since it's a club (witnessed by the constant function $x$).
Assume $\left[f\right]_{U_{Y}}E_{Y}\left[i_{Y}^{X}\right]_{U_{Y}}$, that is, $\left\{ Z\subseteq Y\mid f\left(Z\right)\in Z\cap X\right\} \in U_{Y}$. We need to show $f$ is constant on a set in $U_{Y}$. For that we need to apply normality to some stationary set in $G$ that will project to $Y$. We use genericity. Let $c\in\mathbb{P}_{<\kappa}$ such that $Y\subseteq\cup c$ which forces that $\left[f\right]_{U_{Y}}E_{Y}\left[i_{Y}^{X}\right]_{U_{Y}}$, i.e. $c\leq\left\{ Z\subseteq Y\mid f\left(Z\right)\in Z\cap X\right\}$ and define $f^{*}:\mathcal{P}\left(\cup c\right)\to X$ by $f^{*}(Z)=f\left(Z\cap X\right)$. $f^{*}$ is regressive on $c\mathord{\downarrow}Y$ H by normality there is some $c'\leq c$ and $x\in X$ such that for every $Z\in c'$, $f^{*}\left(Z\right)=x$. So $c'\Vdash\left[f\right]_{U_{Y}}=j_{Y}\left(x\right)$. So by genericity there is in $G$ such $c'$, so indeed $\left[f\right]_{U_{Y}}E_{Y}j_{Y}\left[X\right]$.
For the "in particular", note that $X$ is the pointwise image of $j\left[X\right]$ under the transitive collapse of $j\left[tc(X)\right]$, where also $tc(X)\in V_{\kappa}$ and the collapse can be computed in $M$.
1. $U_{X}=\left\{ A\subseteq\mathcal{P}(X)\mid j\left[X\right]Ej(A)\right\}$
2. $a\in G$ iff $j\left[\cup a\right]\in j\left(a\right)$
3. Any $\gamma<\kappa$ is represented by the function $\operatorname{otp}\left(Z\cap\gamma\right)$
4. If $\gamma\in G$ then it is the critical point of $j$.
1\. For $A\subseteq\mathcal{P}(X)$, $j\left[X\right]Ej(A)$ iff $j_{X}\left[X\right]E_{X}j_{X}(A)$, which is iff $\left[i^{X}\right]_{U_{X}}E_{X}j_{X}(A)$ iff $A\in U_{X}$ as we've seen before.
2\. $a\in G$ iff $a\in U_{\cup a}$ iff $j\left[\cup a\right]\in j\left(a\right)$.
3\. By prop. [\[prop:the-identity\]](#prop:the-identity){reference-type="ref" reference="prop:the-identity"} $\left[i^{\gamma}\right]$ represents $j\left[\gamma\right]$. But $\gamma=\operatorname{otp}j\left[\gamma\right]$ so if $f:\mathcal{P}\left(X\right)\to V$ represents $\gamma$ where $\gamma\subseteq X$ then $f\left(Z\cap\gamma\right)=\operatorname{otp}\left(Z\cap\gamma\right)$ on a large set in $U_{X}$, so $\operatorname{otp}\left(Z\cap\gamma\right)$ indeed represents $\gamma$.
4\. $\gamma\in G$ means $j\left[\gamma\right]\in j(\gamma)$ and $j\left[\gamma\right]$ is of order type $\gamma$, so $\gamma<j(\gamma)$. Also this means that $j\left[\gamma\right]$ is an ordinal which can happen only if $j(\alpha)=\alpha$ for all $\alpha<\gamma$.
Another important consequence is that stationary sets of structures with some property force this property: Take e.g. $V_{\alpha}$ for $\alpha<\kappa$. Then $j\left[V_{\alpha}\right]\cong V_{\alpha}\equiv j(V_{\alpha})$. Now if $a$ is stationary in $V_{\alpha}$ containing elementary substructures satisfying some property (e.g. $X\cap\omega_{1}\in\omega_{1}$ or $\omega_{1}\subseteq X$ etc.), then $a\in G$ iff $j\left[V_{\alpha}\right]\in j(a)$ so $j[V_{\alpha}]$ satisfies the corresponding property, and hence also $j(V_{\alpha})$.
### Precipitousness
Our goal is to prove the following:
If $\delta$ is a Woodin cardinal, then for every $G\subseteq\mathbb{P}_{<\delta}$ generic, the corresponding limit model $\left(M,E\right)$ is closed under sequences of length lt;\delta$ in $V\left[G\right]$, and in particular is well-founded.
Let $\eta<\delta$ be a cardinal, and fix $a_{0}\in\mathbb{P}_{<\delta}$ and a term $\tau$ for an $\eta$-sequence of $\left(M,E\right)$-ordinals. Our first goal is to find some stationary $a\leq a_{0}$ and some function $f:a\to V$ in $V$ representing $\tau$. $\tau^{G}$ is a function from $\eta$ to ordinals. Recall that $j\left[\eta\right]\in M$ and is of order type $\eta$, so we want $\left[f\right]$ to be a function with domain $j\left[\eta\right]$ such that for every $\alpha<\eta$, $\left[f\right]_{G}(j(\alpha))=\tau^{G}(\alpha).$ We've seen that $j\left[\eta\right]$ is represented by the function $g_{\eta}(X)=X\cap\eta$ so for every $X\in a$ we'd like $f(X)$ to be a function with domain $X\cap\eta$.
For any $\alpha<\eta$, there's a maximal antichain $A_{\alpha}\subseteq\mathbb{P}_{<\delta}$ below $a_{0}$ such that for each $b\in A_{\alpha}$ there is some $f_{b,\alpha}:b\to V$ such that $b\Vdash\tau(\check{\alpha})=\left[\check{f}_{b,\alpha}\right]$. So we want that $f(X)(\alpha)=f_{b,\alpha}(X\cap\left(\cup b\right))$ for some $b\in A_{\alpha}$. So we need to choose for every $X,\alpha$ some $b\in A_{\alpha}$ such that $X\cap\left(\cup b\right)\in b$, and we'd like this $b$ to be unique.
To get uniqueness: assume $b_{0},b_{1}\in A_{\alpha}$ are such that $X\cap\left(\cup b_{i}\right)\in b_{i}$. $A_{\alpha}$ is an antichain, which means that the set $\left\{ Y\subseteq\cup b_{0}\bigcup\cup b_{1}\mid Y\cap\left(\cup b_{0}\right)\in b_{0}\land Y\cap\left(\cup b_{1}\right)\in b_{1}\right\}$ is not stationary, i.e disjoint from some $C_{F}$. Now, if $X$ is an elementary substructure of some large enough $V_{\gamma+1}$ and we pick $b_{i}\in X\cap A_{\alpha}$, then there is such an $F\in X$, and $X$ *is* closed under $F$, so we must have $b_{0}=b_{1}$.
For $D\subseteq\mathbb{P}_{<\kappa}$, we say that $Y$ *captures $D$* if there is $d\in Y\cap D$ such that $Y\cap(\cup d)\in d$.
The discussion above shows that if $D$ is an antichain and it is captured by $X$ s.t. $D\in X\prec V_{\gamma+1}$ for large enough strongly inaccessible $\gamma$ (so $X$ knows about some notion of $\mathbb{P}_{<\gamma}$), then the "capturing element" is unique.
Let $a_{0}\in\mathbb{P}_{<\delta}$, $\tau$ a term for an $\eta$-sequence of $\left(M,E\right)$-ordinals, $\left\langle A_{\alpha}\mid\alpha<\eta\right\rangle$ maximal antichains below $a_{0}$ such that for each $b\in A_{\alpha}$ there is some $f_{b,\alpha}:b\to V$ such that $b\Vdash\tau(\check{\alpha})=\left[\check{f}_{b,\alpha}\right]$, $\gamma$ such that $a_{0},\eta\in V_{\gamma}$ and the set $a_{1}=\left\{ X\prec V_{\gamma+1}\Bigg|\begin{array}{l}
\left|X\right|<\gamma\\
X\cap\left(\cup a_{0}\right)\in a_{0}\\
\forall\alpha\in X\cap\eta\,\,X\text{ \emph{captures} }A_{\alpha}
\end{array}\right\}$ is stationary in $\mathcal{P}_{\gamma}(V_{\gamma+1})$. Then $a_{1}\Vdash``\tau\in M"$.
From the discussion above, for every $X\in a_{1}$ and $\alpha\in X\cap\eta$ there is a *unique* $b=b(X,\alpha)\in X\cap A_{\alpha}$ such that $X\cap\left(\cup b\right)\in b$. So define $f(X)(\alpha)=f_{b,\alpha}(X\cap\left(\cup b\right))$ for this unique $b$. We'll be done if we show that for every $\alpha<\eta$, $a_{1}\Vdash\left[\check{f}\right](j(\check{\alpha}))=\tau(\check{\alpha})$. So fix $\alpha<\eta$ and let $G$ be generic such that $a_{1}\in G$. Since $a_{0}\geq a_{1}$ also $a_{0}\in G$, and since $A_{\alpha}$ is a maximal antichain below $a_{0}$, there is some $b^{*}\in G\cap A_{\alpha}$. So we want to find some $a'\in G$ such that for each $X\in a'$ containing $\alpha$, $b(X,\alpha)=b^{*}$.
First let $a_{2}=\left\{ X\in a_{1}\mid\alpha\in X\right\}$ $a_{2}\in G$ since it's the intersection of $a\in G$ with the club of sets containing $\alpha$, which is also in $G$. By the definition of $a_{1}$, the function $X\mapsto b(X,\alpha)$ is regressive on $a_{2}$. So for every $a'\leq a_{2}$ if we apply normality to the function $Y\mapsto b(Y\cap(\cup a_{2}),\alpha)$ we get some $a''\leq a'$ on which it is constant. So by genericity, there is some $a_{3}\in G$, $a_{3}\leq a_{2}$ on which the above function is constant. This means that we have some $b^{**}\in A_{\alpha}\cap\mathbb{P}_{<\gamma}$ such that for every $Y\in a_{3}$, $b^{**}=b(Y\cap(\cup a_{2}),\alpha)$ so in particular $b^{**}\in Y$ and $Y\cap(\cup b^{**})=Y\cap(\cup a_{2})\cap(\cup b^{**})\in b^{**}$ (the equality holds since $b^{**}\in V_{\gamma}$ and $a_{2}$ is stationary in $\mathcal{P}_{\gamma}(V_{\gamma+1})$ so $\cup b^{**}\subseteq\cup a_{2}$).
$b^{*}=b^{**}$.
Since $b^{*},b^{**}\in A_{\alpha}$, if they are different then they are incompatible (if $b^{**}\in G$ we'd be done, but we only know $a_{3}\in G$). So as before there is some function $F:\left[\cup b^{*}\bigcup\cup b^{**}\right]^{<\omega}\to\cup b^{*}\bigcup\cup b^{**}$ such that if $X$ is closed under $F$ then either $X\cap(\cup b^{*})\notin b^{*}$ or $X\cap(\cup b^{**})\notin b^{**}$. Since $a_{3},b^{*}\in G$, there is some $c\in G$ below both of them. There is some $X\in c$ closed under $F\mathord{\uparrow}(\cup c)$ so $X'=X\cap(\cup b^{*}\bigcup\cup b^{**})$ is closed under $F$. But we get a contradiction:
- On one hand, $c\leq b^{*}$ so $X\cap\left(\cup b^{*}\right)\in b^{*}$.
- On the other hand, $c\leq a_{3}$ so $X\cap\left(\cup b^{**}\right)=\underbrace{\left(X\cap(\cup a_{3})\right)}_{\in a_{3}}\cap(\cup b^{**})\in b^{**}$.
To conclude $b^{*}$ was such that $b^{*}\Vdash\tau(\alpha)=\left[\check{f}_{b^{*},\alpha}\right]_{G}$. On the other hand, for every $Y\in a_{3}$ by $b^{*}=b^{**}$ we have that $f(Y)(\alpha)=f_{b^{*},\alpha}(Y\cap\left(\cup b^{*}\right)).$ So since $b^{*},a_{3}\in G$ we get that $\tau^{G}(\alpha)=\left[f_{b^{*},\alpha}\right]_{G}=\left[f\right]_{G}(j(\alpha))$ as required.
So the next step in the proof is to find conditions under which the set $a_{1}=\left\{ X\prec V_{\gamma+1}\Bigg|\begin{array}{l}
\left|X\right|<\gamma\\
X\cap\left(\cup a_{0}\right)\in a_{0}\\
\forall\alpha\in X\cap\eta\,\,X\text{ captures }A_{\alpha}
\end{array}\right\}$ is stationary. The first conditions are easy -- the upward projection of $a_{0}$ to $V_{\gamma+1}$ will give us a stationary set $a$ satisfying $\forall X\in a(X\cap\left(\cup a_{0}\right)\in a_{0})$, in which we can find a stationary subset of elementary submodels of $V_{\gamma+1}$ of size lt;\gamma$. The hard part is the last condition -- $\forall\alpha\in X\cap\eta$ $X$ captures $A_{\alpha}$, i.e $\exists b\in X\cap A_{\alpha}\:$ s.t. $X\cap(\cup b)\in b$. So we would like to take some $X_{0}$ satisfying the first conditions, and then build an elementary chain such that in each successor step we capture one more $A_{\alpha}$. But in order to preserve e.g. the fact that $b$ witnesses the capturing of $A_{\alpha}$, we shouldn't add elements of $\cup b\subseteq V_{\gamma}$, so we want the successor steps to be *end extentions*:
Given sets $X,Y$, we say that $Y$ *end-extends* $X$ if $X\subseteq Y$ and $Y\cap V_{\beta}=X$ for the least $\beta$ such that $X\subseteq V_{\beta}$.
We say that $Y$ is a $\gamma$-end-extension of $X$ (or end-extension below $\gamma$) if $Y\cap V_{\gamma}$ end-extends $X\cap V_{\gamma}$.
So we want the set of end-extensions of $X$ which capture $A_{\alpha}$ to be big enough so we can always find elements there. This motivates the following definition:
Let $\kappa$ strongly inaccessible. For $D\subseteq\mathbb{P}_{<\kappa}$ let $sp(D)$ be the set of $X\prec V_{\kappa+1}$ with $\left|X\right|<\kappa$, $D\in X$, such that there is $Y\prec V_{\kappa+1}$ satisfying:
1. $X\subseteq Y$,
2. $Y\cap V_{\kappa}$ end-extends $X\cap V_{\kappa}$,
3. $Y$ captures $D$.
We call $D$ *semi-proper* if $sp(D)$ is club in $\mathcal{P}_{\kappa}(V_{\kappa+1})$.
Sometimes having a club won't be enough, and we want to be able to extend ANY (appropriate) model to one capturing $D$. For that we need elementary submodels of $V_{\kappa+2}$ rather than $V_{\kappa+1}$:
If $D\subseteq\mathbb{P}_{<\kappa}$ is semi-proper then for ANY $X\prec V_{\kappa+2}$ such that $\left|X\right|<\kappa$ and $D\in X$ there is $Y\prec V_{\kappa+2}$ such that *1-3* above holds.
Assume $D\subseteq\mathbb{P}_{<\kappa}$ is semi-proper and let $X\prec V_{\kappa+2},$ $\left|X\right|<\kappa$ and $D\in X$. The fact that $D$ is semi-proper is expressible in $V_{\kappa+2}$, so $X\vDashquot;$D$ is semi-proper", i.e. there is some $F\in X$, $F:\left[V_{\kappa+1}\right]^{<\omega}\to V_{\kappa+1}$ such that any $Z$ closed under $F$ is in $sp(D)$. In particular, $X\cap V_{\kappa+1}$ is closed under $F$ (since $F\in X\prec V_{\kappa+2}$) so $X\cap V_{\kappa+1}\in sp(D)$. Let $X^{*}$ witness this -- $X\cap V_{\kappa+1}\subseteq X^{*}\prec V_{\kappa+1}$, $X^{*}\cap V_{\kappa}$ end-extends $X\cap V_{\kappa}$ and $X^{*}$ captures $D$. Fix a well order $\leq^{*}$ of $V_{\kappa+2}$ such that $\leq^{*}\mathord{\upharpoonright}V_{\kappa}\in X$[^2]. Let $Y$ be the Skolem hull of $X\cup(X^{*}\cap V_{\kappa})$ in $V_{\kappa+2}$ under $\leq^{*}$ (so (1) trivially holds).
$Y\cap V_{\kappa}=X^{*}\cap V_{\kappa}$
Any $y\in Y\cap V_{\kappa}$ is of the form $y=f(\bar{a},\bar{b})$ for $\bar{a}\in X^{m}$, $\bar{b}\in(X^{*}\cap V_{\kappa})^{n}$ and $f$ a Skolem function of $V_{\kappa+2}$ according to $\leq^{*}$. Fixing $\bar{a}\in X^{m}$, the function $f_{\bar{a}}=\left\{ \left\langle \bar{b},f(\bar{a},\bar{b})\right\rangle \mid\bar{b},f(\bar{a},\bar{b})\in V_{\kappa}\right\}$ is a member of $X\cap V_{\kappa+1}$ since $f\left(\bar{a},\bar{b}\right)$ is defined from $\bar{a}\in X$ and $\leq^{*}\mathord{\upharpoonright}V_{\kappa}\in X$, and is definitely a member of $V_{\kappa+1}$ (note that it is not contained in $X$!). Recall that $X\cap V_{\kappa+1}\subseteq X^{*}$ so $f_{\bar{a}}\in X^{*}$, and since it is a function in $X^{*}$, applying it to $\bar{b}\in(X^{*}\cap V_{\kappa})^{n}$ results in an element of $X^{*}\cap V_{\kappa}$. Hence $y=f_{\bar{a}}(\bar{b})\in X^{*}\cap V_{\kappa}$, as required.
So first we immediatly get (2) - that $Y\cap V_{\kappa}$ end-extends $X\cap V_{\kappa}$, and (3) also follows: $X^{*}$ captured $D$ so there is $d\in X^{*}\cap D\subseteq X^{*}\cap V_{\kappa}$ (so $d\in Y\cap D$) such that $X\cap\left(\cup d\right)\in d$ but $\cup d\subseteq V_{\kappa}$ so $X\cap(\cup d)=Y\cap(\cup d)$.
So now we can state the desired condition:
Given $a_{0}\in\mathbb{P}_{<\delta}$ and $\left\langle A_{\alpha}\mid\alpha<\eta\right\rangle$ maximal antichains below $a_{0}$ for $\eta<\delta$, if there is a strongly inaccesible $\gamma<\delta$ such that $a_{0},\eta\in V_{\gamma}$ and $\forall\alpha<\eta$ $A_{\alpha}\cap\mathbb{P}_{<\gamma}$ is semi-proper, then $a_{1}$ is stationary in $\mathcal{P}_{\gamma}(V_{\gamma+1})$ where $a_{1}=\left\{ X\prec V_{\gamma+1}\Bigg|\begin{array}{l}
\left|X\right|<\gamma\\
X\cap\left(\cup a_{0}\right)\in a_{0}\\
\forall\alpha\in X\cap\eta\,\,X\text{ captures }A_{\alpha}
\end{array}\right\} .$
Fix $H:\left[V_{\gamma+1}\right]^{<\omega}\to V_{\gamma+1}$. By stationarity of $a_{0}$ we can choose some $X_{0}\prec V_{\delta}$, $\left|X_{0}\right|<\gamma$ such that $X_{0}\cap\left(\cup a_{0}\right)\in a_{0}$ and $a_{0},\gamma,H,\left\langle A_{\alpha}\cap V_{\gamma}\mid\alpha<\eta\right\rangle \in X_{0}$ (the set of elementray substructures of $V_{\delta}$ containing the above elements is a club, and the set of sets of size lt;\gamma$ which project to $a_{0}$ is stationary). Define an elementary chain $\left\langle X_{\alpha}\mid\alpha\in X_{0}\cap\eta\right\rangle$. At limit stages we take unions. Assume $X_{\alpha}$ is defined such that $X_{\alpha}\prec V_{\delta}$ and $\left|X_{\alpha}\right|<\gamma$. $A_{\alpha}\cap\mathbb{P}_{<\gamma}$ is semi-proper so by the lemma we can find some $Y\prec V_{\gamma+2}$ $\left|Y\right|<\gamma$ satisfying
1. $X\subseteq Y$,
2. $Y\cap V_{\gamma}$ end-extends $X\cap V_{\gamma}$,
3. $Y$ captures $A_{\alpha}\cap\mathbb{P}_{<\gamma}$.
We set $X_{\alpha+1}=Y$. Set $X=\bigcup\left\{ X_{\alpha}\mid\alpha\in X_{0}\cap\eta\right\}$. Note that $\left|X\right|<\gamma$ since $\eta<\gamma$. By induction we have that for all $\alpha<\beta$ in $X_{0}\cap\eta$, $X_{\beta}\cap V_{\gamma}$ end-extends $X_{\alpha}\cap V_{\gamma}$, so also $X\cap V_{\gamma}$ end extends $X_{\alpha}\cap V_{\gamma}$. In particular $X\cap\eta=X_{0}\cap\eta$. For every $\alpha\in X\cap\eta$, $X_{\alpha+1}$ captures $A_{\alpha}\cap\mathbb{P}_{<\gamma}$ i.e. there is some $b\in X_{\alpha+1}\cap A_{\alpha}\cap\mathbb{P}_{<\gamma}$ such that $X_{\alpha+1}\cap\left(\cup b\right)\in b$. But $\cup b\in V_{\gamma}$, so by the end-extension, $X_{\alpha+1}\cap\left(\cup b\right)=X\cap(\cup b)$, so $X$ captures $A_{\alpha}$. Similarly $X\cap(a_{0})=X_{0}\cap(a_{0})\in a_{0}$. Now since $X\prec V_{\delta}$ and $H\in X$ , $X\cap V_{\gamma+1}$ is closed under $H$ (which is defined on $V_{\gamma+1}$), and we can assume it is closed under the Skolem function of $V_{\gamma+1}$, so all-in-all we get that $X\cap V_{\gamma+1}\in a_{1}$ closed under $H$.
A strongly inaccessible cardinal $\delta$ is called a *Woodin cardinal* if for every function $f:\delta\to\delta$ there is an elementary embedding $j:V\to M$ with critical point $\gamma<\delta$ such that $f\left[\gamma\right]\subseteq\gamma$ and $V_{j(f)(\gamma)}\subseteq M$.
We can assume that the embeddings are derived from extenders in $V_{\delta}$, that $M$ is closed under sequences of length lt;\gamma$ and that $j(f)(\gamma)=f(\gamma)$.
If $\delta$ is a Woodin cardinal, then for each sequence $\left\langle D_{\alpha}\mid\alpha<\delta\right\rangle$ of predense subsets of $\mathbb{P}_{<\delta}$ there is a st. inacc. $\gamma<\delta$ such that for all $\alpha<\gamma$, $D_{\alpha}\cap\mathbb{P}_{<\gamma}$ is predense in $\mathbb{P}_{<\gamma}$ and semi-proper.
First we deal with the predenseness:
The set $E$ of strongly inaccessibles $\gamma<\delta$ such that for all $\alpha<\gamma$, $D_{\alpha}\cap\mathbb{P}_{<\gamma}$ is predense in $\mathbb{P}_{<\gamma}$ is club.
For any $\alpha<\delta$ and $a\in\mathbb{P}_{<\delta}$ let $F(\alpha,a)=\min\left\{ \gamma\mid\exists b\in D_{\alpha}\cap\mathbb{P}_{<\gamma}\text{ compatible with }a\right\}$. $D_{\alpha}$ is predense so $F(\alpha,a)$ is always defined. If $V_{\gamma}$ is closed under $F$ then for every $\alpha<\gamma$ and every $a\in\mathbb{P}_{<\gamma}$ $F(\alpha,a)=\gamma^{*}<\gamma$ so there is a $b\in D_{\alpha}\cap\mathbb{P}_{<\gamma^{*}}\subseteq D_{\alpha}\cap\mathbb{P}_{<\gamma}$ compatible with $a$. So by fixing some bijection $\delta\mapsto V_{\delta}$ we can find a function $h$ such that if $\gamma$ is str. inacc. closed under $h$ then for all $\alpha<\gamma$, $D_{\alpha}\cap\mathbb{P}_{<\gamma}$ is predense in $\mathbb{P}_{<\gamma}$.
Now for the semi-properness. Assume toward contradiction that the lemma fails. So for any str. inacc. $\gamma$ there is some $\alpha(\gamma)<\gamma$ minimal such that $D_{\alpha}\cap\mathbb{P}_{<\gamma}$ is not semi-proper. This means that the set $S_{\gamma}=\left\{ Z\prec V_{\gamma+1}\mid\left|Z\right|<\gamma\text{\ensuremath{\land} there's no \ensuremath{\gamma}-end-extension of \ensuremath{Z} capturing \ensuremath{D_{\alpha(\gamma)}\cap\mathbb{P}_{<\gamma}}}\right\}$ is stationary. By the denseness of $D_{\alpha(\gamma)}$ let $g(\gamma)$ be minimal above $\alpha(\gamma)$ such that there is a condition in $D_{\alpha(\gamma)}\cap\mathbb{P}_{<g(\gamma)}$ compatible with $S_{\gamma}$. For $\alpha$ not st.inacc. let $g(\alpha)=\alpha$.
Now set $f(\alpha)=\max\left\{ h(\alpha),g(\alpha),\alpha\right\}$ ($h$ from the claim). So $f$ is weakly increasing and satisfies that if $\gamma$ is a str.inacc. closed under $f$, then:
1. $D_{\alpha}\cap\mathbb{P}_{<\gamma}$ is predense for every $\alpha<\gamma$,
2. In $D_{\alpha(\gamma)}\cap\mathbb{P}_{<f(\gamma)}$ there is a condition compatible with $S_{\gamma}$.
Apply Woodinness to $f$ to get $j:V\to M$ such that
- $\mathrm{crit}\left(j\right)=\gamma<\delta$
- $f\left[\gamma\right]\subseteq\gamma$
- $V_{j(f)(\gamma)+\omega}\subseteq M$
Note that $j(f)(\gamma)\geq\gamma$, so in particular we have $V_{\gamma+\omega}\subseteq M$. A key point is: $\begin{gathered}
\text{Any \ensuremath{a\in V_{j(f)(\gamma)}=M_{j(f)(\gamma)}} is stationary in \ensuremath{V} iff it is stationary in \ensuremath{M}.}\label{eq:equivalent stationarity}\end{gathered}$
Consider $\gamma$ from the point of view of $M$. In $M$, $\alpha^{M}(\gamma)<\gamma$ is the minimal such that $j(D_{\alpha})\cap\mathbb{P}_{<\gamma}$ is not semi-proper. Since $\mathrm{crit}\left(j\right)=\gamma$, for all $\alpha<\gamma$ we have $j(D_{\alpha})\cap\mathbb{P}_{<\gamma}=D_{\alpha}\cap\mathbb{P}_{<\gamma}$ and it is semi-proper in $M$ iff it is semi-proper in $V$ (since the set $\left\{ Z\prec V_{\gamma+1}\mid\left|Z\right|<\gamma\text{\ensuremath{\land} no \ensuremath{\gamma}-end-extension of \ensuremath{Z} capturing \ensuremath{j(D_{\alpha})\cap\mathbb{P}_{<\gamma}=D_{\alpha}\cap\mathbb{P}_{<\gamma}}}\right\}$ is stationary in $M$ iff it is stationary in $V$), so $\alpha^{M}(\gamma)=\alpha(\gamma)$. Denote $\alpha=\alpha(\gamma)$, so $S_{\gamma}$ above is defined in $M$ in the same way, and is stationary in $M$.
Consider now $j(f)$. $j(f)\mathord{\upharpoonright}\gamma=f$ so $\gamma$ is also closed under $j(f)$, and by elementarity, property (2) above holds for $j(f)$, (note that $j(\alpha)(\gamma)=\alpha(\gamma)$ as well) so we get that: $\begin{gathered}
\text{There is \ensuremath{T\in j(D_{\alpha})\cap\mathbb{P}_{<j(f)(\gamma)}} compatible with \ensuremath{S_{\gamma}} (in \ensuremath{j(\mathbb{P}_{<\delta})})}\end{gathered}$ and by ([\[eq:equivalent stationarity\]](#eq:equivalent stationarity){reference-type="ref" reference="eq:equivalent stationarity"}) $T$ is stationary in $V$ as well, and compatible with $S=S_{\gamma}$. Let $T^{*}$ be the common lower bound, i.e. $T^{*}=\left\{ Z\subseteq(\cup S)\cup(\cup T)\mid Z\cap(\cup S)\in S\land Z\cap(\cup T)\in T\right\} .$ So if we have some nice $X\ni T$ that projects to $T^{*}$, on one hand it captures $j(D_{\alpha})$, and on the other hand its projection to $S$ - $X\cap(\cup S)$ has no $\gamma$-end-extension capturing $D_{\alpha}\cap\mathbb{P}_{<\gamma}$, so $j(X\cap(\cup S))$ has no $j(\gamma)$-end-extension capturing $j(D_{\alpha}\cap\mathbb{P}_{<\gamma})$ (in $M$!). We want to use these two properties to derive a contradiction. So let $\eta>j(\delta)$ be regular, and by stationarity of $T^{*}$ in $V$ let $X\prec V_{\eta}$ such that $S,T,j\mathord{\upharpoonright}V_{\gamma+1},j(V_{\gamma+2})\in X$ and $X\cap(\cup T^{*})\in T^{*}$. Note that $\cup S=V_{\gamma+1}$ so our goal is:
In $M$, $j(X\cap V_{\gamma+1})$ has a $j(\gamma)$-end-extension capturing $j(D_{\alpha}\cap\mathbb{P}_{<\gamma})$.
First note that since $X\cap(\cup T^{*})\in T^{*}$, $X\cap V_{\gamma+1}\in S$ and in particular of size lt;\gamma$, hence $j\left(X\cap V_{\gamma+1}\right)=j\left[X\cap V_{\gamma+1}\right]$ . We assumed $j\mathord{\upharpoonright}V_{\gamma+1}\in X$, i.e. $X$ is closed under $j\mathord{\upharpoonright}V_{\gamma+1}$, so $j\left[X\cap V_{\gamma+1}\right]\subseteq X$. Now we want to extend $j\left[X\cap V_{\gamma+1}\right]$ in $M$, so fix $\leq^{*}\in M\cap X$ a well-order of $j(V_{\gamma+1})$ (this is why we wanted $j(V_{\gamma+2})\in X$) and let $Y$ be the Skolem closure of $\left\{ S,T\right\} \cup j\mathord{\upharpoonright}(X\cap V_{\gamma+1})\cup\left(X\cap(\cup T^{*})\right)$ under this well-order. We claim that this $Y$ works:
- $Y\prec j(V_{\gamma+1})$ since it's a Skolem closure, and $\left|Y\right|<\gamma$ since this holds for $\left(X\cap V_{\gamma+1}\right)$ and $\left(X\cap(\cup T^{*})\right)$.
- Clearly $j\left[X\cap V_{\gamma+1}\right]\subseteq Y$.
- Since $X$ is closed under $j\mathord{\upharpoonright}V_{\gamma+1}$ and is an elementary substructure of $V_{\eta}$ for large $\eta$, and $V_{j(f)(\gamma)+\omega}\subseteq M$, we have that $Y\subseteq X$.
- $Y\cap V_{j(\gamma)}$ end-extends $j\left[X\cap V_{\gamma+1}\right]\cap V_{j(\gamma)}$:
$X\cap V_{\gamma+1}=\left(X\cap V_{\gamma}\right)\cup\left(X\cap V_{\gamma+1}\mathord{\smallsetminus}V_{\gamma}\right)$. $\mathrm{crit}\left(j\right)=\gamma$ so on the first part, $j$ is the identity, while the second part is sent above $j(\gamma)$, i.e. $j\left[X\cap V_{\gamma+1}\mathord{\smallsetminus}V_{\gamma}\right]\subseteq V_{j(\gamma)+1}\mathord{\smallsetminus}V_{j(\gamma)}$. So $j\left[X\cap V_{\gamma+1}\right]\cap V_{j(\gamma)}=X\cap V_{\gamma}.$ Since $X\cap V_{\gamma}\subseteq Y\subseteq X$, $Y\cap V_{j(\gamma)}=(Y\cap V_{\gamma})\cup(Y\cap V_{j(\gamma)}\mathord{\smallsetminus}V_{\gamma})=(X\cap V_{\gamma})\cup(Y\cap V_{j(\gamma)}\mathord{\smallsetminus}V_{\gamma})$ so indeed $Y\cap V_{j(\gamma)}$ end-extends $j\left[X\cap V_{\gamma+1}\right]\cap V_{j(\gamma)}$.
- Recall that $T\in j(D_{\alpha})\cap\mathbb{P}_{<j(f)(\gamma)}$, and since $j(f)(\gamma)<j(\gamma)$, and $T\in Y,$ we have $T\in Y\cap j(D_{\alpha}\cap\mathbb{P}_{<\gamma})$. Since $X\cap(\cup T^{*})\subseteq Y\subseteq X$, $Y\cap(\cup T)=X\cap(\cup T)\in T$ (by the choice of $X$).
So $Y$ is a $j(\gamma)$-end-extension of $j\left[X\cap V_{\gamma+1}\right]$ capturing $j(D_{\alpha}\cap\mathbb{P}_{<\gamma})$. This contradicts the fact that $j\left[X\cap V_{\gamma+1}\right]=j\left(X\cap V_{\gamma+1}\right)\in j(S)$. This concludes the theorem.
Applications
------------
The stationary tower and it's variants have many applications concerning satisfaction of formulas in generic extension. We'll give two such applications, both are based of the following lemma:
If $\delta$ is Woodin and $Q\in V_{\delta}$ is a forcing notion then there is $a\in\mathbb{P}_{<\delta}$ forcing the existence of a generic filter for $Q$.
This is denoted by $V^{Q}\subseteq V^{\mathbb{P}_{<\delta}\mathord{\upharpoonright}a}$, and implies that there is a $Q$-name $\dot{R}$ such that $Q*\dot{R}\cong\mathbb{P}_{<\delta}\mathord{\upharpoonright}a$.
Note that since $\delta$ is strongly inaccessible, $\mathcal{P}(Q)\in V_{\delta}$. Let $a=\mathcal{P}_{\omega_{1}}(\mathcal{P}(Q))=\left\{ X\subseteq\mathcal{P}(Q)\mid\left|X\right|=\aleph_{0}\right\}$ then $a$ is stationary and $a\Vdash2^{\left|Q\right|}=\aleph_{0}$: let $G\subseteq\mathbb{P}_{<\delta}$ be generic with $a\in G$ and $j:V\to M\subseteq V\left[G\right]$. Recall that $a\in G$ iff $j\left[\cup a\right]\in j(a)$. In our case $j(a)$ consists of countable sets, and $\cup a=\mathcal{P}^{V}(Q)$ so $M\vDash\left|\mathcal{P}^{V}(Q)\right|=\aleph_{0}$, so this holds in $V\left[G\right]$ as well. So in $V\left[G\right]$ we can enumerate all dense subsets of $Q$ in $V$ in a countable sequence and get a $V$-generic filter for $Q$.
So $\mathbb{P}_{<\delta}$ "absorbs" any smaller forcing, and if we have proper class of Woodins we can absorb any forcing into a suitable stationary tower forcing. This explains the assumption in the following theorems:
If there is a proper class of Woodin cardinals, then for any $\Sigma_{2}$ sentence of set-theory $\Phi$, if $V\vDash\Phi$ and $\mathbb{Q}\in V$ some forcing notion, then there is a forcing $\mathbb{R}$ such that $V^{\mathbb{Q}*\mathbb{R}}\vDash\Phi$.
For $T\cup\left\{ \phi\right\}$ a set of sentences of set-therory, denote by $T\vDash_{\Omega}\phi$ the statement
> "For every forcing notion $P$ and ordinal $\alpha$, if $V_{\alpha}^{P}\vDash T$ then $V_{\alpha}^{P}\vDash\phi$."
The meaning of $V_{\alpha}^{P}\vDash T$ is that $\Vdash_{P}``V_{\check{\alpha}}\vDash T"$ i.e. every $G\subseteq P$ generic, $\left(V\left[G\right]\right)_{\alpha}\vDash T$.
If there is a proper class of Woodin cardinals then the notion $T\vDash_{\Omega}\phi$ is set- forcing absolute.
Before we prove the theorems, we need a few lemmas and properties of the stationary tower.
Fix $\delta$ Woodin, $j:V\to M\subseteq V\left[G\right]$ the derived generic embedding from $\mathbb{P}_{<\delta}$.
[\[cor:.Woodin-fixed\]]{#cor:.Woodin-fixed label="cor:.Woodin-fixed"}$j(\delta)=\delta$.
First we claim that $j(\delta)=\sup\left\{ j(\gamma)\mid\gamma<\delta\right\}$. Let $\alpha<j(\delta)$. Then there is some $a\in G$ and $f:a\to\delta$ such that $\alpha=\left[f\right]_{G}$. In particular $\left|a\right|<\delta$ so there is some $\eta<\delta$ such that $\operatorname{Range}(f)\subseteq\eta$, and so $\left[f\right]_{G}<j(\eta)$.
Now the claim will follow if we show that there are unboundedly many fixed points below $\delta$. Note that by the definition of Woodin cardinal, there are unboundedly many measurables below $\delta$. We show that unboundedly many of them are fixed points. Let $a\in\mathbb{P}_{<\delta}$, and we want to find some $\kappa$ and $b\leq a$ such that $b\Vdash j(\kappa)=\kappa$. As we've seen before, $\kappa$ is represented by the function $Z\mapsto\operatorname{otp}(Z\cap\kappa)$ on subsets of any $X\supset\kappa$. So let $\kappa<\delta$ be large enough measurable such that $a\in V_{\kappa}$. If the set $b=\left\{ Z\subseteq V_{\kappa}\mid Z\cap(\cup a)\in a\land\operatorname{otp}(Z\cap\kappa)=\kappa\right\}$ is stationary, then it is as required. Let $F:\left[V_{\kappa}\right]^{<\omega}\to V_{\kappa}$, we need to find $M\in b$ closed under $F$. Look at the model $\mathfrak{A}=\left\langle V_{\kappa},F,x\right\rangle _{x\in\cup a}$. Recall Rowbottom's theorem:
If $\kappa$ is measurable [^3] and $\mathfrak{A}\supset\kappa$ is a model for a language of size lt;\kappa$ then it has a set of indiscernibles of order type $\kappa$.
So let $I\subseteq\kappa$ be a set of indiscernibles for $\mathfrak{A}$ of size $\kappa$. Let $\left\langle i_{k}\mid k<\omega\right\rangle$ be the first $\omega$ indiscernibles, and define for each $n,k<\omega$ $G_{k}:\left[\cup a\right]^{<\omega}\to\cup a$ by $G_{k}(a_{0},\dots,a_{n-1})=F(a_{0},\dots,a_{n-1},i_{0},\dots,i_{k-1})$ if it is in $\cup a$ (and some arbitrary value otherwise). By stationarity of $a$ there is some $X\in a$ closed under all the $G_{k}