Proof Let $\delta =(2^\kappa )^V$ . For every $f\!:\delta \leftrightarrow \mathcal P(\kappa )$ in V set

$$ \begin{align*}X_f=\{\alpha<\delta \mid f(\alpha) \in W\}.\end{align*} $$

Clearly, if for some such $f, X_f \in V$ , then also $U \in V$ .

Suppose that $X_f $ never is in $ V$ .

Define, for such f’s, $\alpha _f\leq \delta $ be the least $\alpha $ such that $X_f \cap \alpha \not \in V$ .

Set

$$ \begin{align*}\alpha^*=\min\{\alpha_f \mid f:\delta \leftrightarrow \mathcal P^V(\kappa), f \in V\}.\end{align*} $$

Claim. $\alpha ^*<\kappa $ .

So, $\alpha ^*<\kappa $ . Fix $f:\delta \leftrightarrow \mathcal P(\kappa )$ with $\alpha _f=\alpha ^*$ . So, the set

$$ \begin{align*}X_f \cap \alpha^*=\{\alpha<\alpha^* \mid f(\alpha) \in W\}\not \in V.\end{align*} $$

Work in $V[G]$ . Set $A_0=\bigcap _{\alpha \in X_f \cap \alpha ^*} f(\alpha )$ and $A_1=\bigcap _{\alpha \in \alpha ^* \setminus X_f } \kappa \setminus f(\alpha )$ . Both sets are in W due to $\kappa{-}$ completeness. So, $A=A_0\cap A_1 \in W$ , as well. Pick $\zeta \in A_0\cap A_1$ .

Then

$$ \begin{align*}\{\alpha<\alpha^* \mid \zeta\in f(\alpha)\}=X_f \cap \alpha^*,\end{align*} $$

but clearly, $\{\alpha <\alpha ^* \mid \zeta \in f(\alpha )\}$ is in V.

So, $X_f \cap \alpha ^* \in V$ . Contradiction.

Proof Suppose that W is a $\kappa{-}$ complete ultrafilter which extends U in $V[G]$ . Clearly, then $W\cap V=U$ .

Also,

$$ \begin{align*}U=\{A\subseteq \kappa \mid A\in V \text{ and } [id]_W \in j_W(A)\}.\end{align*} $$

Consider $j_W:V[G]\to M_W$ . Then, $M_W=M[j_W(G)]$ .Footnote ⁶

Define $k:M_U \to M$ by setting $k([f]_U)=[f]_W$ .

Proof Clearly, k respects $\in $ and $=$ .

Let $\varphi (v_1,\ldots ,v_n)$ be a formula and $f_1,\ldots ,f_n$ be functions in V from $\kappa $ .

Suppose that

$$ \begin{align*}M_U\models \varphi([f_1]_U,\ldots,[f_n]_U).\end{align*} $$

Then

$$ \begin{align*}\{\nu<\kappa \mid V\models \varphi(f_1(\nu),\ldots,f_n(\nu)\}\in U\subseteq W.\end{align*} $$

So,

$$ \begin{align*}M_W\models (M\models \varphi([f_1]_W,\ldots,[f_n]_W)).\end{align*} $$

Hence,

$$ \begin{align*}M\models \varphi([f_1]_W,\ldots,[f_n]_W).\\[-30pt]\end{align*} $$

Proof Let $x \in V$ . Then $j_W(x) = [c_x]_W$ , where $c_x$ is the constant function with value x. So,

$$ \begin{align*}j_W(x) = [c_x]_W=k([c_x]_U)=k(j_U(x)),\end{align*} $$

and we are done.

Now, let A be subset of $ \kappa $ in $V[G]$ . Then, there is $p \in G$ such that

Suppose that

Apply k.

Then

Recall that $M_W=M[j_W(G)]$ . So, $j_W(p)\in j_W(G)$ , and then, in $M_W$ ,

Hence, $A \in W$ .

This shows that $U^*=W$ , and we are done.

Proof $(1)\Rightarrow (2)$ .

Let $A\subseteq \kappa $ in $V[G]$ and suppose that there is $p \in G$ such that

. Let, in V,

Then, $B\in U$ , by the Los Theorem. In addition, $B\subseteq A$ , since $p\in G$ .

$(2)\Rightarrow (1)$ . Let $A\subseteq \kappa $ in $V[G]$ and suppose that $A\supseteq B$ , for some $B\in U$ .

Pick $p\in G$ such that

. Then, in $M_U$ ,

By the assumption, U generates an ultrafilter, hence for every $A\subseteq \kappa $ in $V[G]$ , either A or $\kappa \setminus A$ contain an element of U. So, (1) holds.

Question. Suppose that U is a $\kappa{-}$ complete (or normal) ultrafilter over $\kappa $ in V. Let $V[G]$ be a generic extension of V by a forcing P. Is it possible that U generates an ultrafilter in $V[G]$ which is not $\sigma{-}$ complete?

Proof Let us deal with $\kappa{-}$ completeness. Directedness is similar.

We assume $2^\kappa =\kappa ^+$ holds in V. Hence, $U=\{B_\alpha \mid \alpha <\kappa ^+\}$ .

Now let $a\subseteq \kappa ^+, |a|<\kappa $ in $V[G]$ . Pick $f:\kappa \to \sup (a)+1$ in V to be onto. Then, due to regularity of $\kappa $ , there is $\delta <\kappa $ , such that $f"\delta \supseteq a$ . Set $b=f"\delta $ . Clearly, $b\in V$ and $|b|<\kappa $ there. Then

$$ \begin{align*}\bigcap_{\alpha\in a}B_\alpha\supseteq \bigcap_{\alpha\in b}B_\alpha \in U.\\[-35pt]\end{align*} $$

Proof Define $\sigma :M_U \to N$ by setting $\sigma ([f]_U)=i(f)(\delta )$ .

Then $i=\sigma \circ j_U$ and $\sigma ([id]_U)=\delta $ . So, by elementarity of $\sigma $ ,

for every $p \in P$ and a name

. In particular, the assumption of the proposition implies that for every $A\subseteq \kappa $ in $V[G]$ there is $p \in G$ such that

. Now the conclusion follows by 2.6.

Proof $(1)\Rightarrow (2)$ .

Let f be a regressive function on $\kappa $ in $V[G]$ and suppose that there is $p \in G$ such that for some $\alpha <\kappa $ ,

. Let, in V,

Then, $B\in U$ , by the Los Theorem. In addition, $B\subseteq \{\nu <\kappa \mid f(\nu )=\alpha \}$ , since $p\in G$ .

$(2)\Rightarrow (1)$ . Let f be a regressive function on $\kappa $ in $V[G]$ and suppose that for some $\alpha <\kappa $ , the set $A=\{\nu <\kappa \mid f(\nu )=\alpha \}\in W$ .

Then, $A\supseteq B$ , for some $B\in U$ .

Pick $p\in G$ such that

. Then, in $M_U$ ,

So, (1) holds.

Question. Suppose that U is a normal ultrafilter over $\kappa $ in V. Let $V[G]$ be a generic extension of V by a forcing P. Is it possible that U generates a $\kappa{-}$ complete ultrafilter in $V[G]$ which is not normal?

Proof Let be a name which is forced, say by the weakest condition in $P_\kappa $ , to be a dense open subset of . Pick a function F which represents in the ultrapower $M_U$ .

Assume, for simplification of the notation, that for every $\beta \in I$ , $F(\beta )$ is forced by the weakest condition in $P_\beta $ to be a dense open subset of ${\langle } P_\kappa /P_\beta , \leq ^*{\!\rangle }$ .

Fix $p \in P_\kappa $ . We define by induction a $\leq ^*$ -increasing sequence ${\langle } p_i \mid i<\kappa {\rangle }$ of extensions of p and a decreasing sequence ${\langle } C_i \mid i<\kappa {\rangle } $ of closed unbounded subsets of $\kappa $ such that for every $i<\kappa $ the following hold:

1. 1. $C_i \cap {\textrm {supp}}(p_i)=\emptyset $ ,

2. 2. $p\upharpoonright {i} \Vdash p_{i}\setminus {i} \in F({i})$ , if $i \in I\cap \bigcap _{j<i}C_j$ ,

3. 3. $p_i \upharpoonright {i +1}=p_{i'}\upharpoonright {i+1}$ , for every $i<i'<\kappa $ .

Proceed as follows.

Set $p_0=p$ . Let $C_0$ be a club disjoint with ${\textrm {supp}}(p_0)$ . Set $i_0=\min (I\cap C_0)$ .

For every $i<i_0$ , set $p_i=p_0$ and $C_i=C_0$ .

Now let $p_{i_0}\geq ^* p_0$ be such that $p_{i_0}\upharpoonright i_0=p_0\upharpoonright i_0$ and $p\upharpoonright i_0 \Vdash p_{i_0}\setminus i_0 \in F(i_0)$ .

Let $C_{i_0}$ be a proper club subset of $C_0$ which is disjoint from ${\textrm {supp}}(p_{i_0})$ .

Suppose now that $p_j,C_j$ are defined for every $j<i$ . Define $p_i$ and $C_i$ . If there is $i'<i$ such that $C_{i'}=C_j$ , for every $j, i'\leq j<i$ , then we proceed as above.

Suppose that this is not the case. Note that then $i \not \in {\textrm {supp}}(p_j)$ , for every $j<i$ .

Set $C_i'=\bigcap _{j<i}C_j$ . There is a $\leq ^*$ -extension of the sequence ${\langle } p_j \mid j<i{\rangle }$ . Just put together the conditions using the inductive assumption (3) above and the strategic closure. Note that $|P_{\xi }|$ is below the first inaccessible above $\xi $ , so we have enough strategic closure in order to proceed. The only problematic coordinate may be i itself, however, $i \not \in {\textrm {supp}}(p_j)$ , for every $j<i$ , and so we will have the i-th coordinate just empty.

This is actually the crucial point in the use of nonstationary support.

Now, if $i \not \in I$ , then set $C_i = C_i'$ and $p_i=p_i'$ .

If $i \in I$ , we proceed as above using $p_i'$ and $C_i'$ .

This completes the construction.

Now set $C=\Delta _{i<\kappa }C_i$ and put the sequence ${\langle } p_i \mid i<\kappa {\rangle }$ into a single condition $p^*$ using the inductive assumption (3) above.

Then for every $i \in C\cap I$ , $p\upharpoonright i \Vdash p_i\setminus i+1\leq ^* p^*\setminus i+1.$

Hence, $p^*\upharpoonright i \Vdash p^*\setminus i+1 \in F(i)$ .

Finally, in $M_U$ , .

Recall that $p \in P_\kappa $ was arbitrary. So there is such $p^* $ in $G_\kappa $ .

Proof Note that for every $X\subseteq \kappa $ in $V[G_\kappa ]$ , the set

is a $\leq ^*{-}$ dense open in $M_U[G_\kappa ]$ , by the Prikry condition of the forcing $ j_U(P_\kappa )/G_\kappa $ . Now, use Lemma 3.3.

Proof Let U be as above and $\lambda $ be a regular cardinal above $\kappa $ . Fix a normal ultrafilter F over $\mathcal P_\kappa (\lambda )$ .

Consider the embedding $j=j_{j_U(F)}\circ j_U: V\to M$ .

We have the following:

1. A. ,

2. B. $M_U \cap {}^{j_U(\lambda )}M \subseteq M$ ,

3. C. $j_U(\lambda )<j(\kappa )=j_{j_U(F)}(j_U(\kappa ))$ .

Now instead of deciding statements we would like to decide or .

Note that

So, the former decisions will define a uniform $\kappa{-}$ complete ultrafilter over $\lambda $ . Similar, the latter will define a fine $\kappa{-}$ complete ultrafilter over $\mathcal P_\kappa (\lambda )$ , since for every $\alpha <\lambda $ ,

Apply Lemma 3.3. The set $H=\{j_U(p)\setminus \kappa \mid p \in G_\kappa \}$ is $M_U[G_\kappa ]$ -generic for ${\langle } j_U(P_\kappa )/G_\kappa , \leq ^*{\!\rangle } $ . Construct also (in $V[G_\kappa ]$ ) an $M_U[G_\kappa ,H]$ -generic set R for ${\langle } Q_{j_U(\kappa )}, \leq ^*_{Q_{j_U(\kappa )}} {\!\rangle }$ . Note that in case of the Prikry forcings iteration this will be just an atomic forcing, since any two conditions with the same trunk will be compatible.

Next, consider $j_{j_U(F)}:M_U \to M$ . It is an ultrapower embedding by a normal ultrafilter $j_U(F)$ over $\mathcal P_{j_U(\kappa )}(j_U(\lambda ))$ of $M_U$ .

Take . Actually, only H is moved. By supercompactness, it can be made into a single condition s. Just take the union of all supports of this conditions, it is a union of less than $j(\kappa )$ nonstationary subsets of $j(\kappa )$ , and so is nonstationary etc. Also, note that $j_U(\kappa )$ does not belong to ${\textrm {supp}}(s)$ , since $j_U(F)$ is normal.

Now, using the closure, find in $M_U[G_\kappa ,H,R]$ an $M[G_\kappa ,H,R]$ -generic subset S for the forcing ${\langle } j(P_\kappa )/ G_\kappa *H*R, \leq ^*{\!\rangle }$ with $s \in S$ .

Finally, for every $D \in M[G_\kappa ]$ which is a dense open for the forcing ${\langle } j(P_\kappa )/ G_\kappa , \leq ^* {\!\rangle }$ there will be $p \in G_\kappa $ and $q \in R*S$ such that $j_U(p)^\frown q \geq j(p)$ and $j_U(p)^\frown q \in D$ .

Proof (sketch)

Let E be a $(\kappa , \lambda )$ -extender for some regular $\lambda>\kappa $ . Proceed as in the previous lemma, only use E instead of F and $j=j_{j_U(E)}\circ j_U: V\to M$ . We do not construct a $\leq ^*$ -master condition sequence now, but rather use the image of a $\leq ^*$ -generic over $M_{E(\kappa )}$ to generate a $\leq ^*$ -generic over $M_{E}$ , and then over M, where $E(\kappa )=\{X\subseteq \kappa \mid \kappa \in j_E(X)\}$ is the normal measure of E.

Question. Suppose that there is a strongly compact cardinal, does it follow that there is more than one normal ultrafilter?

Proof $U=V\cap W$ is in V by the Hamkins Gap Theorem, since $j_W\upharpoonright V$ is definable in V and

$$ \begin{align*}U=\{X\subseteq \kappa \mid X\in V \text{ and } [id]_W \in j_W\upharpoonright V(X)\}.\end{align*} $$

Now, by Kunen-Paris [Reference Kunen and Paris14], generates a generic subset $H^*$ of $j_U(Cohen(\omega )*Col(\kappa ^+,2^\kappa ))$ over $M_U$ .

Then, in $V[H]$ , is a $\kappa {-}$ complete ultrafilter which extends U. By 2.3, $U^*$ is the unique $\kappa {-}$ complete ultrafilter which extends U and so $W=U^*$ .

Proof The argument for supercompactness is rather standard. Namely, let $\lambda>2^\kappa $ be a regular cardinal and let F be a normal ultrafilter over $\mathcal P_\kappa (\lambda )$ in V. We will extend the embedding $j_F:V\to M_F$ to an elementary embedding $j^*:V[G_\kappa ,H] \to M_F[G_\kappa ^*,H^*]$ .

Thus consider first . Note that $\kappa $ does not appear in supports of its elements due to normality of F. So, we can using the closure of the forcing and the ultrapower, combine elements of this set into a single condition s in $j_F(P_\kappa *Col(\kappa ^+,2^\kappa ))/ G_\kappa *H$ . Finally, build a master condition sequence above s for $M_F[G_\kappa *H]$ .

Now let us deal with second part of the statement.

Let W be a $\kappa{-}$ complete ultrafilter over $\kappa $ in $V[G_\kappa ,H]$ .

Again, as in 4.1, by Hamkins Gap Theorem, $U=V\cap W \in V$ (a gap now, for example, at the second inaccessible below $\kappa $ ). Still, generates a generic subset $G^**H^*$ of $j_U(P_\kappa *Col(\kappa ^+,2^\kappa ))$ over $M_U$ . We can apply 3.3 together with Kunen–Paris for this.

The final stage is as in 4.1.

Proof By Lemma 3.6, $\kappa $ is strongly compact in $V[G_\kappa ]$ . Clearly, it is a unique measurable there.

Let W be a normal ultrafilter over $\kappa $ in $V[G_\kappa ]$ . Note that W concentrates on cardinals which are non-measurable in V. Just otherwise, $\kappa $ will be a measurable in the ground model of $M_W$ , and so, a non-reflecting stationary set S should be added there. But such S will be stationary in $V[G_\kappa ]$ as well, since ${}^\kappa M_W\subseteq M_W$ . However, $\kappa $ is a measurable cardinal in $V[G_\kappa ]$ .

By Hamkins [Reference Hamkins10], no new fresh subsets are added to $\kappa ,\kappa ^+$ . So, by 2.1, $U=W\cap V$ is in V. Hence, U is a normal ultrafilter on $\kappa $ in V which concentrates on non-measurable cardinals.

By our assumption, U is a unique ultrafilter like these.

Now, 2.3 and 3.3 complete the argument.

Proof Suppose otherwise. Let A be a fresh subset of $\kappa $ . Consider $j_W:V[G]\to M_W$ . Then $M_W=M[j_W(G))]$ . By freshness of A, we have then

$$ \begin{align*}A=j_W(A)\cap \kappa \in M.\end{align*} $$

Set $k([f]_U)=[f]_W$ . By 2.4, it is an elementary embedding from $M_U$ to M. The critical point of k (if exists) should be $>\kappa ^+$ , due to the canonical functions from $\kappa $ to $\kappa $ .

Finally, both $M_U$ and M satisfy GCH at $\kappa $ . Hence, $A\in M_U\subseteq V$ . Contradiction.

Proof Suppose otherwise. Let $G\subseteq P $ be a generic and let A be a fresh subset of $\lambda $ in $V[G]$ . For every $\nu \in A$ pick a condition $p_\nu \in G$ which decides . By regularity of $\lambda $ , there is a single $p\in G$ which decides for unboundedly many $\nu $ ’s. Set B to be the union of such decisions made by p. Clearly, $B\in V$ and . So, $A=B\in V$ . Contradiction.

Proof Suppose otherwise. Let $G_\kappa \subseteq P_\kappa $ be a generic and let A be a fresh subset of $\lambda $ in $V[G]$ .

Fix some . For every $\nu \in A$ pick a condition $p^\nu \in G_\kappa , p^\nu \geq s $ which decides . Then there is a finite $a^\nu \subseteq \kappa $ such that at every coordinate $\beta <\kappa $ outside of $a^\nu $ , .

By regularity of $\kappa ^+$ , there will be a single a such that $a=a_\nu $ , for unboundedly many $\nu $ ’s.

By shrinking more if necessary, we can assume that there is a single condition $q \in P_{\max (a)+1}$ such that $q=p^\nu \upharpoonright \max (a)+1$ , for unboundedly many $\nu $ ’s.

Then, by the assumption of the lemma, if $t \in P_\kappa $ , $t\upharpoonright \max (a)+1=q$ and $t\setminus \max (a)+1\geq ^* s\setminus \max (a)+1$ then t is compatible with all such $p^\nu $ ’s.

Set

Clearly, $B\in V$ and $A=B$ . Contradiction.

Proof Assume that is a $ P_{\kappa } $ -name for a fresh function in $ 2^{\kappa ^{+}} $ , and this is forced by the weakest condition in $ P_{\kappa } $ .

At least one of is non-trivial, since otherwise a generic extension will be just V.

Case 1. There exists $ \mu \in I $ which is above the first $\beta <\kappa $ for which

is non-trivial, and a condition $ p^* \in P_{\mu } $ which forces that the following property holds:

(here, $ G_{\mu } $ is the canonical name for the generic set for $ P_{\mu } $ ). By extending $ p^* $ , we can decide the value of $ p $ in the statement above, and thus assume that $ p^* \geq p $ . Let

be a $ P_{\mu } $ -name for $ s $ from the above property, and assume that this is forced by $ p^* $ .

Fix a $ P_{\mu } $ -name for a strategy, , for the second player in the game of length $ \mu +1 $ , which witnesses the $ \mu +1 $ -strategically closure of $ \langle P\setminus \mu , \leq ^* \rangle $ . Note that such a strategy exists since $ \mu \in I$ . Assume that $ p^* $ forces the is such a strategy.

Let us apply the same methods as in the main lemma in [Reference Hamkins10]. We construct, in $ V $ , a binary tree of conditions, and a tree of functions $ \langle b_{\sigma } \colon \sigma \in {}^{\mu>}2 \rangle $ such that , and for every $ \sigma \in {}^{\mu>}2$ :

1. 1. .

2. 2. .

3. 3. If $ \mbox {lh}(\sigma ) $ is limit, then $ p^* $ forces that is an upper bound, with respect to the direct extension order, of .

4. 4. $ b_{\sigma } $ is an end extension of $ b_{\sigma \restriction { \xi }} $ for every $ \xi <\mbox {lh}(\sigma ) $ .

5. 5. For every $ \xi <\mbox {lh}(\sigma ) $ , $ p^* $ forces that is the sequence of moves of the first player in the game, where the second player plays according to the strategy $ \tau $ .

Now assume that $ g\subseteq P_{\mu } $ is generic over $ V $ with $ p^* \in g $ . In $ V\left [g\right ] $ , let $ h\in 2^{<\mu } $ be the characteristic function of a new subset of $\mu $ . Clearly, $ h \notin V $ . $ h $ defines a branch through the binary tree, . The forcing $ P\setminus \mu $ has a direct extension order which is more than $ \mu {-}$ closed, so there exists an upper bound for the conditions in the branch, of the form . It forces that

$$ \begin{align*}b= \bigcup_{\xi<\mu} b_{h\restriction { \xi }}\end{align*} $$

is an initial segment of . Therefore, $ b\in V $ , and thus $ h $ can be defined, in $ V $ , using the binary tree and the set $ b $ . This is a contradiction to the choice of $ h $ .

Case 2. For every $ \mu \in I $ which is above the first $\beta <\kappa $ for which

is non-trivial, every condition in $ P_{\mu } $ forces that

so, for every $ \mu \in I $ , the following set is forced to be $ \leq ^* $ -dense open in $ P\setminus \mu $ :

Note that we used here the fact that $ \left | G_{\mu } \right |$ is below the first inaccessible above $ \mu $ , and $ P\setminus \mu $ has a direct extension order which has enough strategic closure (since $ \mu $ is in I).

Let us apply lemma 3.5. There exists $ p\in P_{\kappa } $ and a club $ C\subseteq \kappa $ such that for every $ \mu \in C\cap I$ ,

$$ \begin{align*}p\restriction{ \mu} \Vdash p\setminus \mu \in e(\mu).\end{align*} $$

Now, assume that $ G\subseteq P_{\kappa } $ is generic over $ V $ and $ p\in G $ . For every $ \xi <\kappa ^{+} $ , there exists $ p_{\xi }\in G $ above $ p $ which decides . For every such $ \xi $ , there exists $ \mu _{\xi } \in C\cap I $ such that coordinates above $ \mu _{\xi } $ in $ p_{\xi } $ are direct extensions of $ p $ . There exists $ \mu ^*<\kappa $ such that, for unboundedly many values of $ \xi <\kappa ^{+}$ , $ \mu _{\xi } = \mu ^* $ . Similarly, there exists $ p^* \in P_{\mu ^*} $ such that for unboundedly many values of $ \xi <\kappa ^{+} $ , $ p_{\xi }\restriction { \mu ^* } = p^* $ . Let $ q = {p^*}^{\frown } p\setminus {\mu ^*} $ . So $ q\in G $ has the following property: For every $ \xi <\kappa ^{+} $ there exists $ r_{\xi }\in P\setminus {\mu ^*} $ such that $ {q\restriction { {\mu ^*} }}^{\frown } r_{\xi } \in G $ decides and $ q\restriction {{\mu ^*}} \Vdash r_{\xi }\geq ^* q\setminus {\mu ^*}$ .

Let $ q' \in G $ be a condition which forces that the property above holds for $ \check {q} $ .

In $ V $ , let $ g \colon \kappa ^{+} \to 2 $ be the union of all the functions which are forced to be an initial segment of , by some direct extension of $ q $ of the form $ {q\restriction { {\mu ^*} }}^{\frown } r $ , where $ q\restriction {{\mu ^*}} \Vdash r\geq ^* q\setminus {\mu ^*} $ . $ g $ is a function since $ q\setminus {\mu ^*}= p\setminus {\mu ^*}\in e({\mu ^*}) $ .

Finally, let us note that $ q' $ forces that , a contradiction.

Proof Note that $|P_\kappa |=\kappa $ , and so, $\kappa ^+$ is preserved and $2^\kappa =\kappa ^+$ in $V[G_\kappa ]$ .

By Lemma 4.6, no fresh subsets are added to $\kappa $ and by Corollary 4.9, no fresh subsets are added to $\kappa ^+$ .

Now, by Proposition 2.1, $U\in V$ .

Proof Note that $P_\kappa $ preserves $\kappa ^+$ and $2^\kappa =\kappa ^+$ in $V[G_\kappa ]$ .

By Lemma 4.6, no fresh subsets are added to $\kappa $ and by Lemma 4.10, no fresh subsets are added to $\kappa ^+$ .

Now, by Proposition 2.1, $U\in V$ .

Proof We repeat the arguments of 4.4, 4.5, only use $I=\{\nu <\kappa \mid \text { the Mitchell order of } \nu <\eta \}$ instead of $I=\{\nu <\kappa \mid \nu \text { is not measurable} \}$ .

By Lemma 3.6, $\kappa $ is strongly compact in $V[G_\kappa ]$ .

Let W be a normal ultrafilter over $\kappa $ in $V[G_\kappa ]$ . Note that W concentrates on cardinals which has the Mitchell order $<\eta $ , in V. Just otherwise, $\kappa $ will be a measurable in the ground model of $M_W$ of the Mitchell order $\geq \eta $ , and so, a non-reflecting stationary set S or a Prikry sequence should be added there. Say that a non-reflecting stationary set S was added. But such S will be stationary in $V[G_\kappa ]$ as well, since ${}^\kappa M_W\subseteq M_W$ . However, $\kappa $ is a measurable cardinal in $V[G_\kappa ]$ .

By Hamkins [Reference Hamkins10], for non-reflecting stationary set or by 4.6, 4.11, for Prikry, no new fresh subsets are added to $\kappa ,\kappa ^+$ . So, by 2.1, $U=W\cap V$ is in V. Hence, U is a normal ultrafilter on $\kappa $ in V which concentrates on measurable cardinals of the Mitchell order $<\eta $ .

Then, by $\kappa{-}$ completeness, there is $\zeta <\eta $ such that U concentrates exactly on measurable cardinals of the Mitchell order $\zeta $ .

By our assumption, U is a unique ultrafilter like these.

Now, 2.3 and 3.3 complete the argument.

Proof (sketch)

We preserve the notation of [Reference Ben-Neria4]: .

Proceed as on p. 382 of [Reference Ben-Neria4].

Let W be a normal ultrafilter over $\kappa $ and $\Gamma \in W$ . $\Gamma \in W$ implies which in turn,implies that in $M_W = M[G_W]$ , $|d^{-1}(\{\kappa \})|= 1$ . Denote $\mu = d^{-1}(\kappa )$ . Consider, in $V[G]$ ,

$$ \begin{align*}W_\mu=\{ X\subseteq \kappa \mid \mu \in j_W(X)\}.\end{align*} $$

It is a $\kappa{-}$ complete ultrafilter over $\kappa $ which is Rudin–Keisler equivalent to W, since d is one to one on . In particular, $j_W=j_{W_\mu }$ .

We have $\Delta \in W_\mu $ and $\mu \in j_W(\Delta )$ .

Set $U_\mu =W_\mu \cap V$ . Then by 2.1 and 4.10, $U_\mu \in V$ .

Proof Proceed as in 3.6 of [Reference Ben-Neria4]. Suppose otherwise. Then there is a regressive f which represents $\kappa $ in the ultrapower $M_{U_\mu }$ .

We will find a condition $q \in G$ such that , which is impossible, since , and so, $d(\mu )>\kappa $ in $M_W$ .

For every $p \in P_\kappa $ define $ p^{-f}\geq ^* p$ by reducing the (name of) every measure one set $X_\nu (p)$ for all $\nu \in \Delta $ such that

$$ \begin{align*}p^{-f}\upharpoonright \nu\Vdash X_\nu(p^{-f} ) = X_\nu(p)\setminus ( f(\nu)+1).\end{align*} $$

The definition of $ p^{-f}$ implies that there is a finite subset $b \subseteq \Delta $ such that for every $\nu \in \Delta \setminus b$ , .

Now, using the density argument, find such $ p^{-f}\in G$ . Then

since $\Delta \in W_\mu $ and $\kappa \leq k(\kappa )=k([f]_{U_\mu })=[f]_{W_\mu }=j_{W_\mu }(f)(\mu )=j_W(f)(\mu )$ , where $k:M_{U_\mu }\to M_{W_\mu }$ defined as usual: $k([h]_{U_\mu })=[h]_{W_\mu }$ .

Let us argue now that $W=U^{\times }_\mu $ . It is sufficient to prove that $W\supseteq U^{\times }_\mu $ . As in [Reference Ben-Neria4], it is enough to show that $d"A \in W$ , for every $A \in U_\mu $ .

Let $A\in U_\mu $ , then $A\in W_\mu \supseteq U_\mu $ , and so, $d"A \in W=d_*W_\mu $ .