Proof Let be an enumeration, in $ V $ , of $ P = P_{\kappa } $ -names, such that every $ X\in \left ( \mathcal {P}(\kappa ) \right )^{V\left [G\right ]} $ has the form for some $ \alpha <\kappa ^{+} $ . Such a list of names exists since $ P = P_{\kappa } $ is $ \kappa -c.c. $ Now, construct, in $ V\left [G\right ] $ , a $ \leq ^{*} $ -increasing sequence of conditions $ \langle q_{\alpha } \colon \alpha <\kappa ^{+}\rangle $ , such that, over $ N\left [G\right ] $ , . Such a sequence exists since $ V\left [G\right ] \vDash \mbox {"}\langle j_U(P)\setminus \kappa , \leq ^{*}\rangle \mbox { is } \kappa ^{+}\mbox -{closed}$ .”

Let be a $ P $ -name for the above sequence. Now, define $ U^{*}\supseteq U $ as follows: For every $ \alpha <\kappa ^{+} $ , if and only if there exist $ p\in G $ and $ \alpha <\kappa ^{+} $ such that

We argue that $ U^{*} $ defined above is a normal measure which extends $ U $ .

Assume that $ \delta <\kappa $ and is a $ P_{\kappa } $ -name for a partition of $ \kappa $ in $ V\left [G\right ] $ . For every $ \alpha <\delta $ , define

Since $ P $ is $ \kappa -c.c. $ , $ \left | Y_{\alpha } \right |<\kappa $ . Denote

$$ \begin{align*}Y = \bigcup_{ \alpha<\delta } Y_{\alpha}.\end{align*} $$

Then $ Y\subseteq \kappa ^{+} $ is a bounded subset. Pick $ \alpha ^{*}<\kappa ^{+} $ high enough which bounds $ Y $ . Let us argue that there exists $ p\in G $ and a unique $ \beta <\delta $ such that

and thus .

Work in $ N\left [G\right ] $ . Note that $ \langle A_{\beta } \colon \beta \in Y \rangle $ covers the sequence $ \langle X_{\alpha } \colon \alpha <\delta \rangle $ . Since $ {q}_{\alpha ^{*}} $ is $ \leq ^{*} $ above any $ q_{\beta } $ for $ \beta \in Y $ ,

Since $ \langle i\left (X_{\xi }\right ) \colon \xi <\delta \rangle $ is a partition of $ i(\kappa ) $ , there exists a unique $ \xi ^{*}<\delta $ such that . Let $ p\in G $ be a condition forcing this. Then , as desired.

A similar argument shows that $ U^{*} $ is normal. Indeed, given a $ P_{\kappa } $ -name for a regressive function , define, for every $ \alpha < \kappa $ ,

$$ \begin{align*}X_{\alpha} = \{ \xi<\kappa \colon f(\xi) = \alpha \}\end{align*} $$

and proceed as before to find a unique $ \alpha <\kappa $ such that $ X_{\alpha }\in U^{*} $ .

Proof

1. 1. We can enumerate, in $ V\left [G\right ] $ , all the maximal antichains in $ \langle i(P)\setminus \kappa , \leq ^{*} \rangle $ with order type $ \kappa ^{+} $ , by $ i(\kappa ) $ -c.c. of the forcing, and since $ V\left [G\right ]\vDash \left | i(\kappa ) \right | = \kappa ^{+} $ . Note that $\kappa \not \in i\left (\Delta \right )$ , so in the sense of $ N\left [G\right ] $ , the forcing $ \langle i(P)\setminus \kappa , \leq ^{*} \rangle $ is more than $ \kappa $ -closed. Moreover, since $ V\vDash {}^{\kappa } N \subseteq N $ , and $ P = P_{\kappa } $ is $ \kappa - $ c.c., $ V\left [G\right ]\vDash {}^{<\kappa }N\left [G\right ]\subseteq N\left [G\right ]$ . Therefore, every sequence of length $ \kappa $ of conditions in $ i(P)\setminus \kappa $ which belongs to $ V\left [G\right ] $ belongs to $ N\left [G\right ] $ as well. Thus, in the sense of $ V\left [G\right ] $ , the forcing $ \langle i(P)\setminus \kappa , \leq ^{*} \rangle $ is $ \kappa ^{+} $ -closed.

   Starting from any condition in $ i(P)\setminus \kappa $ , we can construct (in $ V\left [G\right ] $ ) a sequence of direct extensions of it, meeting every maximal antichain. This sequence generates a $ \leq ^{*} $ -generic over $ N\left [G\right ] $ for $ i(P)\setminus \kappa $ , which belongs to $ V\left [G\right ] $ .

2. 2. First, we prove that $ W = U_{H} $ is a normal, $ \kappa $ -complete ultrafilter on $ \kappa $ which extends $ U $ . It is not hard to verify that $ W $ is a filter. We prove that $ W $ is a $ \kappa $ -complete ultrafilter. Assume that $ \langle X_{\alpha } \colon \alpha <\delta \rangle $ is a partition of $ \kappa $ , for some $ \delta <\kappa $ . Work in $ N\left [G\right ] $ . Let $ D\subseteq i(P)\setminus \kappa $ be the $ \leq ^{*} $ -dense open set of conditions which decide the unique $ \alpha <\delta $ for which . Then such a statement is forced by some $ r\in H $ . Let $ p\in G $ be a condition which forces that $ r $ has this property, and also decides the value of $ \alpha $ . Then $ p^{\frown } r \Vdash \kappa \in i\left ( X_{\alpha } \right ) $ and thus $ X_{\alpha }\in W $ . Normality of $ W $ follows by a similar argument, using the dense set of conditions deciding the value of for a given regressive function $ f\colon \kappa \to \kappa $ . The argument works since we don’t force over $ \kappa $ in $ N $ .

Proof

1. 1. $\{X\subseteq \kappa \colon \kappa \in i(X)\}=U$ : Indeed, let $ X\subseteq \kappa $ in $ V $ with $ \kappa \in i(X) $ . By applying $ k\colon N\to M $ it follows that $ \kappa \in j_W(X) $ and hence $ X\in W $ . Since $ X\in V $ and $ U = W\cap V $ , it follows that $ X\in U $ .

2. 2. $ \left |i(\kappa )\right | = \kappa ^{+} $ : This holds since, in $ V\left [G\right ] $ , $ \left | j_W(\kappa ) \right | = 2^{\kappa } = \kappa ^{+} $ (since, in $ V $ , $ 2^\kappa = \kappa ^{+} $ ), and $ i(\kappa )\leq j_W(\kappa ) $ .

3. 3. $\{i(f)(\kappa ) | f:\kappa \to \kappa \}$ is unbounded in $i(\kappa )$ : Given $ \beta <i(\kappa ) $ , let $ f\in V\left [G\right ] $ be a function such that $ \left [f\right ]_{W} = k\left (\beta \right ) $ . Since $ k\left (\beta \right )<k\left ( i(\kappa ) \right )= j_W(\kappa ) $ , we can assume that $ f\colon \kappa \to \kappa $ . The Easton support ensures that there exists $ g\colon \kappa \to \kappa $ in $ V $ which dominates $ f $ . Thus $ i\left (g\right )(\kappa ) \geq \beta $ (indeed, by applying $ k\colon N\to M $ on both sides, this is equivalent to $ j_W(g)(\kappa ) \geq k\left ( \beta \right ) = \left [f\right ]_W $ , which holds, since $ g $ dominates $ f $ . Note that, when applying $ k $ , we used the fact that $ {\mbox {crit}(k)> \kappa} $ ).

Proof For every $ \alpha <\kappa $ , let $ \tau _{\alpha } <\kappa $ be the least ordinal such that $ P\restriction { \left ( \alpha , \tau _{\alpha } \right )} $ is not $ \alpha -c.c. $ We will argue below that such $ \tau _{\alpha }<\kappa $ exists, but first, let us show that this suffices: Pick an unbounded subset $ X\subseteq \kappa $ , such that, for every $ \alpha , \alpha '\in X $ ,

$$ \begin{align*}\alpha<\alpha' \implies \tau_{\alpha}< \tau_{\alpha^{\prime}}\end{align*} $$

(for instance, let $ X $ be the club of closure points of the function $ \alpha \mapsto \tau _{\alpha } $ ). Enumerate $ X = \langle x_{\alpha } \colon \alpha \in \Delta \rangle $ . For every $ \alpha \in \Delta $ , let $ \langle q_{x_\alpha , \xi } \colon \xi <x_{\alpha } \rangle $ be an antichain in $ P_{ \left ( x_{\alpha }, \tau _{x_{\alpha }} \right ) } $ of cardinality $ x_\alpha $ . Define $ \theta _{\alpha } $ to be the unique ordinal $ \xi <x_{\alpha } $ for which $ q_{x_\alpha , \xi }\in G\restriction { \left ( x_{\alpha }, \tau _{x_{\alpha }} \right ) } $ (if there is no such $ \xi $ , which is possible since the antichain is not necessarily maximal, set $ \theta _\alpha =0 $ ).

Now, given $ \beta <\kappa $ and a condition $ p\in P_{\kappa } $ , pick first $ \alpha \in \Delta $ for which $ x_{\alpha } $ bounds the support of $ p $ . Direct extend $ p $ to $ p^{*} $ such that $ p^{*}\restriction { \left ( x_{\alpha }, \tau _{x_{\alpha }} \right ) } = q_{x_{\alpha }, \beta } $ . ThenFootnote ² by our definition, $ p^{*} $ forces that $ \theta _{\alpha } = \beta $ .

Let us prove now that for every $ \alpha <\kappa $ and condition $ p\in P_{\kappa } $ , there exists $ p^{*}\geq ^{*} p $ which decides the value of $ \theta _{\alpha } $ .

We will direct extend $ p $ in the interval $ \left ( x_{\alpha }, \tau _{ x_{\alpha } } \right ) $ , $ x_{\alpha } $ -many times, to decide whether $ q_{ x_{\alpha }, \xi }\in G\restriction { \left ( x_{\alpha }, \tau _{x_{\alpha }} \right )} $ , for every $ \xi <x_{\alpha } $ . Note that this is possible since $ {\langle } P\restriction { \left ( x_{\alpha }, \tau _{ x_{\alpha } } \right ) } , \leq ^{*} {\rangle } $ is more than $ x_{\alpha } $ -closed. Let $ p^{*}\geq ^{*} p $ be the obtained condition. Then either there exists $ \xi < x_{\alpha } $ such that $ p^{*} $ forces that $ q_{ x_{\alpha }, \xi } $ is in the generic, and then ; or, there is no such $ \xi $ , and then .

Let us argue now that indeed, for every $ \alpha <\kappa $ there exists $ \tau _{\alpha }<\kappa $ such that $ P\restriction { \left ( \alpha , \tau _{\alpha } \right ) } $ is not $ \alpha -c.c. $ : Pick $ \tau _{\alpha } $ such that there are $ \alpha $ -many elements of $ \Delta $ in the interval $ \left ( \alpha , \tau _{\alpha } \right ) $ . Let $ \langle \tau _{\alpha ,\xi } \colon \xi <\alpha \rangle $ be an enumeration of the first $ \alpha $ -many elements in $ \left ( \alpha ,\tau _{\alpha } \right )\cap \Delta $ . For every $ \xi <\alpha $ , let be $ P_{\tau _{\alpha ,\xi }} $ -names, forced by $ 0_{P_{ \tau _{\alpha ,\xi } }} $ to be a pair of incompatible elements of . Such a pair exists since is nontrivial.

Now, for every $ \sigma \in 2^{\alpha } $ , let $ p_{\sigma }\in P\restriction { \left ( \alpha , \tau _{\alpha } \right ) } $ be the condition which satisfies, for every $ \xi <\alpha $ , that

Note that $ \tau _{\alpha } $ is the limit of the first $ \alpha $ many elements above $ \alpha $ in $ \Delta $ , and thus $ \tau _{\alpha } $ is singular, so the support of a condition in $ P = P_{\kappa } $ may be unbounded in $ \tau _{\alpha } $ .

Then $ \langle p_{\sigma } \colon \sigma \in 2^{<\alpha } \rangle $ is an antichain in $ P\restriction {\left ( \alpha , \tau _{\alpha } \right )} $ of cardinality at least $ \alpha $ .

Proof In $ V\left [G\right ] $ , let ${\langle } A_\xi \mid \xi <\kappa ^+{\rangle }$ be an enumeration of maximal antichains in $i(P)$ . Let ${\langle } \beta _\xi \mid \xi <\kappa ^+{\rangle }$ be an enumeration of all the generators of $ i $ below $i(\kappa )$ . Define in $V[G]$ a $\leq ^{*}$ -increasing sequence ${\langle } r_\xi \mid \xi <\kappa ^+{\rangle }$ . Assume that $ {\langle } r_{\xi } \colon \xi < \xi ^{*} {\rangle } $ has been constructed for some $ \xi ^{*}< \kappa ^{+} $ . Pick a condition $ r $ which $ \leq ^{*} $ extends all the conditions $ \langle r_{\xi } \colon \xi < \xi ^{*} \rangle $ constructed so far, and, by extending it, assume that $ r $ extends a condition in $ A_{\xi ^{*}} $ . Finally, let $ \alpha _0 < i(\kappa ) $ be such that for every $ \alpha \geq \alpha _0 $ there exists $ r^{*} \geq ^{*} r $ which forces that $ i\left ( \xi \mapsto \theta _{\xi } \right )\left ( \alpha \right ) = \beta _{\xi ^{*}} $ . Pick any $ \alpha \geq \alpha _0 $ below $ i(\kappa ) $ which has the form $ i\left ( f \right )\left ( \kappa \right ) $ for some $ f = f_{ \beta _{\xi ^{*}} } \in V $ , and let $ r_{\xi ^{*}} \geq ^{*} r $ be a condition which forces that $ i\left ( \xi \mapsto \theta _{\xi } \right )\left ( \alpha \right ) = \beta _{\xi ^{*}} $ .

Finally, let $ H $ be the $ \leq ^{*} $ -generic generated from $ \langle r_{\xi } \colon \xi <\xi ^{*} \rangle $ .

Proof Denote $ W = U_{H} $ . Let us prove that the embedding $ k $ defined above is elementary. Let us prove, for example, that for every $ x,y\in N $ , $ x\in y $ if and only if $ k(x) \in k(y) $ . Fix such $x,y$ and let $ f,g\in V $ , $\beta _1, \ldots , \beta _l $ and $ \alpha _1 < \cdots < \alpha _k< \alpha $ be such that

$$ \begin{align*}x= i(f)\left( \kappa,\beta_1, \ldots, \beta_l \right) , \ y = i(g)\left(\kappa, \beta_1, \ldots, \beta_l \right).\end{align*} $$

Assume now that $ k(x) = k(y) $ , namely

$$ \begin{align*}j_W(f)\left( \kappa, \theta_{ j_W\left( f_{\beta_1} \right)(\kappa) }, \ldots, \theta_{ j_W\left( f_{\beta_l} \right)(\kappa) } \right) \in j_W(g)\left( \kappa, \theta_{ j_W\left( f_{\beta_1} \right)(\kappa) }, \ldots, \theta_{ j_W\left( f_{\beta_l} \right)(\kappa) } \right) .\end{align*} $$

Then

$$ \begin{align*}\{ \xi<\kappa \colon f\left( \xi, \theta_{ f_{ \beta_1 }(\xi) } , \ldots, \theta_{ f_{ \beta_l }(\xi) } \right) \in g\left( \xi, \theta_{ f_{ \beta_1 }(\xi) } , \ldots, \theta_{ f_{ \beta_l }(\xi) } \right) \}\in W\end{align*} $$

and by the definition of $ W $ , there exists $ p\in G $ and $ r\in H $ such that

By extending $ r\in H$ finitely many times, holds for every $ 1\leq m\leq k $ . Thus, the last equation can be replaced with

$$ \begin{align*}p^{\frown} r \Vdash i(f)\left( \kappa, \beta_1, \ldots, \beta_l \right) \in i(g)\left( \kappa, \beta_1 , \ldots, \beta_l\right)\end{align*} $$

but the forced statement above is entirely in $ N $ , and since a condition forces it, it is true in $ N $ . Thus

$$ \begin{align*}i(f)\left( \kappa, \beta_1, \ldots, \beta_l \right)\in i(g)\left( \kappa, \beta_1, \ldots ,\beta_l \right),\end{align*} $$

as desired. The implication in the other direction can be proved similarly.

Clearly $ \mbox {crit}(k)> \kappa $ . We finish the proof by showing that $ j_W\restriction {V} = k\circ i $ . Let $ x\in V $ and let $ c_x \colon \kappa \to V $ be the constant function with value $ x $ . Then

$$ \begin{align*}k\left(i(x)\right) = k\left( i\left( c_x \right)(\kappa) \right) = j_W\left( c_x \right)(\kappa)= j_W(x),\end{align*} $$

as desired.

Proof Fix an ordinal $ \eta $ , and let $ f\in V\left [G\right ] $ be a function such that $ \eta = \left [f\right ]_W $ . We will prove that $ \eta \in \mbox {Im}(k) $ . Indeed, consider the set

It is $\leq $ -dense in $N[G]$ . So, if $\leq =\leq ^{*}$ , then H meets it. Thus, there exists a condition $ q\in H $ , a function $ g\in V $ and generators $ \beta _1, \ldots , \beta _l $ of $ i $ , such that . Thus, by the definition of $ W $ ,

$$ \begin{align*}\{ \xi<\kappa \colon f(\xi) = g\left( \xi, \theta_{ f_{ \beta_1 }(\xi) }, \ldots, \theta_{ f_{ \beta_l }(\xi) } \right) \} \in W,\end{align*} $$

and thus $ \eta = \left [ f \right ]_W = k\left ( i(g)\left ( \kappa , \beta _1, \ldots , \beta _l \right ) \right ) $ .

Question 2.17. Suppose that for unboundedly many $\alpha <\kappa $ , $\leq _\alpha \not =\leq _\alpha ^{*}$ . Is then $M\neq N$ ?

Proof Let $q $ be in H, and let $ p\in G $ be a condition such that (recall that $ H\in V\left [G\right ] $ ). Clearly,

where $ \Gamma $ is the canonical $ i(P) $ -name for the generic set for $ i(P) $ over $ V $ .

Pick $f{\kern-1pt}:[\kappa ]^n\to \kappa , f{\kern-1pt}\in{\kern-1pt} V$ and $\beta _1<\cdots <\beta _n<i(\kappa )$ such that . For every $m,1\leq m\leq n$ , there are $f_m:\kappa \to \kappa ,f_m\in V$ such that $ q_{ i\left ( f_m \right )(\kappa ), \beta _m }\in H $ , namely, $\beta _m =\theta _{i(f_m)(\kappa )}$ .

Let us argue that the set

$$ \begin{align*}A_q=\{\nu<\kappa\mid f(\theta_{f_1(\nu)},\ldots,\theta_{f_n(\nu)})\in G\setminus \nu \}\end{align*} $$

is in W. Pick any $q\leq ^{*} q^{*}\in H$ which $ \leq ^{*} $ which forces that $ \beta _{m} = \theta _{ i\left ( f_{ \beta _m } \right )(\kappa ) } $ , for every $ 1\leq m\leq n $ . Recall that

$$ \begin{align*}q=i(f)(\beta_1,\ldots,\beta_n)=i(f)(\theta_{i(f_1)(\kappa)},\ldots,\theta_{i(f_n)(\kappa)}),\end{align*} $$

and thus .

Proof Suppose otherwise, i.e., $H\not =H'$ , but $U_H = U_{H'}:=W$ .

Let $k:N \to M$ be the elementary embedding defined from H and $k':N \to M$ from $H'$ .

Claim 1. $k\neq k'$ .

Proof Assume for contradiction that $ k= k' $ . Thus, by Lemma 2.18, every pair of elements from $ H, H' $ are $ \leq $ -compatible. We will argue that this implies that $ H = H' $ . It suffices to prove that every pair of conditions $ q\in H , q'\in H'$ are $ \leq ^{*} $ -compatible.

Assume otherwise. Let $ \alpha <\kappa $ be the least ordinal such that there are pair of conditions $ q\in H, q'\in H' $ for which $ q\restriction {\alpha } , q'\restriction {\alpha }$ are $ \leq ^{*} $ -incompatible. $ \alpha $ cannot be limit, since $ \leq ^{*} $ -compatibility of all the initial segments of $ q,q' $ below $ \alpha $ implies that $ q\restriction {\alpha } $ and $ q'\restriction {\alpha } $ are $ \leq ^{*} $ -compatible themselves (if $ \alpha $ is inaccessible, this is clear since the support of $ q,q' $ is bounded in $ \alpha $ ; if the supports of $ q,q' $ are unbounded in $ \alpha $ , direct extend both $q,q'$ coordinate by coordinate to find a common direct extension). Thus $ \alpha = \beta +1 $ is successor, and $ q\left ( \beta \right ) $ , $ q'\left ( \beta \right ) $ are $ \leq ^{*} $ -incompatible. By the property of the forcing $ Q_\beta $ , without loss of generality, $ D_\beta \left ( q \right ) $ is $ \leq ^{*} $ -dense open above $ q'(\beta ) $ . Since $ q'(\beta )\in H'\left ( \beta \right ) $ , $ q' $ can be extended to a condition $r\in H' $ , such that $ r(\beta )\in D_{\beta }(q) $ . In particular, $ q\in H $ , $ r\in H' $ are $ \leq $ -incompatible, which is a contradiction. $ \Box $ $\mbox { of Claim}$ 1.

Claim 2. For every generator $ \beta '< \beta $ of $ i $ , there exists a function $ f_{\beta '}\in V $ such that each generic $ H,H' $ has a condition which forces that .

Proof Let $ f,f' $ be functions such that some condition in $ H $ forces that , and some condition in $ H' $ forces that . Let $ q\in H $ be a condition which decides the statement and assume for contradiction that it is decided negatively. By applying $ k $ , $ k(q)\in j_W(G) $ forces that

namely $ k(\beta ') \neq k'\left ( \beta ' \right ) $ , contradicting the minimality of $ \beta $ . $\Box\ \mbox { of Claim}$ 2.

Proof By Proposition 2.1 in [Reference Gitik and Kaplan6], it suffices to prove that there are no new fresh unbounded subsets of cardinals in the interval $ \left [\kappa , \left (2^{\kappa }\right )^V\right ] = \left [ \kappa , \kappa ^{+} \right ] $ . Thus, it suffices to prove the following pair of claims:

Claim 3. $ P = P_{\kappa } $ does not add fresh unbounded subsets to $ \kappa $ .

Proof The fact that there are no fresh unbounded subsets of $\kappa $ follows essentially from the facts that $ 2^{\kappa } = \kappa ^{+} $ , and that there exists a normal measure on $ \kappa $ in $ V\left [G\right ] $ : Given a normal measure $ U\in V $ with $ \Delta \notin U $ , take any $ U^{*}\in V\left [G\right ] $ which extends it. Given a fresh unbounded $ A\subseteq \kappa $ , $ A = j_{U^{*}}\left ( A \right )\cap \kappa $ and thus, by elementarity, $ A $ belongs to the ground model $ M $ of $ \mbox {Ult}\left ( V\left [G\right ], U^{*} \right ) $ . Now set $ k_U\colon M_{U}\to M $ to be the function which maps $ \left [f\right ]_{U} $ to $ \left [f\right ]_{U^{*}} $ . Then $ k_U $ is a well-defined elementary embedding since $ U\subseteq U^{*} $ , and $ \mbox {crit}\left ( k_U \right )> \kappa $ by normality of $ U^{*} $ . Since $ 2^{\kappa } = \kappa ^{+} $ holds in $ M $ , $ k_U $ maps the sequence of subsets of $ \kappa $ to itself, and thus every subset of $ \kappa $ which belongs to $ M $ , already belongs to $ V $ . So the above set $ A $ belongs to $ V $ , which is a contradiction.

Claim 4. For every measurable (in $ V $ ) $ \lambda \leq \kappa $ , $ P_{\lambda } $ doesn’t add fresh unbounded subsets of $ \lambda ^{+} $ . In particular, $ P_{\kappa } $ does not add fresh subsets to $ \lambda ^{+} $ .

Proof Let $ f\in V\left [G\right ] $ be the characteristic function of a fresh unbounded subset of $ \lambda ^{+} $ . Let be a $ P_{\lambda } $ -name and assume that $ p\in P $ forces that is fresh.

Let $ G\subseteq P_{\lambda } $ be generic over $ V $ . For every $ \xi <\lambda ^{+} $ , let $ p_{\xi } \in G $ be a condition which decides . For every $ \xi <\lambda ^{+} $ there exists $ \alpha _{\xi } <\lambda $ such that the support of $ p_{\xi } $ is bounded by $ \alpha _{\xi } $ . Let $ A\subseteq \lambda ^{+} $ and $ \alpha ^{*}<\lambda $ be such that $ \left |A\right |=\lambda ^{+} $ and $ \alpha _{\xi } =\alpha ^{*} $ for every $ \xi \in A $ .

By shrinking $ A\subseteq \lambda ^{+} $ even further (to a set of cardinality $ \lambda ^{+} $ ), we can assume that there exists $ q^{*} \in P_{\lambda } $ such that, for every $ \xi \in A $ , $ p_{\xi } \restriction { \alpha ^{*}} = {q^{*}}\restriction { \alpha ^{*}} $ , and $ q^{*} \restriction {\left [ \alpha ^{*} , \lambda \right )} $ is trivial.

Let . Clearly, $ h\colon \lambda ^{+} \to 2 $ is a function and .

$\Box $ of Lemma 2.23.

Proof Denote $ \lambda = \mbox {crit}\left ( k \right ) $ . Then for some $ h\in V $ and $ \kappa = \beta _0 < \beta _1 < \cdots < \beta _k $ ,

$$ \begin{align*}\lambda = i(h)\left( \kappa, \beta_1, \ldots, \beta_k \right).\end{align*} $$

By the definition of $ k $ , $ \lambda>\kappa $ .

We first prove that $ \lambda \in i\left ( \Delta \right ) $ . Assume otherwise. We can assume without loss of generality that for every $ \xi , \nu _1, \ldots , \nu _k $ below $ \kappa $ , $ h\left ( \xi , \nu _1, \ldots , \nu _k \right )> \xi $ does not belongs to $ \Delta $ : this can be assumed by replacing the function $ h $ with the function $ h' \colon \left [ \kappa \right ]^{k+1} \to \kappa $ defined as follows: For every $ \xi , \eta _1, \ldots , \eta _k $ , $ h'\left ( \xi , \eta _1, \ldots , \eta _k \right ) $ equals $ h\left ( \xi , \eta _1, \ldots , \eta _k \right ) $ if $ h\left ( \xi , \eta _1, \ldots , \eta _k \right )> \xi $ is not measurable in $ V $ ; and else, $ h'\left ( \xi , \eta _1, \ldots , \eta _k \right ) $ is an arbitrary non-measurable above $ \xi $ . By our assumption,

$$ \begin{align*}i\left( h \right)\left( \kappa, \beta_1, \ldots, \beta_k \right) = i\left( h' \right)\left( \kappa, \beta_1, \ldots, \beta_k \right)\end{align*} $$

so we can replace $ h $ with $ h' $ . Since $ \lambda $ is regular (as a critical point of an elementary embedding), we can assume, using a similar argument, that each $ h\left ( \xi , \nu _1, \ldots , \nu _k \right ) $ is regular.

We can assume that for every $ \xi , \mu _1,\ldots ,\nu _k $ , there are no elements of $ \Delta $ in the interval $ \left ( \xi , h\left ( \xi , \nu _1, \ldots , \nu _k \right ) \right ) $ .

Let $ f\in V\left [G\right ] $ be a function such that $ \left [f\right ]_W = \lambda $ . Then

$$ \begin{align*}\left[f\right]_W=\lambda < k(\lambda) = j_W\left( h \right)\left( \kappa, \theta_{ \left[ f_{\beta_1} \right]_W }, \ldots, \theta_{ \left[ f_{ \beta_k } \right]_W } \right).\end{align*} $$

By the definition of $ W $ , there exist $ p\in G $ and $ r\in H $ such that

Recall that, for every $ 1\leq i \leq k $ , there exists a condition in $ H $ forcing that $ \theta _{ i\left ( f_{ \beta _i } \right )(\kappa ) } = \beta _i $ . Thus by extending $ r $ inside $ H $ ,

Since there are no measurables of $ N $ in the interval $ \left ( \kappa , i(h)\left ( \kappa , \beta _0, \ldots , \beta _k \right ) \right ] $ , we can find $ r'\geq ^{*} r $ inside $ H $ such that

$$ \begin{align*}p\Vdash \exists \alpha < i(h)\left( \kappa, \beta_1, \ldots, \beta_k \right), \ r'\Vdash i(f)(\kappa)< \alpha,\end{align*} $$

and since $ P = P_{\kappa } $ is $ \kappa $ -c.c. and $ i(h)\left ( \kappa , \beta _1, \ldots , \beta _k \right ) $ is regular, there exists $ \alpha < i(h)\left ( \kappa , \beta _1, \ldots , \beta _k \right ) $ such that

Now apply $ k $ on both sides. By Lemma 2.18,

$$ \begin{align*}M\left[j_W(G)\right] \vDash \lambda = \left[ f \right]_W < k\left( \alpha \right)\end{align*} $$

but $ \alpha < i(h)\left ( \kappa , \beta _1, \ldots , \beta _k \right ) = \lambda $ and thus $\lambda < k(\alpha ) = \alpha < \lambda $ , which is a contradiction.