\section{Introduction}
It is clear that if $\kappa \geq \aleph_0$ is an infinite cardinal, then the Cohen forcing
 $$\mathbb{P}_{\kappa}=\{p: \kappa \longrightarrow 2 : |p|< \aleph_0 \}$$
  forces the existence of a set $A \subseteq \kappa$ of size $\kappa$ such that $X \cap A$ and $X \setminus A$ are non-empty for all countable ground model sets
  $X \subseteq \kappa$. It  also forces $2^{\aleph_0}\geq \kappa$, hence for  $\kappa \geq \aleph_2$, the $\GCH$ fails in the extension.
In personal communication with the second author, Moti Gitik asked the following natural question:
 \begin{question}
 \label{q1}
 Suppose that the $GCH$ holds and $\kappa \geq \aleph_2$ is a  cardinal. Is there a cardinal and $GCH$ preserving extension of the universe in which there exists a set $A \subseteq \kappa$ of size $\kappa$, such that for all countable sets $X \in \mathcal{P}(\kappa) \cap V$, $X \cap A$ and $X \setminus A$ are nonempty?
 \end{question}
  In this paper we use Todorcevic's method of forcing with matrices of countable elementary substructures to answer the question for the case $\kappa = \aleph_2$.

  \begin{theorem}
  \label{t1}
($\GCH$) There exists a cardinal and the $\GCH$ preserving generic extension of the universe by a strongly proper forcing notion,
such that in the generic extension, there exists  a set $A \subseteq \omega_2$ of size $\aleph_2$ such that for all countable ground model sets $X \subseteq \omega_2$,  $X \cap A$ and $X \setminus A$ are non-empty.
  \end{theorem}
  \begin{remark}
  It follows from the proof of Theorem \ref{t1} that the set $A$ above  also satisfies the conclusion of the abstract, namely  every
countable subset of $A$ or $\omega_2 \setminus A$ is Cohen generic over the ground model.
  \end{remark}
  The paper is organized as follows. In section \ref{pre}, we recall some basic definitions and results about strongly proper forcing notions,
  and review Todorcevic's matrix $\in$-collapse forcing, then in section \ref{proof}, we present a proof of Theorem \ref{t1}.

\section{Some preliminaries}
\label{pre}
The notion of proper forcing was introduced by Shelah, see \cite{shelah80}, who showed that proper forcing notions preserve
$\aleph_1$ and that their countable support iteration is again proper. In this paper we work a stronger concept, called strongly proper,
which was introduced by Mitchell \cite{mitchell}.
\begin{definition}
\label{d1} Let $\mathbb{P}$ be a forcing notion.
\begin{enumerate}
\item Let $M$ be a countable elementary submodel of $H(\theta)$ with $\mathbb{P} \in M.$
We say that $p \in \mathbb{P}$ is $(M, \mathbb{P})$-strongly generic if every
open dense set $D \subseteq \mathbb{P} \cap M$ is predense below $p$.

\item  We say that $\mathbb{P}$ is strongly
proper if for every large enough $\theta$ and for every countable elementary submodel $M$ of $H(\theta)$ with $\mathbb{P} \in M,$
for every $p \in \mathbb{P} \cap M,$ there is $q \leq p$ such that $q$
is $(M, \mathbb{P})$-strongly generic.
\end{enumerate}
\end{definition}
The following easy lemma gives a characterization of strongly generic conditions.
\begin{lemma}
\label{l1} Let $\mathbb{P}, \theta$ and $M$ be as above and let $p \in \mathbb{P}.$
Then $p$ is $(M, \mathbb{P})$-strongly generic iff
\begin{itemize}
\item[$(*)$:] For every $q \leq p,$ there is $q\mid_M \in \mathbb{P} \cap M$ such that for every $r \in \mathbb{P}  \cap M,$
if $r \leq q\mid_M$, then $r$ and $q$ are compatible.
\end{itemize}
\end{lemma}
The method of forcing with side conditions was introduced by Todorcevic \cite{todorcevic1}, who used it to get several consequences of the proper forcing axiom. He also introduced a variant of his method, where the side conditions form a matrix, and not a chain, see \cite{todorcevic3} and \cite{todorcevic2} for a detailed exposition of the method. We will use matrix side conditions as working parts of our forcing notion to prove Theorem \ref{t1}.


Fix a well-ordering $\lhd$ of $H(\omega_2)$.
Throughout this paper, by $M \prec H(\omega_2)$ we mean $\langle M, \in, \lhd \cap M^2 \rangle$ is an elementary substructure of $\langle H(\omega_2), \in, \lhd \rangle$. Set
$$\mathcal{S}=\{M \in [H_{\omega_2}]^{\aleph_0}: M \prec H_{\omega_2} \}.$$
 Note that $\mathcal{S}$ is a stationary subset of $[H(\omega_2)]^{\aleph_0}$. For every $M, N \in \mathcal{S}$, we write $M\cong N$ if and only if $\langle M, \in\rangle$ is isomorphic to $\langle N, \in \rangle$, and denote the unique isomorphism between them by $\varphi_{M,N}:M \xrightarrow{\simeq} N$.
 For each $M \in \mathcal{S}$ we denote $M \cap \omega_1$ by $\delta_M$, $M\cap \omega_2$ by $\beta_M$, and if $p \subseteq \mathcal{S}$, we let $dom(p)=\{\delta_M: M\in p \}$. Also $p(\delta)$ denotes the set of all $M \in p$ with $\delta_M=\delta$. We are now ready to define
 the matrix $\in$-collapse forcing.

\begin{definition}\label{deftod}
The forcing notion $\mathbb{P}_{\in}^M$ consists of all finite $p\subset \mathcal{S}$ satisfying the following conditions:
\begin{itemize}
\item[$(1)$] If $M,N \in p$ and $\delta_M=\delta_N$, then $ M \cong N$;
\item[$(2)$] If $M \in p$ and $\delta \in dom(p)$ is such that $\delta_M < \delta$, then there exists $N \in p(\delta)$ such that $M \in N$.
\end{itemize}
For  $p, q \in \mathbb{P}_{\in}^M$, we say  $p \leq q$ if $q \subseteq p$.
\end{definition}
We have the following lemma.
\begin{lemma}
\label{l2} ($\GCH$) The forcing notion $\mathbb{P}_{\in}^M$ is strongly proper,
satisfies the $\aleph_2$-c.c., and preserves the $\GCH.$
\end{lemma}
\begin{proof}
See \cite{todorcevic3}
\end{proof}


\section{Proof of Theorem \ref{t1}}
\label{proof}
In this section we prove our main theorem, by introducing a strongly proper forcing notions
which preserves the $\GCH$ and adds a set $A \subseteq \omega_2$ as requested.
Let us start by defining our forcing notion.
\begin{definition}
\label{d2}
A condition $p$ of $\mathbb{P}$ is a pair $\langle \mathcal{M}_p, f_p \rangle$, whenever:
\begin{itemize}
\item[$(i)$] $\mathcal{M}_p \in \mathbb{P}_{\in}^M$;
\item[$(ii)$] $f_p: \omega_2 \longrightarrow 2$ is a finite partial function;
\item[$(iii)$] If $M,N \in \mathcal{M}_p$ with $\delta_M=\delta_N$, then
 \begin{itemize}
\item $\alpha \in (dom(f_p) \cap M) \Rightarrow \varphi_{M,N}(\alpha) \in dom(f_p)$,
\item for each $\alpha$ as above, $f_p(\varphi_{M,N}(\alpha))=f_p(\alpha)$.
\end{itemize}
\end{itemize}
For $p, q \in \mathbb{P}$, we say $p \leq q$ if and only if $\mathcal{M}_q \subseteq \mathcal{M}_p$ and $f_q \subseteq f_p$.
\end{definition}
\iffalse
\begin{lemma}
Let $\theta > \omega_2$ be a regular cardinal, and let $ M \prec H(\theta)$ be countable. If $M\cap H_{\omega_2}=M_0$, then $ M_0 \prec H(\omega_2)$, $\delta_M=\delta_{M_0}$ and $\beta_M=\beta_{M_0}$.
\end{lemma}
\fi
 The following lemma plays a key role
in the verification of strong properness of $\mathbb{P}$.
\begin{lemma}\label{mhatlemma}
Let $\theta > \omega_2$ be a large enough regular cardinal and let $M \prec H(\theta)$ be countable. Let $p=\langle \mathcal{M}_p, f_p \rangle \in \mathbb{P}$ such that $M \cap H(\omega_2)=M_0 \in \mathcal{M}_p$. Then there are $\hat{\mathcal{M}}_p$ and $\hat{f}_p$ which satisfy the following conditions:
\begin{itemize}
\item[$(1)$] $\hat{\mathcal{M}}_p \in \mathbb{P}_{\in}^M \cap M$;
\item[$(2)$] $dom(\hat{\mathcal{M}}_p)= dom(\mathcal{M}_p) \cap M$;
\item[$(3)$] $\mathcal{M}_p \cap M \subseteq \hat{\mathcal{M}}_p$;
\item[$(4)$] If $\alpha \in dom(\hat{\mathcal{M}}_p), N_1 \in \mathcal{M}_p(\alpha)$ and $N_2 \in \hat{\mathcal{M}}_p(\alpha)$, then $N_1 \cong N_2$;
\item[$(5)$] $\hat{\mathcal{M}}_p \cup \mathcal{M}_p$ is $M_0$-full, i.e. for every $N\in(\hat{\mathcal{M}}_p \cup \mathcal{M}_p) \cap M_0$, for every $\delta \in dom(\hat{\mathcal{M}}_p \cup \mathcal{M}_p)$, with $\delta_N < \delta < \delta_{M_0}$, there exists $K\in (\hat{\mathcal{M}}_p \cup \mathcal{M}_p)(\delta)$ such that $N \in K \in M_0$;
  \item[$(6)$] $\hat{f}_p \supseteq f_p \restriction M;$
\item[$(7)$] $\hat{p}= \langle \hat{\mathcal{M}}_p, \hat{f}_p \rangle \in \mathbb{P} \cap M$;
\item[$(8)$] $\hat{p}$ and $p$ are compatible.
\end{itemize}
\end{lemma}
\begin{proof}
 Note that if $\delta \in dom(\mathcal{M}_p) \cap M$
  and $N \in \mathcal{M}_p(\delta)$, then $N\cong N'$ for some $N'\in M_0$. Thus by elemantarity, we can obtain $\hat{\mathcal{M}}_p$,  satisfying items $(1)$-$(4)$. It is then clear that  $\hat{\mathcal{M}}_p \cup \mathcal{M}_p$ is $M_0$-full,
  hence clause $(5)$
  is satisfied as well. Furthermore note that $\hat{\mathcal{M}}_p$ and $\mathcal{M}_p$ are compatible.

 Now set $\hat{f}_p=f_p \restriction M.$ Then  items $(6)$ and $(8)$ are satisfied trivially, so we are left to show that
$\hat{p}= \langle \hat{\mathcal{M}}_p, \hat{f}_p \rangle$ is indeed a condition.
We just need to show that if $N_1, N_2 \in \hat{\mathcal{M}}_p, \delta_{N_1}=\delta_{N_2}$ and $\alpha \in dom(f_p{\restriction M}) \cap N_1$, then $\varphi_{N_1,N_2}(\alpha) \in dom(f_p{\restriction M})$ and $f_p(\varphi_{N_1,N_2}(\alpha))=f_p(\alpha)$.
Fix $N_1, N_2$ and $\alpha$ as above. Note that $\alpha \in M.$

Since $N_1, N_2 \in M, \varphi_{N_1,N_2} \in M$ and hence  $\varphi_{N_1,N_2}(\alpha) \in M.$
 Now $\alpha \in N_1 \in \hat{\mathcal{M}}_p$ implies that there exist $x$, $N_1'$ and $M_1'$ such that:
  \begin{enumerate}
  \item $x \in N_1^{\prime} \in M_1^{\prime} \in \mathcal{M}_p(\delta_{M_0})$;
   \item  $N_1= \varphi_{M_1^{\prime}, M_0}(N_1^{\prime})$;
   \item $\alpha= \varphi_{M_1^{\prime}, M_0}(x)$.
   \end{enumerate}
      Then by Definition \ref{d2}(iii),
       \begin{center}
       $x \in dom(f_p)$ and $f_p(x)=f_p(\alpha)$.
        \end{center}
        Also $\varphi_{N_1,N_2}(\alpha) \in N_2 \in \hat{\mathcal{M}}_p$ implies that there exist
         $y, N_2^{\prime}$ and $M_2^{\prime}$ such that
    \begin{enumerate}
    \item[(4)] $y \in N_2^{\prime} \in M_2^{\prime} \in \mathcal{M}_p(\delta_{M_0})$;

    \item[(5)] $N_2= \varphi_{M_2^{\prime}, M_0}(N_2^{\prime})$;

    \item[(6)] $\varphi_{N_1,N_2}(\alpha)= \varphi_{M_2^{\prime}, M_0}(y)$.
     \end{enumerate}
     Then $\varphi_{M_1^{\prime}, M_2^{\prime}}(x)=y$, so
     by Definition \ref{d2}(iii),
       \begin{center}
       $y \in dom(f_p)$ and $f_p(y)=f_p(x)$.
        \end{center}
        By clause (6), $\varphi_{N_1,N_2}(\alpha) \in dom(f_p)$ and
        \[
        f_p(\varphi_{N_1,N_2}(\alpha) )=f_p(\varphi_{M_2^{\prime}, M_0}(y))=f_p(y)=f_p(x)=f_p(\alpha).
        \]
        The lemma follows.
\end{proof}
The next lemma is standard.
\begin{lemma}\label{xi}
If $N_0$ and $N_1$ are two isomorphic elementary substructures of $H(\omega_2)$ and $\beta \in N_0 \cap N_1 \cap \omega_2$, then for all $\xi < \beta, \xi \in N_0$ if and only if $\xi \in N_1$.
\end{lemma}
\begin{proof}
For each $\xi \in \omega_2$, there is a $1-1$-function from $\xi$ into $\omega_1$. Without loss of generality we can assume that both $N_0$ and $N_1$ contain the same family of mappings $\langle e_{\gamma}: \gamma \in \omega_2 \rangle$ where $e_{\gamma}: \gamma \longrightarrow \omega_1$ is a $1-1$-function. Let $\xi < \beta$ and $\xi \in N_0$, so $e_{\beta}(\xi) \in N_0 \cap \omega_1=N_1 \cap \omega_1$. Hence $\xi=e_{\beta}^{-1}(e_{\beta}(\xi)) \in N_1$.
\end{proof}
The next lemma guarantees the existence of natural $(M, \mathbb{P})$-strongly generic conditions.
\begin{lemma}
\label{l22}
Let $\theta > \omega_2$ be a large enough regular cardinal and let $M \prec H_{\theta}$ be countable. If $p=\langle \mathcal{M}_p, f_p \rangle \in \mathbb{P}$ with $M \cap H(\omega_2) = M_0 \in \mathcal{M}_p$ and $f_p \in M$, then $p$ is an $(M, \mathbb{P})$-strongly generic condition.
\end{lemma}
\begin{proof}
Let $D$ be an open dense subset of $\mathbb{P} \cap M$ and $q=\langle \mathcal{M}_q, f_q \rangle \leq p$. We have to show that $q$ is compatible with some element of $D$. Let $\hat{\mathcal{M}}_q$ be as in Lemma \ref{mhatlemma}. Let $X$ be the set of all $(\hat{\mathcal{M}}_q \cup \mathcal{M}_q)$-paths $w=\{N_0^w, \dots, N_l^w\}$ such that  $N_l^w \in \mathcal{M}_q(\delta_{M_0})$, hence is isomorphic to $M_0$. Set $$\mathcal{M}_{q \restriction M} =\{ \varphi_{N_l^w, M_0}(N_i^w): w=\langle N_0^w, \dots, N_l^w \rangle \in X \land i <l \}.$$

Note that by the construction, $\hat{\mathcal{M}}_q \subseteq \mathcal{M}_{q \restriction M}$, $dom(\mathcal{M}_{q \restriction M})=dom(\mathcal{M}_q) \cap M$ and it is easy to see that $\mathcal{M}_{q \restriction M} \in  \mathbb{P}_{\in}^M \cap M$.
 \iffalse
 Clearly it satisfies clause (1) of Definition \ref{deftod}. To show that clause (2) of the definition is satisfied, let $N_1 \in \mathcal{M}_{q \restriction M}(\alpha)$ and $\alpha < \delta$ for some $\delta \in dom(\mathcal{M}_{q \restriction M})$. Since $\delta < \delta_{M_0}$ is in $dom(\mathcal{M}_p)$, there is a path $w= \langle N_o^w, \dots, N_l^w \rangle$ such that $N_l^w \cong M_0$ and $N_1= \varphi_{N_l^w, M_0}(N_i^w) \in \varphi_{N_l^w, M_0}(N_j^w)$ for some $i < j <l$ such that $\delta_{\varphi_{N_l^w, M_0}(N_j^w)}=\delta_{N_j^w}=\delta$.
Note that by the construction, $\hat{\mathcal{M}}_q \subseteq \mathcal{M}_{q \restriction M}$ and $dom(\mathcal{M}_{q \restriction M})=dom(\mathcal{M}_q) \cap M$.
\fi
By an argument similar to the proof of Lemma \ref{mhatlemma},  $q{\restriction M}=\langle \mathcal{M}_{q \restriction M}, f_q{\restriction \beta_M} \rangle \in  \mathbb{P} \cap M$.
Since $D \subseteq \mathbb{P} \cap M$ is an open dense set, we can find $r \in D \cap M$ such that $r \leq q{\restriction M}$. We now define $\bar{q}=\langle \bar{\mathcal{M}}, \bar{f} \rangle$ where:
\begin{itemize}
 \item $\bar{\mathcal{M}}=\mathcal{M}_r \cup \mathcal{M}_q \cup \{\varphi_{M_0, N}(K): N \in \mathcal{M}_q(\delta_{M_0}) \land K \in M_r \}$;
   \item $\bar{f}=f_r \cup f_q \cup \{\langle \varphi_{N^{\prime}, N^{\prime\prime}}(\alpha), f_r(\alpha) \rangle :\alpha \in dom(f_r), N^{\prime} , N^{\prime\prime} \in \bar{\mathcal{M}} \land \delta_{N^\prime}=\delta_{N^{\prime\prime}}\}.$
\end{itemize}
\begin{claim}
 $\bar{f}:\omega_2 \longrightarrow 2$ is a finite partial function.
  \end{claim}
\begin{proof}  Let $x_1, x_2 \in dom(\bar{f})$ with $x_1=x_2$. Note that $f_q{\restriction {\beta_M}} \subseteq f_r$ and $dom(f_r) \subseteq \beta_M$, so it is enough to consider the following two cases.

{\bf Case 1: $x_1=\varphi_{N_0^{\prime},N_0^{\prime\prime}}(\alpha_1)$ for some $\alpha_1 \in dom(f_r)$ and $x_2=\varphi_{N_1^{\prime}, N_1^{\prime\prime}}(\alpha_2)$ for some $\alpha_2 \in dom(f_r)$ where $\delta_{N_0^{\prime}}=\delta_{N_0^{\prime\prime}}$ and $\delta_{N_1^{\prime}}=\delta_{N_1^{\prime\prime}}$.} Without loss of generality we can assume that $\delta_{N_0^{\prime\prime}}=\delta_{N_1^{\prime\prime}}$, since otherwise, suppose that $\delta_{N_0^{\prime\prime}} < \delta_{N_1^{\prime\prime}}$. So for some $N_2^{\prime}, N_2^{\prime\prime} \in \bar{\mathcal{M}}(\delta_{N_1^{\prime\prime}})$, $N_0^{\prime} \in N_2^{\prime}$, $N_0^{\prime\prime} \in N_2^{\prime\prime}$ and $\varphi_{N_2^{\prime},N_2^{\prime\prime}}\restriction {N_0^{\prime}}=\varphi_{N_0^{\prime},N_0^{\prime\prime}}$.
 Thus $x_1=\varphi_{N_2^{\prime},N_2^{\prime\prime}}(\alpha_0)$, where $\delta_{N_2^{\prime\prime}}=\delta_{N_1^{\prime\prime}}$,
 and we may replace $(N_0^{\prime},N_0^{\prime\prime})$ by $(N_2^{\prime},N_2^{\prime\prime})$.
 Then $\varphi_{N_0^\prime, N_1^\prime}(\alpha_1)=\alpha_2$, and hence
$$\bar{f}(x_1)=f_r(\alpha_1)=f_r(\varphi_{N_0^\prime, N_1^\prime}(\alpha_1))=f_r(\alpha_2)=\bar{f}(x_2).$$

{\bf Case 2: $x_1 \in dom(f_r)$ and $x_2=\varphi_{N_0^{\prime},N_0^{\prime\prime}}(\alpha)$ for some $\alpha \in dom(f_r)$.} Again we can assume that for some $N^{\prime} \in \bar{\mathcal{M}}(\delta_{N_0^{\prime\prime}})$ we have $x_1 \in N^{\prime}$ and $x_2=\varphi_{N^{\prime}, N_0^{\prime\prime}}(x_1)$. Thus
$$\bar{f}(x_1)=f_r(x_1)= f_r(\varphi_{N^{\prime}, N_0^{\prime\prime}}(x_1)) =f_r(x_2)=\bar{f}(x_2).$$
\end{proof}
\begin{claim}
 $\bar{\mathcal{M}} \in \mathbb{P}_{\in}^M$.
 \end{claim}
\begin{proof} It suffices to show that $\bar{\mathcal{M}}$
 satisfies clause (2) of Definition \ref{deftod}. Thus suppose that $\alpha < \beta < \omega_1$ are in $dom(\bar{\mathcal{M}})$ and  $N \in \bar{\mathcal{M}}(\alpha)$. If $\alpha \geq \delta_{M_0}$, then $\bar{\mathcal{M}}(\alpha)=\mathcal{M}_q(\alpha)$ and $\bar{\mathcal{M}}(\beta)= \mathcal{M}_q(\beta)$, hence there is $N^{\prime} \in \mathcal{M}_q(\beta) = \bar{\mathcal{M}}(\beta)$ such that $N \in N^{\prime}$.

Now suppose that $\alpha < \delta_{M_0}$. There are three subcases, depending on the relation between  $\beta$ and $\delta_{M_0}$.

First suppose that  $\beta < \delta_{M_0}$. If $N \in \mathcal{M}_r(\alpha)$, then we can find some   $N^{\prime} \in \mathcal{M}_r(\beta) \subseteq \bar{\mathcal{M}}(\beta)$ such that $N \in N^{\prime}$ and we are done. Otherwise, $N=\varphi_{M_0, N'}(K),$ where $N' \in \mathcal{M}_q(\delta_{M_0})$ and  $K \in M_r$.
$N \in \bar{\mathcal{M}}(\alpha)$. Let $K^\prime \in \mathcal{M}_r(\beta)$ be such that $K \in K^\prime.$ Then $\varphi_{M_0, N'}(K^\prime) \in \bar{\mathcal{M}}(\beta)$
and $N \in \varphi_{M_0, N'}(K^\prime).$

Now let $\beta = \delta_{M_0}$. If $N \in \mathcal{M}_r(\alpha)$, then $N \in M_0 \in \bar{\mathcal{M}}(\beta)$, otherwise, $N=\varphi_{M_0, N'}(K),$ where $N' \in \mathcal{M}_q(\delta_{M_0})$ and  $K \in M_r$. But then $N \in N'$ and we are done again.

Finally suppose that $\beta > \delta_{M_0}$. Then we can find some $N' \in \bar{\mathcal{M}}(\delta_{M_0})$ and $N'' \in \bar{\mathcal{M}}(\beta)$
such that $N \in N' \in N''$. Thus $N \in N''$ and we are done
\end{proof}
\begin{claim}
 $\bar{q}= \langle \bar{\mathcal{M}}, \bar{f} \rangle \in \mathbb{P}$.
  \end{claim}
 \begin{proof}
By the previous claims, $\bar f$ is a finite partial function and $\bar{\mathcal{M}} \in \mathbb{P}_{\in}^M$.  Let $\alpha \in N_1 \cap dom(\bar{f})$ and $N_1 \cong N_2$,
where $N_1, N_2 \in \bar{\mathcal{M}}$. We have to show that $ \varphi_{N_1, N_2}(\alpha) \in dom(\bar{f})$
and $\bar f( \varphi_{N_1, N_2}(\alpha))=\bar f(\alpha)$.


 If $\alpha \in dom(f_r)$, then $\langle \varphi_{N_1, N_2}(\alpha), f_r(\alpha) \rangle \in \bar{f}$, and we are done. If $\alpha \in dom(f_q) \setminus dom(f_r)$, then $\alpha \notin \beta_M$ and we must have $N_1, N_2 \in \mathcal{M}_q \setminus \mathcal{M}_r$. So $\varphi_{N_1, N_2}(\alpha) \in dom(f_q)$ and $f_q(\varphi_{N_1, N_2}(\alpha))=f_q(\alpha)$, which implies $\varphi_{N_1, N_2}(\alpha) \in dom(\bar{f})$ and $\bar{f}(\varphi_{N_1, N_2}(\alpha))=\bar{f}(\alpha)$. Finally if $\alpha=\varphi_{N^{\prime}, N_1}(\beta)$ for some $\beta \in dom(f_r)$ and $N^\prime \in \bar{\mathcal{M}}$ with $N^\prime \simeq N_1$, then
 $ \varphi_{N_1, N_2}(\alpha)=\varphi_{N_1, N_2}\varphi_{N^{\prime}, N_1}(\beta)= \varphi_{N^\prime, N_2}(\beta) \in dom(\bar f)$ and
 \[
 \bar{f}(\varphi_{N_1, N_2}(\alpha))=\bar{f}(\varphi_{N^\prime, N_2}(\beta))=f_r(\beta)=f_r(\alpha)=\bar{f}(\alpha).
 \]
 The claim follows.
  \end{proof}
It is evident that $\bar q$ extends both $q$ and $r$, and hence $q$ and $r$ are compatible.   The lemma follows.
\end{proof}
We have the following easy lemma.
\begin{lemma}
\label{l3}
Let $\theta > \omega_2$ be a large enough regular cardinal and $M\prec H(\theta)$ countable. If $p \in \mathbb{P} \cap M$, then $p^{\prime}=\langle \mathcal{M}_p \cup \{M\cap H(\omega_2)\}, f_p \rangle$ is a condition.
\end{lemma}
Putting all things together, we get the following
\begin{corollary}
\label{l31}
$\mathbb{P}$ is strongly proper
\end{corollary}
\begin{proof}
Let $\theta$ be large enough regular, $M \prec H(\theta)$ with $\mathbb{P} \in M$ and let $p \in \mathbb{P} \cap M$. Set  $p^{\prime}=\langle \mathcal{M}_p \cup \{M\cap H(\omega_2)\}, f_p \rangle$. By Lemma \ref{l3}, $p^\prime$ is a condition and by Lemma
\ref{l22}, $p^\prime$ is an $(M, \mathbb{P})$-strongly generic condition.
\end{proof}
In particular, it follows that forcing with $\mathbb{P}$ preserves $\aleph_1$.
\begin{lemma}
\label{l4}
$\mathbb{P}$ satisfies the $\aleph_2$-c.c.
\end{lemma}
\begin{proof}
Let $\{p_{\alpha}=\langle \mathcal{M}_{\alpha}, f_{\alpha} \rangle: \alpha < \omega_2\}$ be a collection of conditions. For each $\alpha < \omega_2, dom(f_{\alpha})$ is a finite subset of $\omega_2$, so by the  $\Delta$-system lemma, we may assume that   $\{dom(f_{\alpha}):\alpha < \omega_2\}$ forms a $\Delta$-system with root $d \subset \omega_2$, so that for every $\alpha \neq \beta, dom(f_{\alpha}) \cap dom(f_{\beta})=d$. Since there are only finitely many functions $f: d \longrightarrow 2$,
by shrinking the sequence, we may also assume that $f_{\alpha}{\restriction d}=g$   for some fixed $g:d \longrightarrow 2$ and all $\alpha < \omega_2$.

For each $\alpha < \omega_2$ set
$$\bar{\mathcal{M}}_{\alpha}=\{\bar{M}: \exists M \in \mathcal{M}_{\gamma}, \bar{M} \ \text{is the transitive collapse of}\ M\} \in H(\omega_1).$$
Clearly for every $\alpha < \beta < \omega_2$, if $\bar{\mathcal{M}}_{\alpha}=\bar{\mathcal{M}}_{\beta}$, then $\mathcal{M}_{\alpha} \cup \mathcal{M}_{\beta} \in \mathbb{P}_{\in}^M$.

By $\CH,$ $|H(\omega_1)|=\aleph_1$, so by shrinking the sequence of conditions further, we may assume that $\bar{\mathcal{M}}_{\alpha}=\bar{\mathcal{M}}_{\beta}$
for all $\alpha < \beta < \omega_2$.

We now show that for $\alpha < \beta < \omega_2$, the conditions $p_\alpha$ and $p_\beta$ are compatible. Thus fix $\alpha < \beta < \omega_2$.
Let $q=\langle \mathcal{M}_q, f_q \rangle$ where
\begin{itemize}
\item $\mathcal{M}_q=\mathcal{M}_{p_{\alpha}} \cup \mathcal{M}_{p_{\beta}}$;

 \item $f_q=f_{p_{\alpha}} \cup f_{p_{\beta}} \cup \{\langle \varphi_{N,N^{\prime}}(\gamma), (f_{p_{\delta}} \cup f_{p_{\delta^{\prime}}})(\gamma) \rangle : N, N^{\prime} \in \mathcal{M}_q \land N \cong N^{\prime} \land \gamma \in dom(f_{p_{\alpha}} \cup f_{p_{\beta}}) \}$.
      \end{itemize}
      It is easily seen that  $q$ is a condition which extends both $p_{\delta}$ and $p_{\delta^{\prime}}$.
\end{proof}
Using corollary \ref{l31} and Lemma \ref{l4}, we get the following.
\begin{corollary}
\label{c1}
The forcing notion $\mathbb{P}$ preserves all cardinals.
\end{corollary}
We now show that forcing with $\mathbb{P}$ preserves the $\GCH$. We only need to consider the case of $\CH.$
\begin{lemma}
 Suppose $G$ is a $V$-generic filter over $ \mathbb{P}$. Let $M$ and $M^{\prime}$ be countable isomorphic elementary substructures of $H(\theta)$, for a large enough regular cardinal $\theta$, with  $\mathbb{P} \in M \cap M^{\prime}$, and let $p \in \mathbb{P} \cap M$. Set $M_0=M \cap H(\omega_2)$ and $M_0^\prime=M^{\prime} \cap H(\omega_2)$. Then $$p_{MM^{\prime}}=\langle \mathcal{M}_p \cup \{M_0, M_0^{\prime}\}, f_p \cup \{\langle \varphi_{M^{\prime},M}(\alpha), f_p(\alpha) \rangle : \alpha \in dom(f_p) \cap M^{\prime}\} \rangle$$ is a condition, and it forces  $\check{\varphi}_{M, M^{\prime}}[\dot{G} \cap \check{M}]=\dot{G} \cap \check{M^{\prime}}$.
\end{lemma}
\begin{proof}
We can easily check that $p_{M,M^{\prime}}$ is a condition. By contradiction suppose that there is a condition $q \leq p_{M, M^{\prime}}$ and there is  $p'$ such that $q \Vdash$``$ \check{p'} \in \dot{G} \cap \check{M} \ \text{but} \ \check{\varphi}_{M, M^{\prime}}(\check{p'}) \notin \dot{G} \cap \check{M^{\prime}}$''. Since $q \Vdash \check{p'} \in \dot{G}$,  $q$ and $p'$ are compatible, so let $q^{\prime}$ be a common extension of $q, p'$. If for all $r \leq q^{\prime}$, there be some $t \leq r$ such that $t \leq \varphi_{M,M^{\prime}} (p')$, it then follows that the set
$$\{t \in \mathbb{P}: t\Vdash \check{\varphi}_{M, M^{\prime}}(\check{p'}) \in \dot{G} \}$$
 is dense below $q^{\prime}$, which is impossible, because it would imply that $q^{\prime} \Vdash$``$ \check{\varphi}_{M, M^{\prime}}(\check{p'}) \in \dot{G} \cap \check{M^{\prime}}$'', which contradicts our assumption. Hence we can pick some $r \leq q^{\prime}$ such that for all $t \leq r, \neg\left(t \leq \varphi_{M,M^{\prime}}(x)\right)$ i.e. $r$ is incompatible with $\varphi_{M, M^{\prime}}(p')$. Now consider the condition $r{\restriction} M \in M$. we have the following easy claim.
 \iffalse
\begin{claim}
 For all $M \prec H(\theta)$, if $M \cap H(\omega_2) \in \mathcal{M}_r$, then $r$ and $r{\restriction} M$ are compatible.
\end{claim}
\begin{proof}
Let $s=\langle \mathcal{M}_s, f_s \rangle =\langle \mathcal{M}_r \cup \mathcal{M}_r{\restriction}_M, f_r \cup f_r{\restriction}_M \rangle=r  \cup r{\restriction}_M$. Clearly $\mathcal{M}_s \in \mathbb{P}_{\in}^M$ and $f_s=f_r$. Let $\alpha \in dom(f_r) \cap N_0$ and $N_0 \cong N_0^{\prime}$. If $\delta_{N_0} < \delta_M$, then there exist $N_1, N_1^{\prime}\in \mathcal{M}_r(\delta_M)$ such that $\alpha \in N_1$ and $\varphi_{N_0, N_0^{\prime}}(\alpha) \in N_1^{\prime}$ so we done. If $\delta_{N_0} \geq \delta_M$, then $N_0$ and $N_0^{\prime}$ are in $\mathcal{M}_r$ and we done. Hence $s \in \mathbb{P}$ and extends both $r$ and $r{\restriction}_M$.
\end{proof}
\fi
\begin{claim}
$\varphi_{M, M^{\prime}}(r{\restriction} M)=r{\restriction} {M^{\prime}}$.
\end{claim}
\iffalse
$r{\restriction}_M=\langle \mathcal{M}_r{\restriction}_M, f_r{\restriction}_{\beta_M} \rangle$, so $\varphi_{M, M^{\prime}}(r{\restriction}_M)=\langle \varphi_{M, M^{\prime}}(\mathcal{M}_r{\restriction}_M), \varphi_{M, M^{\prime}}(f_r{\restriction}_{\beta_M}) \rangle$ . $\mathcal{M}_r{\restriction}_M= \{\varphi_{N_l^w, M_0}(N_i^w): w=\langle N_0^w, \dots, N_l^w \rangle \in X \land i<l\}$ so $\varphi_{M, M^{\prime}}(\mathcal{M}_r{\restriction}_M)= \{\varphi_{N_l^w, M_0^{\prime}}(N_i^w): w=\langle N_0^w, \dots, N_l^w \rangle \in X \land i<l\}$, also $\mathcal{M}_r{\restriction}_{M^{\prime}}= \{\varphi_{N_l^{w^{\prime}}, M_0^{\prime}}(N_i^{w^{\prime}}): w^{\prime}=\langle N_0^{w^{\prime}}, \dots, N_l^{w^{\prime}} \rangle \in X \land i<l\}$. Let $N \in \varphi_{M, M^{\prime}}(\mathcal{M}_r{\restriction}_M)$. Hence there is $N^{\prime} \in \mathcal{M}_r{\restriction}_M$ such that $N=\varphi_{M, M^{\prime}}(N^{\prime})$. If $N^{\prime} \in \mathcal{M}_r \cap M$, then there is a walk $\langle N_0^{w^{\prime}}, \dots, N_{l-1}^{w^{\prime}}, M_0 \rangle$ such that $N^{\prime}=N_j^{w^{\prime}}$ for  some $j<l$. Hence $\varphi_{M_0, M_0^{\prime}}(N_j^{w^{\prime}})=\varphi_{M_0, M_0^{\prime}}(N^{\prime})=N \in \mathcal{M}_r{\restriction}_{M^{\prime}}$. If $N^{\prime} \in \mathcal{M}_r{\restriction}_M \backslash \mathcal{M}_r$, then for some $w=\langle N_0^w, \dots, N_l^w\rangle, N^{\prime}=\varphi_{N_l^w, M_0}(N_j^w)$. Hence $N=\varphi_{M_0, M_0^{\prime}}(N^{\prime})=\varphi_{M_0, M_0^{\prime}}\varphi_{N_l^w, M_0}(N_j^w) \in \mathcal{M}_r{\restriction}_{M^{\prime}}$. Conversely if $N \in \mathcal{M}_r{\restriction}_{M^{\prime}}$, then either $N \in \mathcal{M}_r \cap M^{\prime}$ or $N \in \mathcal{M}_r{\restriction}_{M^{\prime}} \backslash M^{\prime}$, so $N \in \mathcal{M}_r{\restriction}_M$ as before.
Now let $\alpha \in dom(f_r{\restriction}_{\beta_M})$. So $\alpha \in dom(f_r) \cap M_0$ and $M_0 \cong M_0^{\prime} \in \mathcal{M}_r$. Hence $varphi_{M_0, M_0^{\prime}}(\alpha) \in dom(f_r)$ and $f_r(\varphi_{M_0, M_0^{\prime}}(\alpha))=f_r(\alpha)$. Also $\alpha \in dom(f_r{\restriction}_{\beta_{M^{\prime}}})$ implies that $\alpha= \varphi_{M_0, M_o^{\prime}}(x)$ for some $x \in M_0$. Hence $x=\varphi_{M_0^{\prime}, M_0}(\alpha) \in dom(f_r) \cap M_0$ and $f_r(x)=f_r(\alpha)$. Thus $\varphi_{M,M^{\prime}}(f_r{\restriction}_M)=f_r{\restriction}_{M^{\prime}}$.
\fi
Now since $r \leq p'$ and $p' \in M$, we have $r{\restriction} M \leq p'$. By applying $\varphi_{M, M^\prime}$, we get  $r{\restriction} {M^{\prime}}=\varphi_{M,M^{\prime}}(r{\restriction} M) \leq \varphi_{M, M^{\prime}}(p')$ and hence $r \leq \varphi_{M, M^{\prime}}(p')$. But $r \perp \varphi_{M, M^{\prime}}(p')$,  which is a contradiction.
\end{proof}
The proof of the next lemma is standard, but we present it for completeness.
\begin{lemma}
Forcing with $\mathbb{P}$ preserves the $\CH$.
\end{lemma}
\begin{proof}
By contradiction suppose that $\langle r_{\alpha}: \alpha < \omega_2 \rangle$ is a sequence of pairwise distinct reals in $V[G]$, where $G \subseteq \mathbb{P}$
is $V$-generic. Let $p$ be a condition that force this statement. For each $\alpha < \omega_2$ let $p_{\alpha} \leq p$ forces  ``$\dot{r}_{\alpha} \subseteq \check{\omega}$ is a real''.  Fix $\theta$ large enough regular. For each $\alpha < \omega_2$, let $M_{\alpha}$ be a countable elementary substructure of $H(\theta)$ with $p_{\alpha}, p, \mathbb{P}, \dot{r}_{\alpha} \in M_{\alpha}$.

 By counting arguments, there are $ \alpha < \beta < \omega_2$ such that
 \[
 \langle M_\alpha, \in, \mathbb{P}, p_\alpha, \dot{r}_\alpha \rangle \simeq \langle M_\beta, \in, \mathbb{P}, p_\beta, \dot{r}_\beta \rangle.
 \]
 In particular,
 $\varphi_{M_{\alpha}, M_{\beta}}(\dot{r}_{\alpha})=\dot{r}_{\beta}$ and $\varphi_{M_{\alpha}, M_{\beta}}(p_{\alpha})=p_{\beta}$.
 Set
  \[
  p_{M_{\alpha}, M_{\beta}} = (\mathcal{M}_{p_\alpha} \cup \mathcal{M}_{p_\beta} \cup \{ M_\alpha \cap H(\omega_2), M_\beta \cap H(\omega_2)   \}, f_{p_\alpha} \cup f_{p_\beta}).
  \]
  Then $p_{M_{\alpha}, M_{\beta}}$ is a condition which extends both $p_\alpha$ and $p_\beta$.
Note that for all $n<\omega$, for all $p^{\prime} \in M_{\alpha} \cap \mathbb{P}$ and for all $\xi \in \{0, 1 \}$, $p^{\prime} \Vdash \dot{r}_{\alpha}(\check{n})= \check{\xi}$ if and only if $ \varphi_{M_{\alpha}, M_{\beta}}(p^{\prime}) \Vdash \dot{r}_{\beta} (\check{n})= \check{\xi}$
\begin{claim}
$p_{M_{\alpha}, M_{\beta}} \Vdash$``$ \dot{r}_{\alpha}=\dot{r}_{\beta}$''.
\end{claim}
\begin{proof}
By contradiction assume that there exist  $q \leq p_{M_{\alpha}, M_{\beta}}$ and $n< \omega$ such that $q \Vdash$``$ \dot{r}_{\alpha}(\check{n})=0 \land \dot{r}_{\beta}(\check{n})=1$''. Then by elementarity, there is some $r \in \mathbb{P} \cap M_{\alpha}$ such that $r \leq q{\restriction} {M_{\alpha}}$ and $r \Vdash$``$ \dot{r}_{\alpha}(\check{n})=0$''. We have $\varphi_{M_{\alpha}, M_{\beta}}(q{\restriction} {M_{\alpha}})= q{\restriction} {M_{\beta}}$, so $q{\restriction} {M_{\beta}}$ and $\varphi_{M_{\alpha}, M_{\beta}}(r)$ are compatible. Hence we can conclude that $q|| \varphi_{M_{\alpha}, M_{\beta}}(r)$. But then $\varphi_{M_{\alpha}, M_{\beta}}(r) \Vdash$``$\dot{r}_{\beta}(\check{n})=0$'' which is a contradiction with the fact that $q \Vdash$``$ \dot{r}_{\beta}(\check{n})=1$''.
\end{proof}
We get a contradiction. Thus forcing with $\mathbb{P}$ must preserve the $\CH$
and the lemma follows.
\end{proof}
Now let $G$ be a $\mathbb{P}$-generic filter over $V$ and set
\begin{center}
$A=\{f_p(\alpha): p \in G, \alpha \in dom(f_p)   \}$.
\end{center}
Then $A$ is a subset of $\omega_2$ of size $\aleph_2$.
\begin{lemma}
Suppose $X \in \mathcal{P}(\omega_2) \cap V$ is a countable set. Then the sets $X \cap A$ and $X \setminus A$ are non-empty.
\end{lemma}
\begin{proof}
 Set
 $D_{X}= \{p \in \mathbb{P}: \exists \alpha, \beta \in X \cap dom(f_p) \big(f_p(\alpha)=1 \land f_p(\beta)=0 \big) \}.$ It suffices to show that the set $D_X$ is dense,
 since if $p \in G \cap D_X$ and $\alpha, \beta \in X \cap dom(f_p)$ be such that $f_p(\alpha)=1$ and $f_p(\beta)=0 $, then $\alpha \in X \cap A$ and $\beta \in X \setminus A$.

  To show that $D_X$ is dense, let  $p \in \mathbb{P}$ be an arbitrary condition. As $X$ is infinite and $dom(f_p)$ is finite, we can find $\alpha, \beta \in X \backslash dom(f_p)$ such that for all $N \cong N^{\prime}$ in $\mathcal M_p$, $\varphi_{N, N'}(\alpha)\neq \beta$. Set
\[
q=\langle \mathcal{M}_p, f_p \cup \{\langle \varphi_{N,N^{\prime}}(\alpha), 1 \rangle, \langle \varphi_{N, N^{\prime}}(\beta), 0 \rangle : N, N^{\prime} \in \mathcal{M}_p \land N \cong N^{\prime} \} \rangle.
 \]
 $q$ is easily seen to be a condition. Furthermore, it extends $p$ and belongs to $D_X$, as requested.
\end{proof}
This completes the proof of Theorem \ref{t1}.
