Suppose $\mathbb{P}$ is a forcing and $\theta>2^{\vert\mathbb{P}\vert}$ is regular.

$(1)$ For a countable $X\prec H_\theta$, a condition $q\in\mathbb{P}$ is $(X, \mathbb{P})$-generic iff $X[g]\cap V=X$ whenever $g$ is $\mathbb{P}$-generic with $q\in g$.

$(2)$ $\mathbb{P}$ is proper iff for a club of countable $X\prec H_\theta$, whenever $p\in X\cap\mathbb{P}$ there is some $q\leq p$ which is $(X, \mathbb{P})$-generic. (This does not depend on the choice of $\theta$!)

The following are equivalent:

$(1)$ $\mathrm{Add}(\omega_1, 1)^V$ is proper in $V[c]$.

$(2)$ $V\models\mathrm{CH}$.

Suppose $G$ is generic for a c.c.c. forcing which adds a real. Then in $V[G]$, $[\omega_2]^\omega\setminus V$ is stationary in $[\omega_2]^\omega$.

First, let us assume $\mathrm{CH}$ fails in $V$ and show that $\mathrm{Add}(\omega_1, 1)^V$ is not proper in $V[c]$. Let $\vec x=\langle x_i\mid i<\omega_2\rangle\in V$ be a sequence of pairwise different reals, which exists as $\mathrm{CH}$ fails. By the above fact, we have that $[\omega_2]^\omega\setminus V$ is stationary in $[\omega_2]^\omega$ in $V[c]$. It follows that there are stationarily many countable $X\prec H_\theta^{V[c]}$ with

• $\vec x\in X$ and

• $X\cap\omega_2\notin V$.

For such $X$, there is no $(X, \mathrm{Add}(\omega_1, 1)^V)$-generic condition! First note that $X\cap\mathcal P(\omega)\cap V\notin V$ as $x_i\in X$ iff $i\in X$ for $i<\omega_2$. If $q$ were a $(X, \mathrm{Add}(\omega_1, 1)^V)$-generic condition then the reals in $X\cap V$ are exactly those which appear as a restriction $q\upharpoonright [\alpha,\alpha+\omega)$ for some $\alpha\in X\cap\omega_1$ and hence we cannot have $q\in V$ either, contradiction. This shows that $\mathrm{Add}(\omega_1, 1)^V$ is not proper.

On the other hand, let us now assume $\mathrm{CH}$ in $V$. So let $X\prec H_\theta^{V[c]}$ be a countable elementary substructure. We may assume that additionally $X\cap V\prec H_\theta^V$ as this is true on a club. Now by elementarity, there is an enumeration $$\langle y_i\mid i<\omega_1\rangle\in X\cap V$$ of the reals in $V$. This implies that $X\cap\mathcal P(\omega)\cap V=\{y_i\mid i\in X\cap\omega_1\}$. As $X\cap\omega_1\in\omega_1$, it follows that $V$ knows $X\cap\mathcal P(\omega)\cap V$ and hence $V$ knows $\mathbb{P}_X:=X\cap\mathrm{Add}(\omega_1, 1)^V$ as well. The problem is that in $V$, we do not know about all dense subsets of $\mathrm{Add}(\omega_1, 1)^V$ which exist in $X$, though we may take a sequence of names $(\dot D_n)_{n<\omega}\in V$ for them.

We can now go on and construct a $(X, \mathrm{Add}(\omega_1, 1)^V)$-generic condition in $V$ as follows: The idea is to construct a descending sequence in $\mathbb{P}_X$ by guessing how each $\dot D_n$ looks like by asking a condition in Cohen forcing on its opininion about $\dot D_n$. More precisely: The construction lasts $\omega$-many steps. In each step, we have already constructed some $q_k\in\mathbb{P}_X$ and are handed a pair $(n, p)$ with $n<\omega$, $p$ a condition of Cohen forcing, by some bookkeeping. We then find $q_{k+1}\leq q_{k}$, $q_{k+1}\in\mathbb{P}_X$ and $\bar p\leq p$ so that $\bar p\Vdash \check q_{k+1}\in\dot D_n$. In the end, we set $q=\bigcup_{k<\omega}q_k$.

Since the construction of $q$ takes place in $V$, $q\in V$ is guaranteed so $q$ is really a condition of the forcing. It is then easy to convince oneself that will be $(X, \mathrm{Add}(\omega_1, 1)^V)$-generic.