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$N$ is a forcing extension of $M$ and $M[G]$ is a forcing extension of $N$. … So $a$ is definitely generic over $M[b]$, and the forcing is a quotient of the two-step forcing giving rise to $a*b$. …

This is true because for any notion of forcing $\mathbb{B}$, then the boolean value $[\![\check b\in\dot G]\!]$ is precisely $b$ for every $b\in \mathbb{G}$. … The reason is that the measure algebra forcing is almost homogeneous, and therefore statements $\varphi(\check t_0,\ldots,\check t_n)$ in the forcing language involving only check names $\check t_i$ will …

Indeed, this forcing is the one of the easiest ways to see that class forcing need not necessarily preserve ZFC. … This is a proper class, and one must pay attention to the issues of class forcing, including the definability of the forcing relation. …

\Add(\omega,\omega_1)^W$ is a maximal antichain in $W$ for the first forcing, since this much of $\pi$ is in $W$. … One may cast the argument in terms of forcing over $V$; there is no need to go to countable transitive models. …

Since the question is still not answered, let me at least explain why the full-support $\omega$-product of $\text{Add}(\omega,1)$ collapses $(2^\omega)^V$ to $\omega$ in the extension $V[G]$. The wa …

This forcing has size $\kappa$, and it adds a club in $\kappa$ containing no regular cardinals. … So the collapse forcing cannot add a generic filter for $\mathbb{P}$. …

Axiom A forcing pre-dated and I believe led directly to the concept of proper forcing, which has been extremely important in the development of forcing. … It also seems relevant to mention in regard to question (2) that while classical forcing uses (complete) Boolean algebras, in topos theory one undertakes the forcing construction with a Heyting algebra …

If $\kappa$ is any uncountable cardinal and $\mathbb{P}$ is a notion of forcing in $L_\kappa$, then for any projective statement $\varphi$, the assertion $1_{\mathbb{P}}\Vdash\varphi$ is absolute between …

Conversely, suppose that CH fails in $V$, and consider Namba forcing. … Thus, forcing over $V[G]$ with the Prikry forcing of $V$ has your desired properties. …

No, not every new real in the Silver extension is generic for Silver forcing. To see this, take any new real $x$ in the extension, and form a new real $y$ by doubling every digit of $x$. … The real $y$ cannot be Silver generic, since it is dense in Silver forcing to violate the digits-doubling property. …

Thus, we can also define all kinds of forcing extensions $L_\alpha[G]$, for any particular $\mathbb{P}\in L_\alpha$, since all elements of $L_\alpha$ are definable there without parameters and hence also … By taking $\mathbb{P}$ to be the forcing to collapse $\omega_1$ to $\omega$ (from the perspective of $L_\alpha$), these extensions will also collapse cardinals of $L_\alpha$. …

The answer is no, and there is no condition on $M$ or $I$ that can ensure the desired property. The reason is that the generic filter $G$ itself can contain no infinite ground model set. To see this, …

Thus, the forcing is gradually killing off components of the stationary partition. … But the combined forcing of all $\mathbb{P}_n$ will kill off the entire partition, which collapses the cardinal. …

Namely, $W$ is bedrock for $V$ if $V$ is a forcing extension of $W$ and $W$ satisfies the ground axiom, meaning that it is not a forcing extension of any deeper ground, or in other words, that it is minimal … $W$, but one can always go deeper, and realize $W=W_0[G_0]$ as a forcing extension of a still deeper ground, with no bottom. …

Here is an example with forcing that is atomic. Let $\mathbb{P}$ be the Boolean algebra arising from the full support $\kappa$-product of the two-atom forcing. … We simply modify the forcing as follows, to get an example where the embedding does not lift. …