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Is it possible to have a structure $T$ in some language which is rigid in $V$, but in a cardinal-preserving extension $T$ is homogeneous (in a suitable sense of the word)? If this is not possible, is …

Then forcing works as it does in $\sf ZF$. Under additional assumptions, you can get more. … Therefore there are generic filters for the collapsing forcing in that model, already in $V$. …

It should be added that the various generic reals are used to separate many of the characteristics of the continuum (e.g. iterating Sacks real forcing for $\omega_2$ steps will force the continuum to be …

key observations are that BPI is equivalent to the Stone representation theorem for Boolean algebras, and that for the Rasiowa–Sikorski lemma we can focus on [complete] Boolean algebras, since they are forcing … Now consider $E_n$ to be the dense open set in the forcing whose conditions are sequences of length at least $n$. …

There are two types of "working with forcing": We can develop the theory of forcing, e.g. iterations, where working with canonical forcing notions is somewhat preferable, so dealing with complete Boolean … Finally, a word about the foundations of forcing. When one learns about forcing, it is often confusing. …

In $\sf ZFC$, any forcing of the form $\operatorname{Add}(\kappa,1)$, for any $\kappa$, will destroy the properness of some forcing. Indeed, much more of that is true. Theorem. …

Can we prove that there is always a cardinal $\kappa$, and a forcing $\Bbb P$, such that: $\Bbb P$ does not add sets of rank $\leq\alpha$. $\Bbb P$ adds sets to $\kappa$. … Satisfying all three would violate Foreman's Maximality Principle (any nontrivial forcing adds a real or collapses cardinals). But the consistency of the principle is an open problem. …

But if you had a $W$-generic $G$ (for a set forcing) such that $W[G]$ and $V$ had the same reals, you would be able to extract $A$ and therefore compute the class generic for the now-collapsed cardinals …

The first and foremost forcing that adds a real is the Cohen forcing which approximates a new real number by giving finite approximations to its characteristics function. … For example, do we gain anything by using a Sacks/Laver/some other arboreal forcing for the working part? …

The answer is no. The proof is in the paper cited in my question, Theorem 4.7 G. P. Monro, On Generic Extensions Without the Axiom of Choice. The Journal of Symbolic Logic Vol. 48, No. 1 (Mar., 1983) …

This is difficult to prove using forcing, for one simple reason. If $M\models\sf ZFC$, and $G$ is an $M$-generic filter (for some forcing notion), then $M[G]\models\sf ZFC$. … In order to use forcing to construct models where the axiom of choice fails you need to first use forcing, and then reduce to an inner model, either by relative constructibility arguments, or by a technique …

No, since there is a surjection from $\Bbb R$ onto $\omega_1$, there is always an injection from $\omega_1$ into $\mathcal P(\Bbb R)$; but it's not difficult to arrange that there is no choice functio …

Consider the forcing $\Bbb P$ whose conditions are partial functions $p\colon\omega\to2$ with $\operatorname{dom}(p)$ a co-infinite subset of $\omega$, ordered by reverse inclusion. …

Let $D$ be a class of uncountable regular cardinals, and for every $\alpha$ let $\Bbb Q_\alpha$ be either trivial if $\alpha\notin D$, or forcing with these two properties: $|\Bbb Q_\alpha|=\alpha$, …

The usual examples are Martin's Axiom where $\kappa=2^{\aleph_0}$, and $\cal P$ is the class of ccc forcings; or the Proper Forcing Axiom where $\cal P$ is the class of proper forcings and $\kappa=2^{\ … Is there any work on separating the various forcing axioms for subclasses of ccc forcings? …