Can one, for an infinite set A in ZFC, use forcing to add so many generic subsets of A as to make the collection of all subsets of A a proper class? Consider now a model $M$ of ZFC and use $Add( , )$ to make the collection of all subsets of any given infinite set in the forcing extension $M$[$G$] a proper class. This forcing extension $M$[$G$] would certainly not be a model of ZFC (the Powerset axiom would fail) but what axioms of ZFC would remain? Also, if replacement would remain and one 'replaces' Replacement with Collection in $M$[$G$], how would that effect $M$[$G$]?

1$\begingroup$ What does "replace Replacement with Collection in $M[G]$" mean? $\endgroup$ – Noah Schweber Dec 3 '14 at 19:46

$\begingroup$ @NoahS, I believe that Thomas is likely referring to the fact that without power set, the axioms of replacement and collection are not equivalent (see jdh.hamkins.org/whatisthetheoryzfcwithoutpowerset). But the issue is moot here, since with this forcing, we get both axioms. $\endgroup$ – Joel David Hamkins Dec 3 '14 at 20:54

$\begingroup$ I guess I was confused by "how would that effect $M[G]$," since that implies some sort of change in $M[G]$ itself  I was expecting to see "does collection remain?" instead. $\endgroup$ – Noah Schweber Dec 3 '14 at 21:19
$\newcommand\Ord{\text{Ord}} \newcommand\Add{\text{Add}}$Yes, one may undertake such kind of class forcing constructions. For example, we can force with $\Add(\omega,\Ord)$ to add $\Ord$ many Cohen reals. As you point out, we won't have ZFC in the resulting forcing extension, but we will get a sensible model of some theory. The resulting forcing extension will be a model of set theory without power set, $\text{ZFC}^$, in which the power set of $\omega$ is a proper class.
Let's be a little more specific. To make things easy to understand at first, lets start with a countable transitive model $M\models\text{ZFC}$, and consider forcing with $\Add(\omega,\Ord)$ from the point of view of $M$. Since this model is countable, we may find a filter $G$ in this partial order such that $G$ meets any dense subclass $D\subset\Add(\omega,\Ord)$ that is definable in $M$ using parameters. (There are only countably many such $D$ and so we can meet them one by one in order to construct $G$.) Now, we build $M[G]$ by using names in $M$ and interpreting them by $G$.
Note that the forcing satisfies the countable chain condition by the usual $\Delta$system argument. This in effect gives us access to the Boolean completion of this partial order, from the perspective of $M$, and we can recursively define the Boolean value $[\![\varphi(\tau)]\!]$ of any assertion in the forcing language. We'll have the usual fullness and mixing lemmas for these Boolean values, and using genericity we'll get that $M[G]\models\varphi(\tau_G)$ just in case $[\![\varphi(\tau)]\!]\in G$. (But in fact, we can use the Boolean algebra to form the Boolean ultrapower of this forcing inside any model of ZFC, not just the countable transitive models.)
Thus, the forcing relation will be definable in $M$ just as for set forcing. Once you have the definability of the forcing relation, together with fullness and mixing, then you'll get all the usual ZFC axioms holding in $M[G]$ for this forcing, including collection, replacement, separation, choice and so on, except for the power set axiom. As you mention, we obviously cannot expect get the power set axiom, since we'll have too many reals in $M[G]$.
So $M[G]$ will be a model of $\text{ZFC}^$.
One can similarly add subsets to $\omega_1$ instead of $\omega$ or to any other fixed cardinal, with a similar effect.
There are also interesting situations to arise if one should collapse all cardinals to become countable. This forcing is not c.c.c., but still one gets $\text{ZFC}^$ in the forcing extension, a model where every set is countable and the reals do not form a set.

$\begingroup$ @Prof Hamkins: Regarding the model where one collapses all cardinals to become countablecan the model 'know' that every set is, in fact, countable? $\endgroup$ – Thomas Benjamin Dec 3 '14 at 21:07

$\begingroup$ Yes, to be countable is a firstorder property that is expressible inside the model. $\endgroup$ – Joel David Hamkins Dec 3 '14 at 21:09

$\begingroup$ @Prof. Hamkins: Thanks. Might this then be a possible 'home universe' for the socalled "antiCantor cranks'? $\endgroup$ – Thomas Benjamin Dec 3 '14 at 21:12

$\begingroup$ Not really, since "the reals are uncountable" would still be true in it  it's just that the reals would form a proper class instead of a set. (That's not to say that this kind of model wouldn't appeal to a certain philosophical stance  personally, I find it very attractive.) $\endgroup$ – Noah Schweber Dec 3 '14 at 21:21

2$\begingroup$ The answer's rather long, but the short version is: I do think of every set as "potentially countable," so insofar as I imagine a "real" universe of sets (which I usually don't), it's one with that property. It's also a very useful (for me) picture to have in order to think up counterexamples for things. Also, I'm interested in how ideas around countable objects (ex: computable structure theory) can be "lifted up" to uncountable settings, and I find that such a picture of the universe helps me get the right intuition for things. $\endgroup$ – Noah Schweber Dec 3 '14 at 21:45