TITLE: Does every model of ZF-foundation have an extension, with no new well-founded sets, where every set is bijective with a well-founded set?
QUESTION [15 upvotes]: This question follows up on an issue arising in Peter LeFanu
Lumsdaine's nice question: Does foundation/regularity have any
categorical/structural consequences, in
ZF?
Let me mention first that my view of the role of the axiom of
foundation in the foundation of set theory differs from Peter's. (But see Peter's remarks about this in the comments below; our views may not actually be so different.)
Specifically, Peter explains how to view the axiom of foundation in
ZFC as an axiom of convenience (see his answer
here). Namely, since every
set is well-orderable, all the non-well-founded sets are bijective
with an ordinal and therefore with a set in the well-founded part
of the universe $\bigcup_\alpha V_\alpha$. In particular, any
mathematical structure to be found at all can be found up to
isomorphism in the well-founded realm, where the axiom of
foundation is true.
Peter's follow-up question arose essentially from his observation that this
argument uses the axiom of choice, and my
answer shows that indeed
this is the case, for there are models of ZF-foundation where some
sets are not bijective with any well-founded set.
My view of the axiom of foundation, in contrast, is that it is not
merely an axiom of convenience, but rather expresses a fundamental
truth of the intended realm of sets the theory is trying to
describe. Specifically, the cumulative universe of all sets arises
from the urelements in a well-ordered series of set-building
stages; every set comes into existence at the first stage after all
of its elements exist. This cumulative universe picture leads
immediately to the axiom of foundation. Andreas Blass expresses a
similar view in answer to an earlier
question. (To my way of
thinking, the real step of convenience in ZFC consists instead of
the insight that we don't actually need urelements for any purpose;
and the resulting ZFC theory without urelements is more uniform and
elegant.)
My question here is whether one can resurrect the idea of the
harmlessness of foundation by showing that every model of
ZF-foundation can be extended, without changing the well-founded
part, to one where every set is bijective with a well-founded set.
Question 1. Does every model of ZF-foundation have an
extension, with the same well-founded sets, in which every set is
bijective with a well-founded set?
A weaker version of the question asks about doing this for just one
set at a time:
Question 2. In any model of ZF-foundation, if $A$ is any set,
then is there an extension of the model, without changing the
well-founded part, in which $A$ becomes bijective with a
well-founded set?
For example, perhaps we might hope to make $A$ well-orderable or
even countable. The trouble would be to do so without adding any
well-founded sets. Definitely we cannot hope to always make a set
$A$ well-orderable, if there are non-well-orderable well-founded
sets.
If you can answer the question for countable models, or by making
further assumptions on the models, I would be interested.

REPLY [12 votes]: Well. The question is what do you mean by an extension exactly.
There's a theorem of Paul Larson and Shelah (originally the work started with Eric Hall, then continued with Paul Larson a few years later) that you can construct a permutation model that has no extension to models of choice with the same pure sets. And of course moving from atoms to Quine atoms, pure sets become well-founded sets.
The idea is roughly to sort of code into the structure of the model a collapse of some cardinals, with the pure sets satisfying $V=L$. Then if you well-order the atoms, then you necessarily had to collapse cardinals and thus add new pure sets.
So from a Platonistic point of view, the answer is negative. If your universe has only those pure sets, then there is no possible way to extend this model further. But if you want to allow forcing extensions, then the answer is yes, you can just add the necessary sets and collapse these cardinals.
In general, if you are willing to add new well-founded sets, then simply collapsing more and more cardinals should in principle work. But I suspect that one can modify Larson and Shelah's work to a class setting such that for every $\alpha$, there is a set of atoms with structure that codes collapsing $\omega_\alpha$ to $\omega$. And then the answer is negative. But this would require a model where the atoms form a proper class.