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?

Joel David Hamkins 05/15/2018. 1 answers, 481 views
set-theory lo.logic axiom-of-choice mathematical-philosophy

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.

1 Answers

Asaf Karagila 05/15/2018.

Well. The question is what do you mean by an extension exactly.

There's a theorem of 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 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. - Download Hi-Res Songs

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