How questions similar to the continuum hypothesis can be solved if we work in a set theory that admits urelements and that permit impure sets that are not injective to any pure set, for example $\sf ZCA$ or even a weakened version of $\sf ZFCA$ where replacement between the pure and impure sets is not allowed. So, to be specific how can we solve the question of independence of the Impure Continuum Hypothesis "ICH" from the rest of axioms of these theories? ICH can be stated as: for any impure infinite set $A$ that is not injective to any pure set, there is no set $B$ that is strictly supernumerous to $A$ and strictly subnumerous to $\mathcal P(A)$? Is permutation models relevant to such contexts? Or there is a version of forcing that works with impure sets?
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3$\begingroup$ @MichaelHardy One for which urelements appear as elements or hereditarily as members. $\endgroup$– Joel David HamkinsCommented May 27 at 15:34
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1$\begingroup$ It seems to me that there is a big preliminary question here, before any consideration of GCH, namely, whether we construct a model of ZCA in which the choice-function version of AC holds, but not every set is bijective with a pure set. This seems already difficult, and would seem to be required in any answer to the main question. $\endgroup$– Joel David HamkinsCommented May 27 at 17:19
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1$\begingroup$ I've realized that ZCA proves the well-order principle, by the usual Zermelo argument, and so it doesn't matter which form of AC you use. Nevetheless, without replacement, we can't turn a well-order of $X$ into a bijection with a pure set, and so that part of the question is still open. Is there a model of ZCA with a set that is not bijective with a pure set? $\endgroup$– Joel David HamkinsCommented May 27 at 18:57
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1$\begingroup$ I guess we can still prove comparability of well-orders in ZCA, and so all we would need is a pure set that doesn't inject into the set, since any well order of that pure set would exceed any well order of the given set, and we'd be done. So the question is: does ZCA prove that for every set, there is a pure set that doesn't inject into it? $\endgroup$– Joel David HamkinsCommented May 27 at 19:28
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1$\begingroup$ @JoelDavidHamkins, Begin with a set $A$ of urelements as big as $V_{\omega.2}$, then take $V_{\omega.2}(A)$ $\endgroup$– Zuhair Al-JoharCommented May 27 at 19:30
1 Answer
Let me record an observation that emerged in the comments.
Theorem. There is a model of ZCA in which the GCH holds for all pure sets, but not generally for all sets.
Proof. We work in ZFC, and assume GCH holds up to $\beth_\omega$, but fails at $\beth_{\omega+1}$. Using the methods of my paper
- Joel David Hamkins and Bokai Yao, Reflection in second-order set theory with abundant urelements bi-interprets a supercompact cardinal, arxiv:2204.09766
we can interpret a model of ZFCA as $V[[A]]$, in which $A$ is a set of urelements of size $\beth_{\omega+1}$. Now, chop this model off at rank $\omega+\omega$ to form $W=V_{\omega+\omega}[[A]]$. It is easy to see that this is a model of ZCA. In particular, we have the axiom of choice here, and every set has a well order.
The pure sets of $W$ are exactly the sets in $V_{\omega+\omega}$. And since the GCH holds up to $\beth_\omega$ in the original model $V$, we have the GCH holding in $V_{\omega+\omega}$. So the GCH holds in the pure sets of $W$.
But the GCH fails between $A$ and $P(A)$, since in $V[[A]]$ there are failures of GCH at $\beth_{\omega+1}$. So this is a model of ZCA where the GCH holds in the inner model of pure sets, but fails generally. $\Box$
Since this method proceeds from a ZFC model with a failure of GCH higher up, it doesn't really explain how one might have used urelements to find a violation of GCH directly, which to my understanding might be part of the motivation of the question. But still, I find the situation identified in the theorem interesting and worthwhile to explore. The proof here shows a general method of transferring behavior from high rank in a ZFC model to a occur in a model of ZCA, which does not have that behavior in the pure sets.
