Timeline for How to settle continuum hypothesis like questions for impure sets?
Current License: CC BY-SA 4.0
16 events
| when toggle format | what | by | license | comment | |
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| May 28 at 5:21 | history | edited | Zuhair Al-Johar | CC BY-SA 4.0 |
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| May 27 at 22:27 | history | became hot network question | |||
| May 27 at 19:52 | vote | accept | Zuhair Al-Johar | ||
| May 27 at 19:49 | answer | added | Joel David Hamkins | timeline score: 8 | |
| May 27 at 19:37 | comment | added | Zuhair Al-Johar | @JoelDavidHamkins, Ah! I see. So, we also turn matters back to comparisons with pure sets albeit this time matters are coming from the external ambience. Nice! So, also we used forcing for the failure of GCH above $\beth_{\omega+\omega}$. I thought there can be another way by using some kind of permutation models to settle matters for impure sets without getting through this external detour. | |
| May 27 at 19:32 | comment | added | Joel David Hamkins | So this is a model of ZCA where we have AC and WOP, but not every set is bijective with a pure set. Indeed, if we start in the right kind of model, where GCH fails above $\beth_{\omega+\omega}$, then we can make GCH hold for pure sets and fail generally. | |
| May 27 at 19:31 | comment | added | Joel David Hamkins | Ah, we had the same idea. | |
| May 27 at 19:30 | comment | added | Joel David Hamkins | And for this, perhaps we can make a model of the form $V_{\omega+\omega}[[A]]$, where $A$ is a set of urelements of size larger than $\beth_{\omega+\omega}$. In other words, take a model of ZFCA with a huge set of urelements, but then chop off at rank $\omega+\omega$. I guess we'll still have ZCA. | |
| May 27 at 19:30 | comment | added | Zuhair Al-Johar | @JoelDavidHamkins, Begin with a set $A$ of urelements as big as $V_{\omega.2}$, then take $V_{\omega.2}(A)$ | |
| May 27 at 19:28 | comment | added | Joel David Hamkins | 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? | |
| May 27 at 18:57 | comment | added | Joel David Hamkins | 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? | |
| May 27 at 17:19 | comment | added | Joel David Hamkins | 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. | |
| May 27 at 15:34 | comment | added | Joel David Hamkins | @MichaelHardy One for which urelements appear as elements or hereditarily as members. | |
| May 27 at 15:26 | comment | added | Michael Hardy | What is an impure set? | |
| May 27 at 15:06 | comment | added | Joel David Hamkins | Just to be clear, you want to keep the axiom of choice? But obviously not the well order principle, since this would make every set bijective with a pure set. Exactly which form of choice do you want to keep? And what do you mean by "replacement between the pure and impure sets is not allowed"? What would be the reason someone might care about this set theory? If you drop AC, then it is easy to make examples with symmetric models of ZFA+$\neg$GCH, since if $X$ is infinite Dedekind finite, then X<P(X) with many infinitely many cardinals in between, including X+n for any finite n, also $X^2$. | |
| May 27 at 14:26 | history | asked | Zuhair Al-Johar | CC BY-SA 4.0 |