Timeline for Independence of CH and permutation models?
Current License: CC BY-SA 4.0
19 events
| when toggle format | what | by | license | comment | |
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| May 25 at 18:50 | comment | added | Zuhair Al-Johar | @NoahSchweber, for the GCH that you've presented, can permutation models have a rule in solving it? | |
| May 11 at 19:23 | comment | added | Zuhair Al-Johar | @NoahSchweber, but can PMs' solve the $\sf HF$ related question (the one you said to be possibly too large) , or we even expect them to solve it? | |
| May 11 at 19:12 | comment | added | Noah Schweber | @bof Yup! They do indeed (and a lot of other things too). | |
| May 11 at 19:11 | comment | added | bof | @NoahSchweber Don't Dedekind-finite infinite sets already violate the "loosely phrased" GCH? | |
| May 11 at 18:52 | vote | accept | Zuhair Al-Johar | ||
| May 11 at 18:51 | comment | added | Zuhair Al-Johar | @PeterLeFanuLumsdaine, Thank you. What Noah wanted to say is that any statement equivalent to a statement about pure sets then it would be unchanged by permutation models, and so cannot be solved using them. I understand now. | |
| May 11 at 18:48 | comment | added | Peter LeFanu Lumsdaine | @GabeGoldberg’s comment is the key point here. As Zuhair is arguing, you can perfectly reasonably define CH in a form that’s not directly in terms of pure sets (and that can happen quite naturally — e.g. if $\mathbb{N}$ is taken axiomatically, not assumed modelled as a von Neumann ordinal). But it will remain equivalent to a statement about pure sets, and so unchanged by taking permutation models. | |
| May 11 at 18:46 | comment | added | Zuhair Al-Johar | @NoahSchweber, yes it would be equivalent. But it is not itself a statement about pure sets. Because that set is not pure. | |
| May 11 at 18:44 | comment | added | Noah Schweber | @ZuhairAl-Johar And then no different from a pure set, so your version of CH will be equivalent to the usual one which is determined by the pure part and so unaffected by FM-methods. | |
| May 11 at 18:43 | comment | added | Zuhair Al-Johar | @NoahSchweber, You can easily take the set of all equivalence classes of finite subsets of $\sf HF$, this would be countable. | |
| May 11 at 18:39 | comment | added | Zuhair Al-Johar | @GabeGoldberg, Ah I see. This is better phrased. So, what you and Noah are saying is that if a statement is equivalent to a statement about pure sets, then it cannot be solved by a permutation model, right! | |
| May 11 at 18:36 | comment | added | Gabe Goldberg | The issue is not whether you can formulate CH in terms of atoms; the issue is that CH (however you rephrase it) is equivalent to a statement about pure sets. So its truth value cannot be changed without changing the pure part of the universe. | |
| May 11 at 18:35 | comment | added | Noah Schweber | @ZuhairAl-Johar In ZFA, your version of HF may be very very large since it will contain each of the atoms. So your version of HF is no longer guaranteed to be countably infinite. I would not consider a version of CH stated for your version of HF to be a faithful rephrasing of CH, to put it mildly. | |
| May 11 at 18:34 | comment | added | Zuhair Al-Johar | For example the set of all hereditarily finite sets, call it $\sf HF$, where finite set is defined as being bi-well-orderable, i.e. there is a well ordering on it whose converse relation is a well ordering too. Then phrase the hypothesis as I did above. | |
| May 11 at 18:29 | comment | added | Noah Schweber | @ZuhairAl-Johar These are infinitely movable goalposts. Why don't you write down a precise version of CH you think isn't determined by the pure set part of a ZFA-structure, and then I'll address it? | |
| May 11 at 18:28 | comment | added | Zuhair Al-Johar | Not necessarily, you can define countable by reference to some non-pure set, that's easy. | |
| May 11 at 18:27 | comment | added | Noah Schweber | @ZuhairAl-Johar Yes, that statement is about pure sets - the issue is the word "countable," meaning "in bijection with $\omega$" (which is a pure set). | |
| May 11 at 18:25 | comment | added | Zuhair Al-Johar | But why one should stick to this pure set exposition of the problem? Why not if $A$ is a countable set, then there is no set $B$ that is strictly supernumerous to $A$ and strictly subnumerous to $\mathcal P(A)$. This statement is not restricted to pure sets, and it seems to be the heart of the issue? Can a permutation model solve this? | |
| May 11 at 17:50 | history | answered | Noah Schweber | CC BY-SA 4.0 |