Timeline for Is it known whether or not $\aleph_\alpha=\beth_\alpha$ can be proven by ZFC?
Current License: CC BY-SA 3.0
12 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Oct 10, 2017 at 4:21 | vote | accept | Zetapology | ||
| Oct 5, 2017 at 3:22 | comment | added | Zetapology | @Asaf Are you saying that ZFC does not prove "there is an $\alpha$ such that $\beth_1<\aleph_\alpha$ or that given any $\alpha$ ZFC does not prove that $\beth_1<\aleph_\alpha$. Because if you were stating the former, ZFC proves that $\beth_1+1$ exists, and $\beth_1<\aleph_{\beth_1+1}$. | |
| Oct 5, 2017 at 3:19 | vote | accept | Zetapology | ||
| Oct 5, 2017 at 3:19 | |||||
| Oct 5, 2017 at 3:03 | comment | added | Zetapology | Ah, I see what you said. It was phrased oddly in my head, but I see now. I was thinking $\alpha$ not as an ordinal set, but as a formula that defines an $\alpha$. However, my question should really have been worded as "is it true that $ZFC\models\forall X(\phi(X)\Leftrightarrow X=\beth_\alpha)$ given a formula $\phi$ for which $ZFC\models\forall X(\phi(X)\Leftrightarrow X=\aleph_\alpha)$." | |
| Oct 4, 2017 at 21:07 | comment | added | David Roberts♦ | @Asaf my goodness! Or is it the case that there not not exists an alpha such that... | |
| Oct 4, 2017 at 14:35 | comment | added | Stefan Mesken | Zetapology, that's actually not an issue. It's just that $\beth_{1}$ represents different ordinals in different models. If we fix the ordinal $\alpha$ - not the definition $\alpha$ may satisfy - as I did, everything works. | |
| Oct 4, 2017 at 14:32 | comment | added | Zetapology | This answer should be modified, because this doesn't hold for $\alpha=\beth_1+1$, for which $\aleph\alpha>\beth_1$. | |
| Oct 4, 2017 at 14:16 | comment | added | Asaf Karagila♦ | Martin @Goldstern: I think that this can be mitigated by stating that ZFC does not prove that there is an $\alpha$ such that $\beth_1<\aleph_\alpha$. | |
| Oct 4, 2017 at 13:58 | comment | added | Stefan Mesken | @Goldstern You're right, of course. I'm not entirely sure how to elegantly state what I have in mind but something like "for every $\alpha > 0$ there is some forcing $\mathbb P_{\alpha}$ such that $1 \Vdash_{\mathbb P_\alpha}^L \neg \mathrm{CH}(\aleph_{\check{\alpha}})$" would work. However, I don't think that focusing on this formality is very helpful to OP which is why, grudgingly, I consider keeping my statement as is. | |
| Oct 4, 2017 at 13:42 | comment | added | Goldstern | The proof is quite clear, but the claim could easily be misunderstood. A background theory for claims of provability is often a rather weak theory, such as PA (Peano arithmetic, or even weaker theories), augmented by a necessary assumption (such as "Con(ZFC)"). But in PA and even its second order relatives you cannot talk about arbitrary ordinals $\alpha$. What is your background theory? -- I would not know how to state your claim syntactically at all; my version would say "For all ZFC-models M and all ordinals alpha in M there is..." | |
| Oct 4, 2017 at 9:16 | comment | added | Stefan Mesken | So while $\mathrm{ZFC}$ proves that $\mathrm{CH}(\aleph_\alpha)$ occurs on a club class, it can't guarantee this for any particular $\alpha$. | |
| Oct 4, 2017 at 9:11 | history | answered | Stefan Mesken | CC BY-SA 3.0 |