Timeline for Is it consistent with ZFC that $\mathrm{dv}(\kappa) = \kappa$ for all infinite cardinal numbers $\kappa$?
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
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| May 14, 2014 at 6:21 | comment | added | Sam Roberts | I don't know. I'd be inclined to give the following proof. Assume GCH below $\kappa$. Then $2^{<\kappa} = sup\{2^{\lambda}: \lambda<\kappa\} = sup\{\lambda^+:\lambda<\kappa\} = \kappa$. | |
| May 13, 2014 at 22:43 | comment | added | Noah Schweber | Isn't that the standard proof that $2^{<\kappa}=\kappa$ assuming GCH holds below $\kappa$? | |
| May 13, 2014 at 8:21 | comment | added | Sam Roberts | Fair enough - it just looks to me a little over complicated to merely establish that $2^{<\kappa} = \kappa$. | |
| May 13, 2014 at 8:18 | history | edited | Noah Schweber | CC BY-SA 3.0 |
added 7 characters in body
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| May 13, 2014 at 8:17 | comment | added | Noah Schweber | That's essentially what I've done - the proof is exactly the same. Rather than just say that's a consequence of GCH, though, I wanted to write it out. | |
| May 13, 2014 at 8:15 | comment | added | Sam Roberts | Isn't it simpler to note that under GCH $2^{<\kappa} = \kappa$? | |
| May 13, 2014 at 8:15 | comment | added | goblin GONE | No worries, I adjoined it to my question. | |
| May 13, 2014 at 7:40 | comment | added | Noah Schweber | I'd be interested in the proof that ZFC allows $dv(\kappa)>\kappa$, if you don't mind sharing it. | |
| May 13, 2014 at 7:39 | comment | added | goblin GONE | I think you're right. For some reason I kept coming to the conclusion that GCH implies that for successor $\kappa$, we have that $|B(\kappa)| = \kappa^+.$ However, your argument looks impeccable. | |
| May 13, 2014 at 7:36 | vote | accept | goblin GONE | ||
| May 14, 2014 at 1:02 | |||||
| May 13, 2014 at 7:30 | history | answered | Noah Schweber | CC BY-SA 3.0 |