Timeline for Are there models of ZF in which all uncountable sets are super/hyper/ultra-singular?
Current License: CC BY-SA 4.0
9 events
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| Nov 8 at 7:02 | comment | added | Zuhair Al-Johar | @JoelDavidHamkins, So, your argument does depend on well-orderability of $\omega_1$, this opens the window wide for those singularities to hold over all non-well orderable sets in some model of ZF in which the reals are non-well orderable. Also Calliope Ryan-Smith also use limits so it pertains to well-orderability. | |
| Nov 8 at 6:54 | comment | added | Zuhair Al-Johar | @CalliopeRyan-Smith, can you show that argument. On the face of it, your answer, is not about super-singularity, since it is enough to have $\omega_2$ be equinumerous to an $\omega_1$ sized set of $\omega_1$ or less sized sets. Anyhow $\omega_2$ is not a limit cardinal, so obviously you are saying "like" argument, so can you post a reference to it. | |
| Nov 8 at 0:37 | comment | added | Joel David Hamkins | I just meant that if $|A|=|\cup x|$, meaning they are equinumerous, then $A$ is equinumerous with $\cup x$, and so we can replace $x$ with $x'$ where $A=\cup x'$. So that part of your change didn't actually change anything. Meanwhile, the rest of my answer here does use that $\omega_1$ is well-orderable, so that all the cardinals below it are also well-orderable, and this is how I concluded that $x$ must be countable and that the elements of $x$ must be countable as well, and indeed finite for the hypersingular case. | |
| Nov 7 at 20:28 | vote | accept | Zuhair Al-Johar | ||
| Nov 8 at 9:15 | |||||
| Nov 7 at 20:21 | comment | added | Zuhair Al-Johar | @JoelDavidHamkins, was that because $\omega_1$ is well-orderable? So, the result must be restricted to non-well orderable sets? | |
| Nov 7 at 19:13 | comment | added | Calliope Ryan-Smith | Also limit cardinals are never super-singular by a Specker-type '$\omega_2$ is not a countable union of countable sets' argument. | |
| Nov 7 at 19:03 | comment | added | Joel David Hamkins | I take same-cardinality to mean equinumerous, so we can assume $\omega_1=\cup x$ without loss. | |
| Nov 7 at 17:56 | comment | added | Zuhair Al-Johar | why $\omega_1=\bigcup x$? | |
| Nov 7 at 17:27 | history | answered | Joel David Hamkins | CC BY-SA 4.0 |