Timeline for What is the relationship between non-existence of those kinds of singular sets and AC?
Current License: CC BY-SA 4.0
12 events
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| Nov 7 at 16:09 | comment | added | Zuhair Al-Johar | @JoelDavidHamkins, I've further refined this definition to include $|A|=|\bigcup x|$ instead of $A=\bigcup x$, because singularity is a cardinality issue. | |
| Nov 7 at 16:08 | history | edited | Zuhair Al-Johar | CC BY-SA 4.0 |
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| Nov 6 at 15:39 | history | edited | Zuhair Al-Johar | CC BY-SA 4.0 |
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| Nov 6 at 15:10 | comment | added | Joel David Hamkins | But with your edit, the definition is now not well defined, since the property depends on $x$ and not just on $\cup x$. What I had meant was for you to say that a set $A$ is supersingular, if there is an $x$ for which $A=\cup x$ and so forth. | |
| Nov 6 at 4:47 | comment | added | Zuhair Al-Johar | @JoelDavidHamkins, thanks! I've edited. | |
| Nov 6 at 4:46 | history | edited | Zuhair Al-Johar | CC BY-SA 4.0 |
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| Nov 6 at 1:45 | history | became hot network question | |||
| Nov 5 at 22:56 | comment | added | Joel David Hamkins | To match the familiar terminology, one should have the adjectives "supersingular", "ultrasingular" etc. apply to the set $\cup x$, not $x$. After all, $\aleph_\omega$ is singular, because it is the union of $x=\{\aleph_0,\aleph_1,\aleph_2,\ldots\}$, but we don't say that that set is singular. A cardinal is singular, when it is the size of a set that is the union of a small number of small sets. | |
| Nov 5 at 22:31 | history | edited | Zuhair Al-Johar | CC BY-SA 4.0 |
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| Nov 5 at 17:53 | answer | added | Joel David Hamkins | timeline score: 10 | |
| Nov 5 at 17:46 | history | edited | LSpice | CC BY-SA 4.0 |
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| Nov 5 at 17:42 | history | asked | Zuhair Al-Johar | CC BY-SA 4.0 |