Timeline for Who introduced the notation for $\beth$ numbers and when?
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 25, 2022 at 18:16 | comment | added | Asaf Karagila♦ | And if you want to preserve inclusions, any non-decreasing cardinal functions preserves inclusions, trivially. Since $\aleph(X)\subseteq\aleph(Y)$ whenever $X\subseteq Y$. If we also assume that AC holds, then there is an injection from $X$ into $\aleph(X)$, and so you can also close the diagram given any $f\colon X\to Y$ by finding a suitable $g\colon\aleph(X)\to\aleph(Y)$ for any $X,Y$ and $f$. Of course, that $g$ is not unique, so it's not quite functorial (if I understand things correctly). | |
| Jan 25, 2022 at 18:09 | comment | added | Asaf Karagila♦ | The $\beth$ function is not the power set, because it is a cardinal function. So it too will destroy automorphisms, I think. | |
| Jan 25, 2022 at 18:08 | comment | added | Paul Taylor | I mean acting on functions (or at least inclusions) not just cardinality order. Powerset acts contravariantly on functions as inverse image and covariantly by either of the quantifiers. The Hartogs construction destroys automorphisms. Can you do better than that? | |
| Jan 25, 2022 at 17:59 | comment | added | Asaf Karagila♦ | I'm not sure what you mean. $\aleph(X)\leq\aleph(Y)$ whenever $|X|\leq|Y|$, so in some sense it is functorial, no? | |
| Jan 25, 2022 at 17:55 | comment | added | Paul Taylor | Tangential question: $\beth(-)$ is a functor (covariant powerset). Can you devise a way of making $\aleph(-)$ (the Hartogs construction) into a functor? | |
| Jan 25, 2022 at 15:24 | vote | accept | Asaf Karagila♦ | ||
| Jan 25, 2022 at 12:53 | answer | added | Carlo Beenakker | timeline score: 14 | |
| Jan 25, 2022 at 12:10 | history | asked | Asaf Karagila♦ | CC BY-SA 4.0 |