Timeline for Tarski's caracterisation of inaccessible cardinals
Current License: CC BY-SA 2.5
Post Revisions
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 19, 2011 at 21:47 | vote | accept | Gérard Lang | ||
| Jan 19, 2011 at 21:47 | vote | accept | Gérard Lang | ||
| Jan 19, 2011 at 21:47 | |||||
| Jan 19, 2011 at 21:47 | vote | accept | Gérard Lang | ||
| Jan 19, 2011 at 21:47 | |||||
| Jan 19, 2011 at 19:45 | comment | added | Andrés E. Caicedo | Is Tarski using the axiom of choice? | |
| Jan 19, 2011 at 19:43 | comment | added | Andrés E. Caicedo | There is no need to write another question, it was already answered here. | |
| Jan 19, 2011 at 18:58 | comment | added | Gérard Lang | YES,I am VERY SORRY, I did not think "inaccessible", but "regular !!! I will write the correct question; Gérard lang | |
| Jan 19, 2011 at 16:40 | answer | added | Amit Kumar Gupta | timeline score: 2 | |
| Jan 19, 2011 at 16:38 | comment | added | Noah Schweber | I'm confused by the statement of the theorem. There are $(2^{\aleph_0})^{\aleph_0}=2^{\aleph_0}=\aleph_1$ many countable subsets of $\aleph_1$, assuming $CH$, so this would imply that $\omega_1$ is strongly inaccessible. What am I missing? | |
| Jan 19, 2011 at 16:34 | comment | added | Andreas Blass |
Are you sure you quoted the theorem correctly? The continuum hypothesis implies that a set of size $\aleph_1$ has only $\aleph_1$ countable subsets.
|
|
| Jan 19, 2011 at 16:09 | history | asked | Gérard Lang | CC BY-SA 2.5 |