Talk:Zero sharp: Difference between revisions
Latest comment: 9 months ago by RhubarbJayde in topic "Totally stable"
Content added Content deleted
(→"Totally stable": new section) |
RhubarbJayde (talk | contribs) No edit summary |
||
Line 2: | Line 2: | ||
Does "totally stable" in "every uncountable cardinal is totally stable" mean \(L_\kappa\prec L\) for every uncountable cardinal \(\kappa\), or \(L_\kappa\prec_{\Sigma_1}L\)? I think the former is proven in Barwise (I'd have to check), and since using "totally stable" to refer to \(L_\kappa\prec_{\Sigma_1}L\) is pretty much only confined to googology I'm not sure [[User:C7X|C7X]] ([[User talk:C7X|talk]]) 22:57, 31 August 2023 (UTC) |
Does "totally stable" in "every uncountable cardinal is totally stable" mean \(L_\kappa\prec L\) for every uncountable cardinal \(\kappa\), or \(L_\kappa\prec_{\Sigma_1}L\)? I think the former is proven in Barwise (I'd have to check), and since using "totally stable" to refer to \(L_\kappa\prec_{\Sigma_1}L\) is pretty much only confined to googology I'm not sure [[User:C7X|C7X]] ([[User talk:C7X|talk]]) 22:57, 31 August 2023 (UTC) |
||
: It means \(L_\kappa \prec L\) for every uncountable cardinal \(\kappa\). Interestingly, the case for all \(\kappa\) follows from a single case: \(0^\sharp\)'s existence is equivalent to \(L_{\aleph_\omega} \prec L\). [[User:RhubarbJayde|RhubarbJayde]] ([[User talk:RhubarbJayde|talk]]) 14:12, 9 September 2023 (UTC) |
Latest revision as of 14:12, 9 September 2023
"Totally stable"[edit source]
Does "totally stable" in "every uncountable cardinal is totally stable" mean \(L_\kappa\prec L\) for every uncountable cardinal \(\kappa\), or \(L_\kappa\prec_{\Sigma_1}L\)? I think the former is proven in Barwise (I'd have to check), and since using "totally stable" to refer to \(L_\kappa\prec_{\Sigma_1}L\) is pretty much only confined to googology I'm not sure C7X (talk) 22:57, 31 August 2023 (UTC)
- It means \(L_\kappa \prec L\) for every uncountable cardinal \(\kappa\). Interestingly, the case for all \(\kappa\) follows from a single case: \(0^\sharp\)'s existence is equivalent to \(L_{\aleph_\omega} \prec L\). RhubarbJayde (talk) 14:12, 9 September 2023 (UTC)