Talk:Zero sharp

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Latest comment: 9 months ago by RhubarbJayde in topic "Totally stable"
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"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)Reply[reply]

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)Reply[reply]